May 25, 2010 3:20 PM Subscribe

"Gary Foshee, a collector and designer of puzzles from Issaquah near Seattle walked to the lectern to present his talk. It consisted of the following three sentences: "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?""

"The event was the Gathering for Gardner earlier this year, a convention held every two years in Atlanta, Georgia, uniting mathematicians, magicians and puzzle enthusiasts. The audience was silent as they pondered the question."

[Previous Metafilter thread on the recently deceased Martin Gardner.]

"The event was the Gathering for Gardner earlier this year, a convention held every two years in Atlanta, Georgia, uniting mathematicians, magicians and puzzle enthusiasts. The audience was silent as they pondered the question."

[Previous Metafilter thread on the recently deceased Martin Gardner.]

OK, I fully understand Monty Hall (including how the right answer works intuitively) but I admit I'm having trouble getting this one. It's definitely going to give me some food for thought on the drive home. Thanks!

posted by kmz at 3:34 PM on May 25, 2010

posted by kmz at 3:34 PM on May 25, 2010

This makes me lament being smart enough to recognize that these folks are having fun because they're a lot smarter than I am.

posted by maxwelton at 3:35 PM on May 25, 2010 [6 favorites]

posted by maxwelton at 3:35 PM on May 25, 2010 [6 favorites]

I clicked the link to see this word and was not disappointed.

posted by I_pity_the_fool at 3:37 PM on May 25, 2010

That is, I see how the math works and I don't disagree with it, but I'm not getting the intuition on *why* it works like that. It also reminds me of an old logic puzzle (also controversial) that involves a conversation providing way more information about one person's kids than you would think, but I don't remember exactly how it goes. Anybody have a clue what I'm talking about?

posted by kmz at 3:39 PM on May 25, 2010

posted by kmz at 3:39 PM on May 25, 2010

Ah, here we go, problem #1. I'm not sure if that's the exact puzzle that I remember, but it worked similarly.

posted by kmz at 3:42 PM on May 25, 2010

posted by kmz at 3:42 PM on May 25, 2010

Goddamnit! If you have one boy, and one other kid, then there's a 50% chance your other kid is a boy. My brain will brook no argument about this.

posted by mrnutty at 3:43 PM on May 25, 2010 [16 favorites]

posted by mrnutty at 3:43 PM on May 25, 2010 [16 favorites]

I don't like his question. The tuesday part bugs me in particular, and the explanation in the article seems wrong. Mainly because in the simplified version (without the Tuesday part), the answer should be 1 in 2, not 1 in 3, since the question did not specify whether the boy/girl or girl/boy ordering mattered. The question was the probability of two boys. You have a total set of 2, the number of combinations are Boy/Boy or Boy/Girl, since semantically Boy/Girl is the same as Girl/Boy in the context of the stated question. The born on a Tuesday part is irrelevant, given that the day of the week in which a child is born does nothing to determine the gender of the child.

I can understand how maybe you could construe a difference between Boy/Girl and Girl/Boy as being relevant to the question, only if it was stipulated as a concern. His mention that "Tuesday is the most relevant part" only confuses the logical steps to create a formula. The main reason being that the day of the week that a child is born on does not determine the gender of the child.

Am I missing something here?

Plate of beans?

posted by daq at 3:44 PM on May 25, 2010 [14 favorites]

I can understand how maybe you could construe a difference between Boy/Girl and Girl/Boy as being relevant to the question, only if it was stipulated as a concern. His mention that "Tuesday is the most relevant part" only confuses the logical steps to create a formula. The main reason being that the day of the week that a child is born on does not determine the gender of the child.

Am I missing something here?

Plate of beans?

posted by daq at 3:44 PM on May 25, 2010 [14 favorites]

Nope. If you look at the sample of all people who have 2 children, one of whom is a boy, one third will have two boys, because whether the original boy is the older or younger is not specified.

You're thinking about a case where you two children, the older of whom is a boy, and trying to calculate the probability the younger child is also a boy. That's extra information.

posted by kmz at 3:47 PM on May 25, 2010 [15 favorites]

Heh. I brought up the very word probelmn Foshee posed on another internet forum and it produced the kind of polarization and dogged stubbornness that I usually associate with threads on vegetarianism or Israel/Palestine. Good luck, all!

posted by ricochet biscuit at 3:48 PM on May 25, 2010 [1 favorite]

posted by ricochet biscuit at 3:48 PM on May 25, 2010 [1 favorite]

"To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3."

GG is of course rightly eliminated but since the problem didn't state the order that the children were born in then GB and BG both describe the same situation. If you want to use GB and BG then you should also use something like bB and Bb which would mean that likelihood of the gender of the other child is equally divided between male and female. Which is abundantly reflected by reality.

The error lies in the construction of the problem.

posted by vapidave at 3:48 PM on May 25, 2010 [13 favorites]

GG is of course rightly eliminated but since the problem didn't state the order that the children were born in then GB and BG both describe the same situation. If you want to use GB and BG then you should also use something like bB and Bb which would mean that likelihood of the gender of the other child is equally divided between male and female. Which is abundantly reflected by reality.

The error lies in the construction of the problem.

posted by vapidave at 3:48 PM on May 25, 2010 [13 favorites]

How about this, daq:

The set of all families with two kids consists of 25% bb, 25% gg, 25% bg, and 25% gb. I think I buy the difference between bg and gb when stated like that.

posted by mrnutty at 3:49 PM on May 25, 2010 [1 favorite]

The set of all families with two kids consists of 25% bb, 25% gg, 25% bg, and 25% gb. I think I buy the difference between bg and gb when stated like that.

posted by mrnutty at 3:49 PM on May 25, 2010 [1 favorite]

God damnit, the plane will take off.

posted by Threeway Handshake at 3:50 PM on May 25, 2010 [16 favorites]

posted by Threeway Handshake at 3:50 PM on May 25, 2010 [16 favorites]

Fascinating, awesome. Now, how can I get a Jigazo, which is mentioned in the article?

The Jigazo is a jigsaw puzzle that can be customised to display any picture you want. It contains 300 identically shaped pieces that are all a different shade of blue, and all have a unique symbol on the back. If you take a portrait of yourself and email the image to the Jigazo website, you will receive an email back with a map of the arrangement of the symbols such that when the pieces are assembled in this way the jigsaw shows your portrait.

posted by nevercalm at 3:51 PM on May 25, 2010

The Jigazo is a jigsaw puzzle that can be customised to display any picture you want. It contains 300 identically shaped pieces that are all a different shade of blue, and all have a unique symbol on the back. If you take a portrait of yourself and email the image to the Jigazo website, you will receive an email back with a map of the arrangement of the symbols such that when the pieces are assembled in this way the jigsaw shows your portrait.

posted by nevercalm at 3:51 PM on May 25, 2010

daq, vapidave: Consider somebody doing this activity 1000 times: flipping a fair coin twice, and recording the results. *Among the results with at least one head*, how many results are both heads?

posted by kmz at 3:51 PM on May 25, 2010 [1 favorite]

posted by kmz at 3:51 PM on May 25, 2010 [1 favorite]

He could be choosing the first or second child from the sample space of GG, BB, GB, BG. If you choose any one of those with a B, then only B, G, G remain, or 1 out of 3 for a boy.

posted by Brian B. at 3:54 PM on May 25, 2010 [1 favorite]

You live in China so the probability is zero. HA!

posted by GuyZero at 3:56 PM on May 25, 2010 [10 favorites]

posted by GuyZero at 3:56 PM on May 25, 2010 [10 favorites]

OK, so the maths does seem to work, not arguing with that. It doesn't look like it works in the real world, though, which is where I'm getting confused.

Consider the situation where Alice tells Bob, "I have two children, at least one of whom is a boy." Bob can then calculate the probability that Alice has two boys (and I'm happy to accept that that's 1/3, although I originally went for 1/2). But if Alice then says "Oh, and that boy was born on a Tuesday," this new information isn't relevant, because (unlike in the Monty Hall problem) it doesn't touch on any of the factors involved in the first calculation (probability of any one birth being a boy, number of children in total). So it's still 1/3.

To my mind, that's an exact restatement of the original problem. If the answers are different, I suppose it can't be exactly the same, but I'm stumped if I can see the difference. Anyone got any bright ideas?

posted by ZsigE at 4:03 PM on May 25, 2010

Consider the situation where Alice tells Bob, "I have two children, at least one of whom is a boy." Bob can then calculate the probability that Alice has two boys (and I'm happy to accept that that's 1/3, although I originally went for 1/2). But if Alice then says "Oh, and that boy was born on a Tuesday," this new information isn't relevant, because (unlike in the Monty Hall problem) it doesn't touch on any of the factors involved in the first calculation (probability of any one birth being a boy, number of children in total). So it's still 1/3.

To my mind, that's an exact restatement of the original problem. If the answers are different, I suppose it can't be exactly the same, but I'm stumped if I can see the difference. Anyone got any bright ideas?

posted by ZsigE at 4:03 PM on May 25, 2010

What about cultures that value boys over girls? Having a boy first makes it less likely that a second child will be born. The majority of two-child families will be GB or GG, and BB would be less common than mathematics would suggest.

posted by bgrebs at 4:04 PM on May 25, 2010 [1 favorite]

posted by bgrebs at 4:04 PM on May 25, 2010 [1 favorite]

Where you're getting hung up is the difference between a "generic" boy, and a "specific" boy. The more specifically the boy is identified, the closer to 50% the answer becomes, as you can get closer to being able to "factor out" a specific boy, which turns it into a simple "what are the odds a single child is a boy or a girl" problem -- 50%.

I have two children. One of them is a boy. How likely is the other child to be a boy? 1/3.

I have two children. One of them is a boy born on Tuesday. How likely is the other a boy? 13/27.

I have two children. One of them is a boy named Sue who walks with a limp, has three nipples and is majoring in tirebiting at the school of hard knocks. How likely is the other a boy? 1/2.

In the generic case, there's three of four cases where there's "a boy" existent -- and it could be either the younger or older sibling being discussed, but only one of those three cases has ANOTHER boy.

Okay, I need MOAR COFFEE.

posted by seanmpuckett at 4:04 PM on May 25, 2010 [18 favorites]

I have two children. One of them is a boy. How likely is the other child to be a boy? 1/3.

I have two children. One of them is a boy born on Tuesday. How likely is the other a boy? 13/27.

I have two children. One of them is a boy named Sue who walks with a limp, has three nipples and is majoring in tirebiting at the school of hard knocks. How likely is the other a boy? 1/2.

In the generic case, there's three of four cases where there's "a boy" existent -- and it could be either the younger or older sibling being discussed, but only one of those three cases has ANOTHER boy.

Okay, I need MOAR COFFEE.

posted by seanmpuckett at 4:04 PM on May 25, 2010 [18 favorites]

I'm with daq ... the question is semantically flawed if the purpose was to create a brainteaser.

*"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"*

It doesn't mention birth order. Everything else being brought up is being inferred*into *the question *by the listener.*

I have two children. One is a boy. What're the odds on the other one? Fifty-fucking-fifty. There's only one answer, not math gymnastics.

posted by Cool Papa Bell at 4:04 PM on May 25, 2010 [5 favorites]

It doesn't mention birth order. Everything else being brought up is being inferred

I have two children. One is a boy. What're the odds on the other one? Fifty-fucking-fifty. There's only one answer, not math gymnastics.

posted by Cool Papa Bell at 4:04 PM on May 25, 2010 [5 favorites]

Yeah, I just figured out what seanmpuckett mentions above. Born on Tuesday narrows the sample space, leading to a closer to 1/2 chance.

I wonder if this thread will be like Wikipedia talk threads about Monty Hall or .999... = 1.

posted by kmz at 4:11 PM on May 25, 2010

I wonder if this thread will be like Wikipedia talk threads about Monty Hall or .999... = 1.

posted by kmz at 4:11 PM on May 25, 2010

Err - yes, even stated as it is:

*""I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?""*

This does not exclude the children being twins. Both born on a Tuesday. Nobody said the other child was NOT born on the same Tuesday. He's just saying that one boy was born on a Tuesday... and could have followed it up with "and the other boy - or girl - was born on the same Tuesday". This also says nothing about birth order... what if it was a Caesarean simultaneous?

posted by VikingSword at 4:13 PM on May 25, 2010

This does not exclude the children being twins. Both born on a Tuesday. Nobody said the other child was NOT born on the same Tuesday. He's just saying that one boy was born on a Tuesday... and could have followed it up with "and the other boy - or girl - was born on the same Tuesday". This also says nothing about birth order... what if it was a Caesarean simultaneous?

posted by VikingSword at 4:13 PM on May 25, 2010

This seems like an annoying trick, not a puzzle. The difference is between the cases where order is important and where it is not. By specifying a distinguishing characteristic for the boy, you are forcing people to consider order and BB becomes bB and Bb.

But really nothing has changed. GB and GB are not equally probable to BB or GG. They are exactly half as probable.

posted by Nothing at 4:17 PM on May 25, 2010 [2 favorites]

But really nothing has changed. GB and GB are not equally probable to BB or GG. They are exactly half as probable.

posted by Nothing at 4:17 PM on May 25, 2010 [2 favorites]

For those who are looking to confirm this for themselves: think of the gender and birthday of the two children as independent events: what happens with the older one has no effect on the younger. Now, we can make a chart of all possible combinations, and cross out the ones where no boys are born on a Tuesday. The remaining combinations can be filled in with 1s (for one boy) and 2s (for two boys):

It's also interesting that if no information is given about the birthday, it goes back to 1/3 chance of two boys:

posted by Upton O'Good at 4:18 PM on May 25, 2010 [28 favorites]

bM .2............ bT 22222221111111 bW .2............ Y bTh .2............ o bF .2............ u bS .2............ n bSu .2............ g gM .1............ e gT .1............ r gW .1............ gTh .1............ gF .1............ gS .1............ gSu .1............ bbbbbbbggggggg MTWTFSSMTWTFSS h u h u Older14 ones, 13 twos, 27 in total: 13/27 chance of two boys.

It's also interesting that if no information is given about the birthday, it goes back to 1/3 chance of two boys:

bM 22222221111111 bT 22222221111111 bW 22222221111111 Y bTh 22222221111111 o bF 22222221111111 u bS 22222221111111 n bSu 22222221111111 g gM 1111111....... e gT 1111111....... r gW 1111111....... gTh 1111111....... gF 1111111....... gS 1111111....... gSu 1111111....... bbbbbbbggggggg MTWTFSSMTWTFSS h u h u Older

posted by Upton O'Good at 4:18 PM on May 25, 2010 [28 favorites]

I'll gladly pay you Tuesday for a hamburger today.

posted by hellojed at 4:18 PM on May 25, 2010 [2 favorites]

posted by hellojed at 4:18 PM on May 25, 2010 [2 favorites]

It's easier to conceptualize if you place both the boy and the other child on an infinite, frictionless plain, in a vacuum. Just make sure they have spacesuits.

posted by Ritchie at 4:20 PM on May 25, 2010 [3 favorites]

posted by Ritchie at 4:20 PM on May 25, 2010 [3 favorites]

My favorite follow-up to this puzzle is, "You have a brother. Is it more likely that you're male or female?" The answer is left as an exercise to the reader.

posted by LSK at 4:20 PM on May 25, 2010 [2 favorites]

posted by LSK at 4:20 PM on May 25, 2010 [2 favorites]

Yeah, it's completely irrelevant that the one of the children is a boy, and it's certainly irrelevant what day it was born on, or how far it shot out. Without detailed genetic information and family history data pushed through a mathputer, the probability of one child being a boy is 50%, and the probability of the other child being a boy is also 50%.

posted by turgid dahlia at 4:21 PM on May 25, 2010 [1 favorite]

The reasoning in the article is sound, but they overlooked the 0.2 percent chance that the boy born on tuesday is one of two identical twins. So, it should be slightly higher than 13/27.

posted by esprit de l'escalier at 4:21 PM on May 25, 2010 [2 favorites]

posted by esprit de l'escalier at 4:21 PM on May 25, 2010 [2 favorites]

Birth order is a red herring. The key point is that they're two distinct independent events. You can't mix them up.

Nothing: Are you saying 100 coin tosses doesn't give 25 each of HH, TH, HT, and TT? You might want to rethink that.

posted by kmz at 4:23 PM on May 25, 2010 [1 favorite]

Nothing: Are you saying 100 coin tosses doesn't give 25 each of HH, TH, HT, and TT? You might want to rethink that.

posted by kmz at 4:23 PM on May 25, 2010 [1 favorite]

Eponysterical. See above :)

posted by VikingSword at 4:24 PM on May 25, 2010

The coins are clearer.

#1. If you have two coins, toss one of them and it comes up heads, then the chance of the second one also being heads is 1/2.

#2. If you know a randomly selected pair of coins have been tossed, and that one of them, randomly selected, came up heads, then the chance of the other one being heads as well is 1/3.

He's worded a #2 instance to look like a #1 instance to make himself look smarter than you.

posted by imperium at 4:26 PM on May 25, 2010 [19 favorites]

#1. If you have two coins, toss one of them and it comes up heads, then the chance of the second one also being heads is 1/2.

#2. If you know a randomly selected pair of coins have been tossed, and that one of them, randomly selected, came up heads, then the chance of the other one being heads as well is 1/3.

He's worded a #2 instance to look like a #1 instance to make himself look smarter than you.

posted by imperium at 4:26 PM on May 25, 2010 [19 favorites]

*sigh* Note to self: don't try to argue about conditional probability on the Internet. It's not worth the trouble. (Probably.)

posted by kmz at 4:27 PM on May 25, 2010 [4 favorites]

posted by kmz at 4:27 PM on May 25, 2010 [4 favorites]

What if you grew up in Hong Kong?

posted by polymodus at 4:30 PM on May 25, 2010

And twins don't matter a whit to the calculations. Ok, now I'm really done.

posted by kmz at 4:32 PM on May 25, 2010 [1 favorite]

posted by kmz at 4:32 PM on May 25, 2010 [1 favorite]

If they had two boys born on a tuesday, then it could be either one they're talking about, so we only need to count it once in the sample space, which results in one less than 28.

posted by Brian B. at 4:35 PM on May 25, 2010

No, it isn't. That's exactly the first point that changes the result.

In your example, the options are binary: boy or girl. Therefore, 50%.

In the original example (minus Tuesday), the options include: 2 boys, 1 boy + 1 girl, 2 girls, for 3 possible combinations, and therefore any given child having a 1 in 3 chance of being a boy.

For each additional factor, the odds change. This is what odds are all about. While it remains true that (all other factors being equal) the odds of any given result of two possible combinations is 50%, it's also true that adding factors changes the possible combinations, which changes the odds.

In other words, your different answers are talking about different things. You're kind of all right (if you ignore strict definitions). Or you're kind of all wrong. Or exactly one of you is right.

posted by It's Raining Florence Henderson at 4:36 PM on May 25, 2010 [2 favorites]

Do you have any reasoning to refute the math in the other answer?

posted by OmieWise at 4:37 PM on May 25, 2010

Oh, oh, me, pick me!

posted by OmieWise at 4:38 PM on May 25, 2010 [1 favorite]

So, I’m aware of the reasoning for why the Tuesday-free version of the question has the answer 1/3, and I do follow why the answer for this version is 13/27. But the problem I have is that questions worded like this are exceptionally hard to read in the intended way — which in this case is something along these lines:

Assuming that the independent probabilities of a child being a boy and being born on any particular day of the week areposted by Tetch at 4:38 PM on May 25, 2010 [1 favorite]exactly1/2 and 1/7 respectively, what is the probability that, in families with exactly two (child × day-of-birth) pairs of which at least one is (boy × Tuesday), the other is (boy ×D) for any weekdayD?”

I choo choo choose you!

posted by It's Raining Florence Henderson at 4:39 PM on May 25, 2010 [1 favorite]

Perhaps they do but this "same situation" is twice as likely as 2 boys.

posted by Obscure Reference at 4:42 PM on May 25, 2010

Isn't the problem here that there are two interpretations of the original statement? English uses "one" as both a pronoun and a number:

**1. ** [This *One*, whom I am specifically referring to] is a boy born on a Tuesday.

**2. ** [*One*, or maybe more, but definitely at least one] is a boy born on a Tuesday.

These are two completely different questions. It's easy to see how people can be confused by the non-50% answers since*they're answering a different question*. Grade-school word trickery masquerading as collegiate-level probability.

posted by 0xFCAF at 4:42 PM on May 25, 2010 [2 favorites]

These are two completely different questions. It's easy to see how people can be confused by the non-50% answers since

posted by 0xFCAF at 4:42 PM on May 25, 2010 [2 favorites]

Having discovered kmz's rule long ago myself, I will content myself just to muddy the waters further with this curious paradox [pdf] of Smullyan.

posted by Wolfdog at 4:46 PM on May 25, 2010 [1 favorite]

posted by Wolfdog at 4:46 PM on May 25, 2010 [1 favorite]

This version clears up some of the ambiguity in the original problem.

posted by Obscure Reference at 4:50 PM on May 25, 2010 [6 favorites]

posted by Obscure Reference at 4:50 PM on May 25, 2010 [6 favorites]

I'm having trouble parsing your comment, OxFCAF. The explanation in the article does account for the possibility that both children are boys born on Tuesday. No word-trickery involved.

posted by roll truck roll at 4:51 PM on May 25, 2010

posted by roll truck roll at 4:51 PM on May 25, 2010

I normally love riddles, but this is seriously confusing and aggravating me. I don't "get" it. Is the 'trick' that by implying one boy is born on a Tuesday, the other child is not? Is that what this is about?

*In the original example (minus Tuesday), the options include: 2 boys, 1 boy + 1 girl, 2 girls, for 3 possible combinations, and therefore any given child having a 1 in 3 chance of being a boy.*

But the question already mentions that one is a boy, so isn't GG ruled out...? I still think the probability is 50%!

Gah!

posted by Gordafarin at 4:51 PM on May 25, 2010

But the question already mentions that one is a boy, so isn't GG ruled out...? I still think the probability is 50%!

Gah!

posted by Gordafarin at 4:51 PM on May 25, 2010

Wow, the amazing thing, and I swear I'm not just making this up to make myself seem all smart, but my first answer to the Tuesday's Boy question was 1/2.

In all probability, I was simply calculating something the wrong way, and incorrectly at that, and it's just a coincidence.

posted by Civil_Disobedient at 4:56 PM on May 25, 2010

In all probability, I was simply calculating something the wrong way, and incorrectly at that, and it's just a coincidence.

posted by Civil_Disobedient at 4:56 PM on May 25, 2010

Okay, after reading the blog post I think I understand it marginally better. So thanks for that. However, my brain is still telling me that any number but 50% makes no sense.

posted by Gordafarin at 4:58 PM on May 25, 2010

posted by Gordafarin at 4:58 PM on May 25, 2010

To properly analyze the problem, don't we need to know the odds of him saying he has a boy over saying he has a girl?

For example, if the man is equally likely to tell us about a boy as a girl and ignoring the Tuesday bit, we have 2 girls at a 25% chance with a 100% chance of being told about a girl, a boy and a girl at a 50% chance with a 50% of being told he has a boy and a 50% chance of telling us he has a girl, and a 25% chance of having two boys with a 100% chance of telling us about a boy.

So in the end, weeding out the cases where we are informed he has a girl, as that did not happen, we have a .25*1 out of .5*.5+.25*1 probability of the second being a boy, or 50/50 as one would naively expect.

Though if the man did value boys more highly than girls, the probability the other child would be a boy would be lower, as he would have been more likely to brag about having a boy when only one child was male.

posted by Zalzidrax at 4:59 PM on May 25, 2010

For example, if the man is equally likely to tell us about a boy as a girl and ignoring the Tuesday bit, we have 2 girls at a 25% chance with a 100% chance of being told about a girl, a boy and a girl at a 50% chance with a 50% of being told he has a boy and a 50% chance of telling us he has a girl, and a 25% chance of having two boys with a 100% chance of telling us about a boy.

So in the end, weeding out the cases where we are informed he has a girl, as that did not happen, we have a .25*1 out of .5*.5+.25*1 probability of the second being a boy, or 50/50 as one would naively expect.

Though if the man did value boys more highly than girls, the probability the other child would be a boy would be lower, as he would have been more likely to brag about having a boy when only one child was male.

posted by Zalzidrax at 4:59 PM on May 25, 2010

Agh! Martin Gardner died 2 days ago?! I literally just finished Calculus Made Easy which shockingly did actually make calculus quite easy.

.

posted by Damienmce at 5:02 PM on May 25, 2010

.

posted by Damienmce at 5:02 PM on May 25, 2010

Ask yourself why do they introduce both GB and BG. I don't know other than to confuse the situation because there is no difference between them. One is a G and the other is a B is not different from one is a B and the other a G.

I first read a version of this problem in Marilyn vos Savant's column in the late 90's and it has been confusing people for a long time now. In any case the reality is that the split for the gender* of any other child are in fact 50-50 and if the reality doesn't match then the phrasing must be in error.

*Used here of course in an oversimplified sense.

posted by vapidave at 5:06 PM on May 25, 2010 [1 favorite]

This is easy easy easy.

*"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"*

The probability that you have two boys is either exactly zero or exactly one, because the outcome has already taken place and the only possible probabilities for realized events are zero and one.

posted by ROU_Xenophobe at 5:07 PM on May 25, 2010 [9 favorites]

The probability that you have two boys is either exactly zero or exactly one, because the outcome has already taken place and the only possible probabilities for realized events are zero and one.

posted by ROU_Xenophobe at 5:07 PM on May 25, 2010 [9 favorites]

I think I get it (I didn't at first, even after reading the FA).

The probability of ONE child (either child) being a boy is 50%. And would continue to be so, no matter how many kids you have, or what you know about them, if you consider them individually, in sequence.

But the question is about the SET of BOTH children. Of all the possible gender combinations available in a set of two, what are the odds for the two-boys combination? And further, what are the odds of the SET being BB, knowing one is B? And further further, what are the odds of the set being BB, knowing one of them is a boy born on a Tuesday (here, we get into counting handshakes, and the odds of two people having the same birthday, and all that other stuff that I'm really bad at thinking about)?

posted by Mister Moofoo at 5:16 PM on May 25, 2010

The probability of ONE child (either child) being a boy is 50%. And would continue to be so, no matter how many kids you have, or what you know about them, if you consider them individually, in sequence.

But the question is about the SET of BOTH children. Of all the possible gender combinations available in a set of two, what are the odds for the two-boys combination? And further, what are the odds of the SET being BB, knowing one is B? And further further, what are the odds of the set being BB, knowing one of them is a boy born on a Tuesday (here, we get into counting handshakes, and the odds of two people having the same birthday, and all that other stuff that I'm really bad at thinking about)?

posted by Mister Moofoo at 5:16 PM on May 25, 2010

See, this is why I did so badly in statistics. Too often my answer to a proof is, "I don't believe you." This is from someone who did well in differential equations in high school.

I mean, I can't deny hundreds of years of thought about this, but I can' shake the feeling that it's just a math game that doesn't cross over into reality either. I wish I could be educated away from under this burden.

posted by cmoj at 5:18 PM on May 25, 2010 [1 favorite]

I mean, I can't deny hundreds of years of thought about this, but I can' shake the feeling that it's just a math game that doesn't cross over into reality either. I wish I could be educated away from under this burden.

posted by cmoj at 5:18 PM on May 25, 2010 [1 favorite]

Probability doesn't lend well to intuition. Which is why casinos, insurance companies, etc. are so profitable.

posted by Blazecock Pileon at 5:19 PM on May 25, 2010 [5 favorites]

Wow, the article on the Gathering includes a link to the blog of the 'youngest attendee', Nicholas Bickford. He sounds like an awesome 12 year old!

posted by jacalata at 5:30 PM on May 25, 2010 [1 favorite]

posted by jacalata at 5:30 PM on May 25, 2010 [1 favorite]

Yes. It's called Math Of The Gut, or Stomach-Reasoning, learned only at the School Of Hard Knocks, and not some fancy ivy league institute.

posted by turgid dahlia at 5:32 PM on May 25, 2010 [2 favorites]

Are you just throwing out Schrodinger's Box, cat unseen?

posted by Lemurrhea at 5:35 PM on May 25, 2010

"Nothing" had it correct above. Let's take the first instance:

you have two kids and one is a boy - what is the chance that both are boys?

The probability is 1/3: GB, BG, BB -- I think everyone agrees with that.

Now, when you add that one is born on tuesday watch how birth order sneaks into the probabilities. Now you have:

GBTu, BTuG, BTuB, BBTu

Now the probability is 1/2 as everyone thinks it should be intuitively.

The calculation in the article is wrong, and is playing a really sneaky trick. The final tabulation should be 14/28. If the boy we don't know about is also born on Tuesday, there are in fact two ways that could happen. He could have been born first, or born after the boy we already know about. Therefore you can't drop that from the tabulation.

posted by spaceviking at 5:35 PM on May 25, 2010 [2 favorites]

you have two kids and one is a boy - what is the chance that both are boys?

The probability is 1/3: GB, BG, BB -- I think everyone agrees with that.

Now, when you add that one is born on tuesday watch how birth order sneaks into the probabilities. Now you have:

GBTu, BTuG, BTuB, BBTu

Now the probability is 1/2 as everyone thinks it should be intuitively.

The calculation in the article is wrong, and is playing a really sneaky trick. The final tabulation should be 14/28. If the boy we don't know about is also born on Tuesday, there are in fact two ways that could happen. He could have been born first, or born after the boy we already know about. Therefore you can't drop that from the tabulation.

posted by spaceviking at 5:35 PM on May 25, 2010 [2 favorites]

Oh, poppycock. The child had to have been born on

posted by Sys Rq at 5:41 PM on May 25, 2010 [1 favorite]

Oh, well, here's a figure that might be of interest. If you want my own explanation, there it is: the chart shows 2*7*2*7 possibilites for the sex and birthday of someone's first- and second-born children. If all we know is that someone has two children, any one of those squares is equally likely. (If you don't agree to that, you're thinking about a different or more complicated one that was intended and you can work on that problem with whatever assumptions you like on your own.) When the speaker says "I have a boy born on tuesday", we know his case belongs to one of the - count them - 27 highlighted green boxes which include a Boy born on Tuesday. Any one of those 27 is still equally likely. Of the 27, 13 have two boys (bolded). This is pretty much a direct translation of what Tetch has written above.

(If you look carefully at the diagram, you can probably refine you sense of "what's really going on here." A temptation in thinking this through casually is to think that there are 4*7 = 28 squares which include "B,Tue" -- the thinking might be, "2 choices for which position has B,Tue; 2 choices for the other sex; 7 choices for the other day". But that's a slight overcount as it counts the {{B,Tue},{B,Tue}} pair twice, incorrectly. Spaceviking has just demontstated precisely the temptation to doublecount that square.)

*Ask yourself why do they introduce both GB and BG*

You have two options in dealing with a problem like this: you can work with three cases ("2B, 2G, and 1B1G") which have different probabilities of occurring (1B1G is twice as likely as either of the other two cases, as has already been explained to death); or you can temporarily introduce an element of order - which is unnecessary but convenient, like an extra construction line in a geometry problem - and work with four cases (BB, BG, GB, GG) which are all equally likely. Both approaches, if worked through carefully, give the same answer. The advantage to the latter is that often reduces the calculation to simply counting boxes in a grid.

posted by Wolfdog at 5:41 PM on May 25, 2010 [3 favorites]

(If you look carefully at the diagram, you can probably refine you sense of "what's really going on here." A temptation in thinking this through casually is to think that there are 4*7 = 28 squares which include "B,Tue" -- the thinking might be, "2 choices for which position has B,Tue; 2 choices for the other sex; 7 choices for the other day". But that's a slight overcount as it counts the {{B,Tue},{B,Tue}} pair twice, incorrectly. Spaceviking has just demontstated precisely the temptation to doublecount that square.)

You have two options in dealing with a problem like this: you can work with three cases ("2B, 2G, and 1B1G") which have different probabilities of occurring (1B1G is twice as likely as either of the other two cases, as has already been explained to death); or you can temporarily introduce an element of order - which is unnecessary but convenient, like an extra construction line in a geometry problem - and work with four cases (BB, BG, GB, GG) which are all equally likely. Both approaches, if worked through carefully, give the same answer. The advantage to the latter is that often reduces the calculation to simply counting boxes in a grid.

posted by Wolfdog at 5:41 PM on May 25, 2010 [3 favorites]

The question doesn't address birth order. It simply asks what are the odds my two children are both boys.

If we know nothing, then the possible, and equally likely, scenarios are that I have two boys, that I have two girls, or that I have one of each. If we know one child is a boy, we've eliminated the two girls scenario, leaving two equally likely possibilities.

posted by Naberius at 5:48 PM on May 25, 2010

If we know nothing, then the possible, and equally likely, scenarios are that I have two boys, that I have two girls, or that I have one of each. If we know one child is a boy, we've eliminated the two girls scenario, leaving two equally likely possibilities.

posted by Naberius at 5:48 PM on May 25, 2010

Aren't we ignoring the sex ratio? We wouldn't want to get the math wrong after all that work.

posted by ecurtz at 5:52 PM on May 25, 2010

posted by ecurtz at 5:52 PM on May 25, 2010

Naberius: "*The question doesn't address birth order. It simply asks what are the odds my two children are both boys.*"

Reread Wolfdog's comment. You're right that birth order doesn't matter, but it's a way of helping yourself understand that**1b1g and 2b don't have the same probability**. If you don't want to call it birth order, then call it alphabetical order, or call it "which one sits on the left in the family photo."

posted by roll truck roll at 5:52 PM on May 25, 2010

Reread Wolfdog's comment. You're right that birth order doesn't matter, but it's a way of helping yourself understand that

posted by roll truck roll at 5:52 PM on May 25, 2010

No. Take an analogous situation like coin flips. The probability of getting two tails is not the same as getting only one tail for two flips. in the set of possible outcomes, {TT, HT, TH, HH} there is a very real difference between them.

posted by Avelwood at 5:56 PM on May 25, 2010

Reality DOES work like this! It is not some word puzzle.

We can illustrate it with coins. Lets play a game. This is something you can try at home:

You and I and a friend. The friend has two coins.

1. He will secretly flip both coins, under a hat or behind a screen, and take a look at them.

2. When at least one of the coins is heads he will say "OK. AT LEAST ONE OF THE COINS is HEADS!"

3. Now, he will grab one of the coins with heads up, show it to us, and HAND IT TO US.

Now, there is one coin sitting there behind the screen. It is either HEADS or TAILS.

That is, my OTHER COIN/CHILD is either BOY or GIRL. Both are equally probable, right? RIGHT?

If you try this at home and repeat this a 100 times, I guarantee that 1/3 OF THE TIME THAT OTHER COIN is HEADS.

NOT 1/2.

1/3!!!

There is no word game here.

In fact, it is probably easy to construct a casino game based on the above. Then people can continue to argue that it is semantics - that that other child is equally likely to be a boy or girl - as they lose the shirt off their back.

posted by vacapinta at 5:56 PM on May 25, 2010 [3 favorites]

*gropes for calculator*

OK...

*frowns*

*boop*

*click*

posted by zarq at 5:59 PM on May 25, 2010 [7 favorites]

OK...

*frowns*

1 Boy 1 X? Carry the Tuesday...*beep*

*boop*

*click*

Total: Ghostbusters 2I'd better get back to you...

posted by zarq at 5:59 PM on May 25, 2010 [7 favorites]

So the question boils-down to...

A: 1/2.

posted by ZenMasterThis at 6:02 PM on May 25, 2010 [2 favorites]

Ok the problem here is that in the article they double count the BB combination giving them equal weight with the BG combos. It should just be:

BtuG, GBtu, BBtu (BtuB)

The order in which the two boys are born shouldn't matter and shouldn't create another set of probabilities. As vacapinta so clearly described.

So there are 3 combos with 7 possibilities each with only 7 satisfying the two boy requirement, so 7/21 = 1/3

posted by spaceviking at 6:05 PM on May 25, 2010

BtuG, GBtu, BBtu (BtuB)

The order in which the two boys are born shouldn't matter and shouldn't create another set of probabilities. As vacapinta so clearly described.

So there are 3 combos with 7 possibilities each with only 7 satisfying the two boy requirement, so 7/21 = 1/3

posted by spaceviking at 6:05 PM on May 25, 2010

My stats prof would likely tell me that the probability is either 0 or 1, for while we do not know if the other child is a boy, the child's sex/gender is already set, thus no chance. How rare it is could be gaged by probability, IMHO.

If the question was something like "I have a child, soon to have another one which is yet conceived. What is the probability that this one will be a male?" then we would have something, for nothing is set in stone yet.

Sorry if this has already been said upthread, I skimmed and really did not see this solution popping out.

posted by JoeXIII007 at 6:05 PM on May 25, 2010

Actually, you point out the problem here right in your step two.

In the boy on Tuesday puzzle, this is only going to similar if the guy has some reason to talk about the boy born on Tuesday over the girl born on Tuesday or the boy born on Wednesday.

A proper formulation where all this tricky probability work would actually apply would go something like this:

"I was talking to my friends recently who just had a boy--he was born on Tuesday--which is quite the coincidence since I also have a boy who was born on Tuesday. Knowing that, what is the probability that the other of my two children is a boy?"

posted by Zalzidrax at 6:12 PM on May 25, 2010

Well, according to this genealogy site, Gary Foshee has FOUR children, and only one of them is a boy. So, I say the probability is 0%, and I SAY HE IS A LIAR.

Seriously, though, I was taught that when you work on probabilities, you make charts to help, and my chart would look like this:

B G

1 1

0 2

2 0

And then I would cross out the 0, 2 solution, since I already know that one of the children is a boy.

So I am left with the possibility of one boy and one girl, or two boys.

So I would have said the answer is 50%, and all the math people would have laughed and pointed their fingers and I'd have been shamed in front of the world.

posted by misha at 6:17 PM on May 25, 2010

Seriously, though, I was taught that when you work on probabilities, you make charts to help, and my chart would look like this:

B G

1 1

0 2

2 0

And then I would cross out the 0, 2 solution, since I already know that one of the children is a boy.

So I am left with the possibility of one boy and one girl, or two boys.

So I would have said the answer is 50%, and all the math people would have laughed and pointed their fingers and I'd have been shamed in front of the world.

posted by misha at 6:17 PM on May 25, 2010

vacapinta, the point is that a reasonable interpretation of "One is a boy" is more like a coin game where your friend flips a nickel and a quarter and says "The nickel was heads. What are the odds the quarter was heads?". In code:

**Intepretation 1:** "One" means "a specific one of them":

**Interpretation 2:** "One" means "one or more of them":

posted by 0xFCAF at 6:24 PM on May 25, 2010 [3 favorites]

int boys = 0, girls = 0; for (int i = 0; i < 1000; i++) { int gender1 = random.Next(2), gender2 = random.Next(2); if (gender1 == 0) { if (gender2 == 0) { boys++; } else { girls++; } } } Console.WriteLine("{0} / {1}", boys, girls);Output: 255 / 256, i.e. 50/50

int boys = 0, girls = 0; for (int i = 0; i < 1000; i++) { int gender1 = random.Next(2), gender2 = random.Next(2); if (gender1 == 0 || gender2 == 0) { if (gender2 == 0) { boys++; } else { girls++; } } } Console.WriteLine("{0} / {1}", boys, girls);Output: 517 / 209, i.e. 1/3

posted by 0xFCAF at 6:24 PM on May 25, 2010 [3 favorites]

For those who want to see it more intuitively, here's a pretty clear illustration of the sample space (which I now see somewhat resembles Wolfdog's). Just count up the area in the BB section and divide it by the rest -- either yellow for the BB probability, or the orange for the BB conditional on one boy having been born on a Tuesday. You can see that, as the width of the orange stripes shrinks (eg, if you were specifying the day of the year, rather than the day of the week), the ratio of the stripe inside the BB square to the stripe outside the BB square approaches 1/2 (ie, the intersection at the center of the cross goes to 0).

So I'd say I understand it mathematically, and with the picture even understand it intuitively -- but I admit, something in me still balks. Not at the logic or truth of the result, but at some ill-defined conflict between my intuitions and the result. I do think it's worth thinking about why, for those of us who are perfectly fine with the 1/3 result, the Tuesday 13/27 bit is still unsettling. Variants show this even more clearly: I have two dice, which may be either black or red; one of them is red. What color is the other? Oh, and I just rolled a 3 with that red one: now what color is the other one? Weird. (Or, "oh, and that red one was in Reagan's pocket when he was shot. Now what color is the other one?")

posted by chortly at 6:24 PM on May 25, 2010 [6 favorites]

So I'd say I understand it mathematically, and with the picture even understand it intuitively -- but I admit, something in me still balks. Not at the logic or truth of the result, but at some ill-defined conflict between my intuitions and the result. I do think it's worth thinking about why, for those of us who are perfectly fine with the 1/3 result, the Tuesday 13/27 bit is still unsettling. Variants show this even more clearly: I have two dice, which may be either black or red; one of them is red. What color is the other? Oh, and I just rolled a 3 with that red one: now what color is the other one? Weird. (Or, "oh, and that red one was in Reagan's pocket when he was shot. Now what color is the other one?")

posted by chortly at 6:24 PM on May 25, 2010 [6 favorites]

This thread makes me want to open an online casino.

posted by esprit de l'escalier at 6:34 PM on May 25, 2010 [3 favorites]

posted by esprit de l'escalier at 6:34 PM on May 25, 2010 [3 favorites]

That's a lovely, well-organized illustration, chortly. Now what happens next is someone will look at it briefly and say "But the other child could be a boy or a girl! So it's 50%, obviously!" and walk away shaking their head at how gosh-darned silly and complicated you're making it.

posted by Wolfdog at 6:34 PM on May 25, 2010

posted by Wolfdog at 6:34 PM on May 25, 2010

Dammit, and I was so proud that I got all those arithmetic questions right. I understand why the answer is not 1/2 even though my instinct wants to bludgeon anyone who says otherwise.

So now I have a better idea of how mathematicians can see the world...and it seems so*fun*. Magic? Balloons? Having your hallways tiled in sequences? How can I get my brain operating on that level so I can get in on that action?

posted by zix at 6:35 PM on May 25, 2010

So now I have a better idea of how mathematicians can see the world...and it seems so

posted by zix at 6:35 PM on May 25, 2010

Ok fuck, I'm wrong. 13/27 is correct. Damnnit.

Ok I'm trying to make this make sense and I think this is the closest I've come.

The question should be worded: What are the chances that I have two boys and at least one is a boy born on tuesday?

You have 2 chances with 14 different possibilities for each chance (7 days boy + 7 days girl). Giving 196 total possible outcomes.

Your chances of having at least one boy born on tuesday is 27/196 = 13.8%

Your chances of having two boys: 25%

Your chances of having two boys and at least one born on tues is 13/196 = 6.6%

This only helps me a little... time to drink

posted by spaceviking at 6:37 PM on May 25, 2010 [2 favorites]

Ok I'm trying to make this make sense and I think this is the closest I've come.

The question should be worded: What are the chances that I have two boys and at least one is a boy born on tuesday?

You have 2 chances with 14 different possibilities for each chance (7 days boy + 7 days girl). Giving 196 total possible outcomes.

Your chances of having at least one boy born on tuesday is 27/196 = 13.8%

Your chances of having two boys: 25%

Your chances of having two boys and at least one born on tues is 13/196 = 6.6%

This only helps me a little... time to drink

posted by spaceviking at 6:37 PM on May 25, 2010 [2 favorites]

Tuesday! That's where I'm a Viking!

posted by magnificent frigatebird at 6:50 PM on May 25, 2010 [3 favorites]

posted by magnificent frigatebird at 6:50 PM on May 25, 2010 [3 favorites]

For those of you that still don't buy the 1/3 explanation, consider it like this:

In the year 2000 the parents have their first child which is either a boy or a girl. That's two possible outcomes. Then later in 2004 they have a second child, which also has equal chances of being a boy or a girl. That's another two possible outcomes. Those two events were not related in any way, which means in 2010 there are FOUR possible configurations:

1. a ten year old boy and a six year old girl

2. a ten year old boy and a six year old boy

3. a ten year old girl and a six year old girl

4. a ten year old girl and a six year old boy

Now, can you see that #4 and #1 are**completely different scenarios**? They both involve having one of each gender, but there were two ways of getting to that result which means it was **twice as likely**. If you know that one of the kids is a boy you know that #3 was not possible, leaving three possible equally likely scenarios, each with a 1/3 chance.

Or another way of stating this: when the problem says "I have two kids, one of which is a boy, what the the probability that the other is a boy", it is**not the same thing** as asking "my younger child is a boy, what are the chances that my older child is a boy?"

posted by Rhomboid at 7:00 PM on May 25, 2010 [4 favorites]

In the year 2000 the parents have their first child which is either a boy or a girl. That's two possible outcomes. Then later in 2004 they have a second child, which also has equal chances of being a boy or a girl. That's another two possible outcomes. Those two events were not related in any way, which means in 2010 there are FOUR possible configurations:

1. a ten year old boy and a six year old girl

2. a ten year old boy and a six year old boy

3. a ten year old girl and a six year old girl

4. a ten year old girl and a six year old boy

Now, can you see that #4 and #1 are

Or another way of stating this: when the problem says "I have two kids, one of which is a boy, what the the probability that the other is a boy", it is

posted by Rhomboid at 7:00 PM on May 25, 2010 [4 favorites]

Try to think of it this way: The reason you are not double-counting the "two boys, both Tuesday" case is that there is no other ordering. For another case, let's say "boy-Tuesday, girl-Friday", you are also counting "girl-Friday, boy-Tuesday". You can't do that when they are both boys born on the same day. The "boy-Tuesday, boy-Tuesday" case is same no matter how you order it. That overlap seems to be confusing the probability issue.

posted by Avelwood at 7:03 PM on May 25, 2010

Whoops -- the math is fine and the picture is solid, but my words were a bit sloppy: I didn't mean "divide it by the rest" but "divide it by the total". Similarly, the ratio of orange inside the BB square to *total* orange goes to 1/2 as the stripe's width shrinks.

posted by chortly at 7:05 PM on May 25, 2010

posted by chortly at 7:05 PM on May 25, 2010

Let's change the question a bit.

I have 2 children, one of them was born on a particular day of the week and is a boy. What are the odds that the other child is a boy?

To my mind, this is the same question, no? The days of the week are arbitrarily named. So let's call the day of the week the child was born on A, the next day B and so on. Fill in the chart the same way.

The length of the week is arbitrary, too. So let's say "I have a child who was born on a particular day in a temporal cycle of length x" What's the probability that the other child is a boy as X approaches infinity? 1/2

But this is not information that is any different from knowing that one child is a boy to begin with. Every child is born on a particular day.

So either A) the day the child is born is irrelevant and shouldn't be taken into account or B) knowing that one child is a boy, the odds the other child is a boy is 50/50.

I'm leaning towards A, personally.

posted by empath at 7:05 PM on May 25, 2010

I have 2 children, one of them was born on a particular day of the week and is a boy. What are the odds that the other child is a boy?

To my mind, this is the same question, no? The days of the week are arbitrarily named. So let's call the day of the week the child was born on A, the next day B and so on. Fill in the chart the same way.

The length of the week is arbitrary, too. So let's say "I have a child who was born on a particular day in a temporal cycle of length x" What's the probability that the other child is a boy as X approaches infinity? 1/2

But this is not information that is any different from knowing that one child is a boy to begin with. Every child is born on a particular day.

So either A) the day the child is born is irrelevant and shouldn't be taken into account or B) knowing that one child is a boy, the odds the other child is a boy is 50/50.

I'm leaning towards A, personally.

posted by empath at 7:05 PM on May 25, 2010

It makes me sad that mathematicians are still missing the gap in logic here.

Possible options: (1) bb / (2) bg / (3) gg / (4) gb

If he reveals that the*first *child is a boy that eliminates TWO options (3 and 4), not one. This leaves two options (1 and 2). We thus have a 50% chance that he has two boys.

If he reveals that the*second *child is a boy that eliminates TWO options (2 and 3). This leaves two options (1 and 4). We again have a 50% chance that he has two boys.

The flaw in all these puzzles is that we assume that the reveal only eliminates one option rather than two. The flaw is easier to visualize in the similar Monty Hall puzzle, and rather obvious if you sketch it out on a chalkboard.

posted by kanewai at 7:17 PM on May 25, 2010 [2 favorites]

Possible options: (1) bb / (2) bg / (3) gg / (4) gb

If he reveals that the

If he reveals that the

The flaw in all these puzzles is that we assume that the reveal only eliminates one option rather than two. The flaw is easier to visualize in the similar Monty Hall puzzle, and rather obvious if you sketch it out on a chalkboard.

posted by kanewai at 7:17 PM on May 25, 2010 [2 favorites]

Ok I think I understand this intuitively now.

Pose the question a bit different: What percentage of families with two children have a boy born on tuesday?

It makes sense that families with two boys will be more likely to have boys born on a tuesday than other families.

Now, of those families how many have two boys?

It makes sense that a much greater proportion of families with two boys will have boys born on a tuesday. In fact, if you think about it the total number of boys in families with two boys is the same as the total number of boys in families with a boy and a girl.

Therefore if you take some selection factor on the boys, it will be half as strong on the families with two boys than the families with a boy and girl because the families with two boys have two chances at it.

posted by spaceviking at 7:19 PM on May 25, 2010 [6 favorites]

Pose the question a bit different: What percentage of families with two children have a boy born on tuesday?

It makes sense that families with two boys will be more likely to have boys born on a tuesday than other families.

Now, of those families how many have two boys?

It makes sense that a much greater proportion of families with two boys will have boys born on a tuesday. In fact, if you think about it the total number of boys in families with two boys is the same as the total number of boys in families with a boy and a girl.

Therefore if you take some selection factor on the boys, it will be half as strong on the families with two boys than the families with a boy and girl because the families with two boys have two chances at it.

posted by spaceviking at 7:19 PM on May 25, 2010 [6 favorites]

Suppose A knows I have a son B, and A also knows I have two children, but A does not know the sex of my other child C. Suppose I do not want A to know the sex of C. Should I then avoid telling A the day of the week that B was born? Because then A will be able to guess C's sex easier? Is that what the math is saying?

posted by gubo at 7:20 PM on May 25, 2010 [1 favorite]

posted by gubo at 7:20 PM on May 25, 2010 [1 favorite]

He doesn't say which child is a boy, only that one of them is.

posted by empath at 7:22 PM on May 25, 2010 [1 favorite]

holy shit people the plane takes off the conveyor belt

posted by chinston at 7:22 PM on May 25, 2010 [3 favorites]

posted by chinston at 7:22 PM on May 25, 2010 [3 favorites]

spaceviking, you know that actually makes sense to me.

I retract my previous comment.

posted by empath at 7:23 PM on May 25, 2010

I retract my previous comment.

posted by empath at 7:23 PM on May 25, 2010

It makes sense that families with two boys will be more likely to have boys born on a tuesday than other families.

DUDE. You rock. I can fricking sleep tonight without grinding my teeth over this.

posted by Durn Bronzefist at 7:26 PM on May 25, 2010

But that's ok ... as there is a 50% chance that he selects either path A or path B, so the odds remain the same.

posted by kanewai at 7:28 PM on May 25, 2010

I don't even understand the introduction.

*"To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3."*

Yes, GB is the same as BG, it is the same outcome. Hence it is twice as likely to have it compared to GG or BB. How can it be 1/3? Why is BB not 1/4? How can you "eliminate" this?

posted by yoyo_nyc at 7:35 PM on May 25, 2010

Yes, GB is the same as BG, it is the same outcome. Hence it is twice as likely to have it compared to GG or BB. How can it be 1/3? Why is BB not 1/4? How can you "eliminate" this?

posted by yoyo_nyc at 7:35 PM on May 25, 2010

Reality really honestly does work like this. Consider this:

If I have two children, one is a boy, and the other is of unknown gender. If as many people have stated, the gender of the second child is a 50/50 split, then of the people who have one male child, half should have two male children, and half should have one son one daughter. This is due to people making the assumption that the having of a son is a confirmed (past) event, and the other event is an unknown event. It certainly makes sense for intuition to tell us this, because that's the situation that will be encountered in reality. In this situation the possibilities:

BB - Still Possible

GG - Impossible, (all will agree)

**GB - Impossible to those making mistake, but in questioner's opinion, STILL POSSIBLE**

BG - Still Possible.

There used to be a game show in which you had to pick one of three doors. One door had a good prize, the others had nonsense. So if you picked truly "at random" you had a 1/3 shot at picking the right door. After you had chosen, one of the remaining "bad" doors was eliminated, and you were left with the option to remain with your chosen door, or switch to the remaining mystery door. Although it is counterintuitive, switching is always right.

You picked your door with a 1/3 chance of success. However when the "bad" door was eliminated, the probability that the unchosen door is correct grew. This is because the "correct" door can never be eliminated, and your door can never be eliminated, correct or not. (Shamefully, I've forgotten how to calculate exactly how much better it is).

The situations you can eliminate or still have to consider matter a huge amount in these problems, and thinking chronologically is a big disadvantage. I know just enough about probability to know how horribly likely it is that I'm even making a mistake now, but this is what I understand at the moment.

posted by SomeOneElse at 7:36 PM on May 25, 2010 [1 favorite]

If I have two children, one is a boy, and the other is of unknown gender. If as many people have stated, the gender of the second child is a 50/50 split, then of the people who have one male child, half should have two male children, and half should have one son one daughter. This is due to people making the assumption that the having of a son is a confirmed (past) event, and the other event is an unknown event. It certainly makes sense for intuition to tell us this, because that's the situation that will be encountered in reality. In this situation the possibilities:

BB - Still Possible

GG - Impossible, (all will agree)

BG - Still Possible.

There used to be a game show in which you had to pick one of three doors. One door had a good prize, the others had nonsense. So if you picked truly "at random" you had a 1/3 shot at picking the right door. After you had chosen, one of the remaining "bad" doors was eliminated, and you were left with the option to remain with your chosen door, or switch to the remaining mystery door. Although it is counterintuitive, switching is always right.

You picked your door with a 1/3 chance of success. However when the "bad" door was eliminated, the probability that the unchosen door is correct grew. This is because the "correct" door can never be eliminated, and your door can never be eliminated, correct or not. (Shamefully, I've forgotten how to calculate exactly how much better it is).

The situations you can eliminate or still have to consider matter a huge amount in these problems, and thinking chronologically is a big disadvantage. I know just enough about probability to know how horribly likely it is that I'm even making a mistake now, but this is what I understand at the moment.

posted by SomeOneElse at 7:36 PM on May 25, 2010 [1 favorite]

Kanewai, think of it in terms of information. You can't just say that he's telling you path A or B with 50% probability. What information is he giving you? Technically, he's just saying (A OR B). Or, equivalently, NOT ((NOT A) AND (NOT B)) - this is the only choice eliminated, which corresponds to option 3, gg.

posted by Lemurrhea at 7:42 PM on May 25, 2010

posted by Lemurrhea at 7:42 PM on May 25, 2010

I think the problem most of us are having with this isn't the math so much as the fact that the reveal isn't that mindblowing:

"What are the chances I have two boys?"

*Oh, I love these! Well, let me see...you would think that it would be 50%, but I bet there's an awesome non-intuitive answer!*

"[does math] Well, as you can see, one logical answer is 1/3..."

*That's what I'm talking about!*

"[does more math] ...but the real answer is 13/27, or 48.1%!"

*Oh. Okay. Cool trick, bro.*

posted by Ian A.T. at 7:46 PM on May 25, 2010

"What are the chances I have two boys?"

"[does math] Well, as you can see, one logical answer is 1/3..."

"[does more math] ...but the real answer is 13/27, or 48.1%!"

posted by Ian A.T. at 7:46 PM on May 25, 2010

Yes. Sorry, I misread it. I know about selective choice in mathematics. I thought he eliminated GB instead of GG.

posted by yoyo_nyc at 7:53 PM on May 25, 2010

posted by yoyo_nyc at 7:53 PM on May 25, 2010

Allright to restate what I said above more clearly.

In all families with 2 children the total number of boys in families with a boy and a girl is equal to the total number of boys in families with two boys.

Let's say you had a total of 56 boys. 28 of the boys will be in families with two boys, and 28 of them will be in families with a boy and a girl.

Let's say you pick all the boys who are born on a Tuesday.

So it stands to reason that (if you ignore small sample size) 4 of the boys in two boy families will be born on tuesday and 4 of the boys in boy girl families will be born on tuesday.

So if you are a boy and you are born on Tuesday you have about a 50% chance to be in a two boy family. It's actually less because you have to ignore the case where both are born on tuesday as I was corrected somewhere up there.

posted by spaceviking at 8:01 PM on May 25, 2010 [1 favorite]

In all families with 2 children the total number of boys in families with a boy and a girl is equal to the total number of boys in families with two boys.

Let's say you had a total of 56 boys. 28 of the boys will be in families with two boys, and 28 of them will be in families with a boy and a girl.

Let's say you pick all the boys who are born on a Tuesday.

So it stands to reason that (if you ignore small sample size) 4 of the boys in two boy families will be born on tuesday and 4 of the boys in boy girl families will be born on tuesday.

So if you are a boy and you are born on Tuesday you have about a 50% chance to be in a two boy family. It's actually less because you have to ignore the case where both are born on tuesday as I was corrected somewhere up there.

posted by spaceviking at 8:01 PM on May 25, 2010 [1 favorite]

I have to work this out as a thought problem:

1. I have two coins, coin A and coin B. I toss coin A, it comes up heads. Now I intend to toss coin B. Given the fact that coin A came up heads, what is the probability that coin B will now come up heads? Answer: 1/2. The outcome of tossing coin A doesn't affect the outcome of tossing coin B.

2. I have two coins, coin A and coin B. I toss coin A, it comes up heads.*Also, I find out that coin A was minted in 1993.* Now I intend to toss coin B. Given the fact that coin A came up heads *and was minted in 1993*, what is the probability that coin B will now come up heads? Answer: 1/2. The outcome of tossing coin A doesn't affect the outcome of tossing coin B, *nor does the year it was minted*.

posted by gubo at 8:03 PM on May 25, 2010

1. I have two coins, coin A and coin B. I toss coin A, it comes up heads. Now I intend to toss coin B. Given the fact that coin A came up heads, what is the probability that coin B will now come up heads? Answer: 1/2. The outcome of tossing coin A doesn't affect the outcome of tossing coin B.

2. I have two coins, coin A and coin B. I toss coin A, it comes up heads.

posted by gubo at 8:03 PM on May 25, 2010

It kind of seems like the 1/3 answer hinges on not "using" your knowledge that one child is a boy, that is, counting the G/G option when calculating the probability.

And the 13/27 answer does use that knowledge, by not including any G/G options.

I still feel like the option where both boys are born on a Tuesday should be double counted because they are still different boys, the one we know about and the one we don't, if birth order is going to matter, but I realise I am probably very wrong.

This makes my head hurt. I need coffee.

posted by lwb at 8:06 PM on May 25, 2010

And the 13/27 answer does use that knowledge, by not including any G/G options.

I still feel like the option where both boys are born on a Tuesday should be double counted because they are still different boys, the one we know about and the one we don't, if birth order is going to matter, but I realise I am probably very wrong.

This makes my head hurt. I need coffee.

posted by lwb at 8:06 PM on May 25, 2010

You've made your argument worse here, because you've changed the question.

If you're a boy in a 2 child family, you always have a 50/50 chance of having a sister or brother in a two child family.

The question was if one of 2 children is a boy, what are the odds that both are boys, which is a 1/3rd chance.

The debate is over whether knowing the boy was born on tuesday makes a difference.

posted by empath at 8:07 PM on May 25, 2010

I am totally ignoring the Tuesday thing atm as well. Another thing to consider.

If you already have one male child, and are expecting a second, there is a 50/50 chance that second child would be a boy or a girl.

Let's say instead you are out and about and (this is why it is hard for us ppl who think) let's say a gypsy tells you. "You will have two children, and one of them will be a boy". Now, you have your first child. Let's say you have a 50/50 shot on this one. If it is a girl,**then at this point** you have reduced the problem to a 100% chance, if the gypsy is right, (and this is one irrationally reliable gypsy). If it is a boy, then when your second child comes, you have another 50/50 shot at boy or girl. This is not the same as assuming that the first child HAS to be a boy.

Now lets consider the insane confluence of events scenario. So let's say the gypsy tells you you will have two children, and one will be a boy who will have three eyes. (Let's consider this to be ridiculously rare). Being a male child almost becomes irrelevant except for its attachment to the hyperirregularity. So my options become:

B3E

G

B3E

B

B

B3E

G

B3E

and because it CAN happen:

B3E

B3E

B3E

G3E

Your first child. 50% chance boy 50% chance girl, .00~1% chance 3 eyes. So essentially, 50/50 girl boy. If it isn't a 3-eyed boy, your next child WILL be a boy with 3 eyes because of that darned gypsy. If it IS a 3 eyed boy, your next child has a 50/50 shot at being boy/girl, and ANOTHER vanishingly small shot at having 3 eyes.

Because most all of us think in terms of Already Having The X in the past, and Having a Chance at the y in the future we look at Boy Child or Boy Child on Tuesday or Boy Child with Three eyes as a confirmed past event and think that it doesn't affect the future event. But that isn't what these problems are asking! And I consider this stuff so challenging I want to point out that I am still probably wrong!

Basically, my advice is:

1. If a person like this asks you a question like this, the answer isn't the obvious answer! (mom's got a probability doctorate and siblings are math students. Really the correct answer is to glare and talk about online videogames in a droning monotonous voice.)

2. Think of these problems as Ides of March style predictions unless order is explicitly given.

posted by SomeOneElse at 8:08 PM on May 25, 2010 [1 favorite]

If you already have one male child, and are expecting a second, there is a 50/50 chance that second child would be a boy or a girl.

Let's say instead you are out and about and (this is why it is hard for us ppl who think) let's say a gypsy tells you. "You will have two children, and one of them will be a boy". Now, you have your first child. Let's say you have a 50/50 shot on this one. If it is a girl,

Now lets consider the insane confluence of events scenario. So let's say the gypsy tells you you will have two children, and one will be a boy who will have three eyes. (Let's consider this to be ridiculously rare). Being a male child almost becomes irrelevant except for its attachment to the hyperirregularity. So my options become:

B3E

G

B3E

B

B

B3E

G

B3E

and because it CAN happen:

B3E

B3E

B3E

G3E

Your first child. 50% chance boy 50% chance girl, .00~1% chance 3 eyes. So essentially, 50/50 girl boy. If it isn't a 3-eyed boy, your next child WILL be a boy with 3 eyes because of that darned gypsy. If it IS a 3 eyed boy, your next child has a 50/50 shot at being boy/girl, and ANOTHER vanishingly small shot at having 3 eyes.

Because most all of us think in terms of Already Having The X in the past, and Having a Chance at the y in the future we look at Boy Child or Boy Child on Tuesday or Boy Child with Three eyes as a confirmed past event and think that it doesn't affect the future event. But that isn't what these problems are asking! And I consider this stuff so challenging I want to point out that I am still probably wrong!

Basically, my advice is:

1. If a person like this asks you a question like this, the answer isn't the obvious answer! (mom's got a probability doctorate and siblings are math students. Really the correct answer is to glare and talk about online videogames in a droning monotonous voice.)

2. Think of these problems as Ides of March style predictions unless order is explicitly given.

posted by SomeOneElse at 8:08 PM on May 25, 2010 [1 favorite]

You've changed the question.

Flip both coins, but don't look at either.

Someone looks at the coins for you and tells you that at least one coin is heads. What are the odds the other is also heads? 1/3.

posted by empath at 8:09 PM on May 25, 2010 [2 favorites]

Actually I just realised 1/3 is after eliminating G/G (I really do need that coffee) so I retract my comment.

But it still seems wrong somehow. *glares at statistics in general*

posted by lwb at 8:13 PM on May 25, 2010

But it still seems wrong somehow. *glares at statistics in general*

posted by lwb at 8:13 PM on May 25, 2010

Okay I get this scenario. Now what if someone looks at the coins and says one coin is heads, and that very same coin was minted in 1993? Is the probability that the other coin is heads still 1/3?

posted by gubo at 8:15 PM on May 25, 2010 [2 favorites]

Wait, but why is G/G in the original set of options but no G/G options in the second? I retract my retraction, I'm still confused, somebody please explain to me.

COFFEE.

posted by lwb at 8:15 PM on May 25, 2010

COFFEE.

posted by lwb at 8:15 PM on May 25, 2010

Okay I get this scenario. Now what if someone looks at the coins and says one coin is heads, and that very same coin was minted in 1993? Is the probability that the other coin is heads still 1/3?

Excellent fucking question. IMO, it's still 1/3rd. This guy would say 1/2, probably. I don't see how its any different from the question in the fpp.

posted by empath at 8:20 PM on May 25, 2010

Actually, gubo has 100% convinced me that the answer in the article is wrong wrong wrong.

posted by empath at 8:21 PM on May 25, 2010

posted by empath at 8:21 PM on May 25, 2010

The problem is "What is the probability that I have two boys" is a bad question. If you interpret it as meaning "what is the chance that my other child is a boy", then it's 50%. But if you interpret it as "of the people who have two children and at least one is a boy, what percentage have two boys", it's 33.3%.

posted by 445supermag at 8:24 PM on May 25, 2010

posted by 445supermag at 8:24 PM on May 25, 2010

445, it's the same question.

You are reading the question as:

"If*this* child is a boy, what are the odds that *that* child is a boy", which is 50%. But he is not specifying this or that child.

posted by empath at 8:26 PM on May 25, 2010

You are reading the question as:

"If

posted by empath at 8:26 PM on May 25, 2010

If you are one of 56 boys from two kid families. You have a 33% chance of having a brother and a 66% chance of having a sister.

Of you and your friends, 28 of them will be in 2 boy families and 28 of them will be in Boy-girl families. However there will be 14 two boy families and 28 boy-girl families.

1/7 of the boys will be born on tuesday: leaving 8.

4 of the two boy families will have boys born on tuesday.

4 of the boy-girl families will have boys born on tuesday.

So if you are a boy born on tuesday, you will have about a 50% chance of being in a two boy family.

posted by spaceviking at 8:32 PM on May 25, 2010

Of you and your friends, 28 of them will be in 2 boy families and 28 of them will be in Boy-girl families. However there will be 14 two boy families and 28 boy-girl families.

1/7 of the boys will be born on tuesday: leaving 8.

4 of the two boy families will have boys born on tuesday.

4 of the boy-girl families will have boys born on tuesday.

So if you are a boy born on tuesday, you will have about a 50% chance of being in a two boy family.

posted by spaceviking at 8:32 PM on May 25, 2010

I would like one of the 13/27th'ers to answer Gubo's question and explain what the difference is.

To restate.

*I have flipped two coins and placed them under a box in front of me.*

I look in the box, and tell you that one of the coins is heads and it was minted in 1993

What are the odds that both are heads?

And actually, I'll add another one.

*I have flipped two coins and hid them under a box.*

I look under the box and tell you that one of the coins is heads and was minted in a year ending in 3.

What are the odds that both coins are heads?

posted by empath at 8:34 PM on May 25, 2010 [1 favorite]

To restate.

I look in the box, and tell you that one of the coins is heads and it was minted in 1993

What are the odds that both are heads?

And actually, I'll add another one.

I look under the box and tell you that one of the coins is heads and was minted in a year ending in 3.

What are the odds that both coins are heads?

posted by empath at 8:34 PM on May 25, 2010 [1 favorite]

I know that the simple case is 1/3.

spaceviking, here is why I am not convinced by your appealing logic:

you say:

4 of the two boy families will have boys born on tuesday.

4 of the boy-girl families will have boys born on tuesday.

and conclude:

So if you are a boy born on tuesday, you will have about a 50% chance of being in a two boy family.

yet you also say:*28 of them will be in 2 boy families and 28 of them will be in Boy-girl families.*

but fail to conclude from here that if you are a boy, you will have about a 50% chance of being in a two boy family. (no day-of-week mention). In our problem, we'd like an answer of 1/3 here. So, your chain of reasoning doesn't apply.

posted by milestogo at 8:44 PM on May 25, 2010

spaceviking, here is why I am not convinced by your appealing logic:

you say:

4 of the two boy families will have boys born on tuesday.

4 of the boy-girl families will have boys born on tuesday.

and conclude:

So if you are a boy born on tuesday, you will have about a 50% chance of being in a two boy family.

yet you also say:

but fail to conclude from here that if you are a boy, you will have about a 50% chance of being in a two boy family. (no day-of-week mention). In our problem, we'd like an answer of 1/3 here. So, your chain of reasoning doesn't apply.

posted by milestogo at 8:44 PM on May 25, 2010

Let's say you flip 56 different coins in pairs so that there are 28 pairs of two coins.

14 times you will have two heads

14 times you will have two tails

28 times you will have one head and one tail

Let's say that 1/7 coins are minted in Denver (to make it more plausible)

That means that of all of the heads that were flipped (56 in total) 8 of them were minted in Denver.

If you flipped two heads you have 28 total heads with 4 of them minted in denver

If you flipped a head and a tail you have 28 total heads with 4 of them minted in denver.

However, it is much more likely to have flipped two heads AND had one of them minted in Denver because there were 4 minted in denver out of 14 pairs.

Likewise you will have 4 heads minted in denver out of 28 Heads/tails pairs.

posted by spaceviking at 8:48 PM on May 25, 2010

14 times you will have two heads

14 times you will have two tails

28 times you will have one head and one tail

Let's say that 1/7 coins are minted in Denver (to make it more plausible)

That means that of all of the heads that were flipped (56 in total) 8 of them were minted in Denver.

If you flipped two heads you have 28 total heads with 4 of them minted in denver

If you flipped a head and a tail you have 28 total heads with 4 of them minted in denver.

However, it is much more likely to have flipped two heads AND had one of them minted in Denver because there were 4 minted in denver out of 14 pairs.

Likewise you will have 4 heads minted in denver out of 28 Heads/tails pairs.

posted by spaceviking at 8:48 PM on May 25, 2010

Yeah it seems to come down to the wording. If you say, "Take all sets of two coins where one coin is heads-up and also has quality X (where the general distribution of X is known). What fraction of these sets have two heads up?" then you move away from one-third territory, no? But the way empath and I have framed it, one-third seems to be the answer.

posted by gubo at 8:53 PM on May 25, 2010

posted by gubo at 8:53 PM on May 25, 2010

I'm skipping the discussion to put in my NO MATH answer and let the hive mind check my work ;) Please PM me if I don't respond to the thread

So I went with 1/2 at first two because I was thinking the "Gamblers Fallacy" problem of, all the previous information (order of kids, day of week) is irrelevant. But there are two scenarioes being worked with here:

S1: "my first child is a boy, what is the probability that my second child is also a boy?" (A: 1/2)

S2: "one of my children is a boy, what is the probably that they both are boys" (A: 1/3)

The difference between S1 and S2 seems nit-picky or semantics until you realize the situations that you would say S1 vs S2.

lets say my first child was a boy... and my wife is pregnant but I don't know what my second child will be yet. I would HAVE to say something like*S1* and get the answer of %50 that my future child will be a boy. The only tricky bit is the first clause of the sentence ("my first child is a ____") is actually irrelevant, all I am really saying is "what is the probability that my NEXT child will be a boy?" which (if gender assignment is like coin flips*) is always 50%. Even if I have had 100 previous children that were all girls, the chance that my next child might be a boy is still 50%.

The tricky part is you could still ask S1 while knowing the outcome and the true state of affairs (we painted the room blue already lets say) but the answer is still 50% to someone who has not seen the room or the baby and is thus answering a general probability question; say I asked them to guess what the sex of my new baby was. In their mind they would have a 50% chance of being right, the fact of my first child being a boy doesn't matter.

But to get back to the question... lets look at S2: It still may seem identical to S1 so lets extend it to show how it tells us more. Lets Take a species of S2, say but morphed and call it*"S2b"*

S2b: - "9 of my children are boys, what is the probability that all 10 are boys"

Same question as S2, different information. In order for this to make any sense you must assume that the person saying S2b (or S2 for that matter) is not trying to refer to a future event, IE not sarcastically saying, while his wife is still in labor "I've had 9 boys, whats the chance that I might*finally *get a girl?" that is really just S1 all over again (the gamblers dilemma)

NO NO, now the 9 and the 1 unknown are linked information. Hypothetically we can go to the census buro (of our fantasy world where people routinely have 10 kids... or it may be better to think about this as coin flips at this point) and look at the breakdown and see "of all the people who had 10 kids of which 9 are boys, it is more likely that, at least one of them was a girl than that all were boys" (10x more likely; or that case is 1 in 11).

You could do a combination tree to see this for yourself. Incidentally the logic tree is what takes the place of our census beuro. But for a logic tree the order matters because (taking the order of children 1st - 10th)

b b b b b b b b b b - is only one outcome where the other child is a boy whereas

G b b b b b b b b b

b G b b b b b b b b

b b G b b b b b b b

b b b G b b b b b b

b b b b G b b b b b

b b b b b G b b b b

b b b b b b G b b b

b b b b b b b G b b

b b b b b b b b G b

b b b b b b b b b G

is 10 outcomes where at least one is a girl.

(All above is from the much greater list of possible outcomes that includes:

G G G G G G G G G

and

G b G b b b b b G b and every other iteration which I won't list here because they do not fit "have 9 boys" criteria)

But you see, that order matters. Otherwise you are asking S**1**b : "my first **9** children are boys, what is the probability that my second 10th child is also a boy?

which the graph comparison is simple:

b b b b b b b b b ? (question S1b)

b b b b b b b b b G (outcome 1)

b b b b b b b b b b (Outcome 2)

Answer = 50%

So, ya order matters is what I am driving at, but it doesn't mean anything in relation to the question, just how the logic works out .

To give a sort of betting example. You visit your friend at a reunion after years of not seeing him. You never really kept up with his Christmas cards, correspondence etc of all the kids names who married who, what jobs they took, colleges, or even their sexes and ages, but you remember he had 10 (all adults now and so you cannot easily tell age, so you cannot guess who is oldest/youngest etc). He introduces you to his 9 grown sons sitting at a table. Then you note "hey, I thought you had 10 kids, I've only met 9" and he says "AHH, Pat is by the punch-bowl, go say hi". Well you see a man and a woman by the punch bowl. Who do you introduce yourself to? One is Pat, the other is a stranger. And you don't want to ask your "supposedly" good friend that you don't know the sex of one of his children...

So you think through the birth announcements you've received over the years (and promptly thrown away without opening them). And you wish you could go through each of those unopened letters to see if one of them was pink or all of them was blue. Now, you think "If I only had those letters now...". You would open them one at a time until you saw either 1 that was pink (AHH beautiful Patricia, problem solved) or ALL of them were blue (ok, Patrick it is, mother must be Irish...).**Notice**, you would only have to find 1 pink notice before stopping... if the first one was pink you wouldn't even have to look at the other 9... you KNOW they are all guys. But if Pat is Patrick then you will only KNOW that he is so after opening ALL of those unopened letters.

AH-HAH! well I might not have the letters with me, but while walking over to the punch bowl this little thought experiment has yielded something! Just like you wouldn't bet that it is very likely I could flip a coin 10 times and get all heads... I can also wager that it isn't likely I would open all 10 letters and see blue each time... (GO breakdown of the wave function) Because opening those letters all those years ago vs opening them now is the same... to me**now**. Each is it's own %50 chance.

So with some confidence you walk over and say hi to the (beautiful by the way) woman... "You must be Patricia" you have ~91% chance of being right (Is my math right I hesitate with percentages?)

As for the Days of the week... I'm not going to get into that, but again it is assuming that by saying that 1 was born on a Tuesday that you are similarly cutting down the lists of the combination/tree... but you can see how that would make a difference now in a similar way.

*gender assignment is not 50/50, it is like 50/49/1 (intersex) but for purposes of mathemagical fun... lets forget about the hermaphrodites and b vs g disparity :)

posted by DetonatedManiac at 8:59 PM on May 25, 2010 [1 favorite]

So I went with 1/2 at first two because I was thinking the "Gamblers Fallacy" problem of, all the previous information (order of kids, day of week) is irrelevant. But there are two scenarioes being worked with here:

S1: "my first child is a boy, what is the probability that my second child is also a boy?" (A: 1/2)

S2: "one of my children is a boy, what is the probably that they both are boys" (A: 1/3)

The difference between S1 and S2 seems nit-picky or semantics until you realize the situations that you would say S1 vs S2.

lets say my first child was a boy... and my wife is pregnant but I don't know what my second child will be yet. I would HAVE to say something like

The tricky part is you could still ask S1 while knowing the outcome and the true state of affairs (we painted the room blue already lets say) but the answer is still 50% to someone who has not seen the room or the baby and is thus answering a general probability question; say I asked them to guess what the sex of my new baby was. In their mind they would have a 50% chance of being right, the fact of my first child being a boy doesn't matter.

But to get back to the question... lets look at S2: It still may seem identical to S1 so lets extend it to show how it tells us more. Lets Take a species of S2, say but morphed and call it

S2b:

Same question as S2, different information. In order for this to make any sense you must assume that the person saying S2b (or S2 for that matter) is not trying to refer to a future event, IE not sarcastically saying, while his wife is still in labor "I've had 9 boys, whats the chance that I might

NO NO, now the 9 and the 1 unknown are linked information. Hypothetically we can go to the census buro (of our fantasy world where people routinely have 10 kids... or it may be better to think about this as coin flips at this point) and look at the breakdown and see "of all the people who had 10 kids of which 9 are boys, it is more likely that, at least one of them was a girl than that all were boys" (10x more likely; or that case is 1 in 11).

You could do a combination tree to see this for yourself. Incidentally the logic tree is what takes the place of our census beuro. But for a logic tree the order matters because (taking the order of children 1st - 10th)

b b b b b b b b b b - is only one outcome where the other child is a boy whereas

G b b b b b b b b b

b G b b b b b b b b

b b G b b b b b b b

b b b G b b b b b b

b b b b G b b b b b

b b b b b G b b b b

b b b b b b G b b b

b b b b b b b G b b

b b b b b b b b G b

b b b b b b b b b G

is 10 outcomes where at least one is a girl.

(All above is from the much greater list of possible outcomes that includes:

G G G G G G G G G

and

G b G b b b b b G b and every other iteration which I won't list here because they do not fit "have 9 boys" criteria)

But you see, that order matters. Otherwise you are asking S

which the graph comparison is simple:

b b b b b b b b b ? (question S1b)

b b b b b b b b b G (outcome 1)

b b b b b b b b b b (Outcome 2)

Answer = 50%

So, ya order matters is what I am driving at, but it doesn't mean anything in relation to the question, just how the logic works out .

To give a sort of betting example. You visit your friend at a reunion after years of not seeing him. You never really kept up with his Christmas cards, correspondence etc of all the kids names who married who, what jobs they took, colleges, or even their sexes and ages, but you remember he had 10 (all adults now and so you cannot easily tell age, so you cannot guess who is oldest/youngest etc). He introduces you to his 9 grown sons sitting at a table. Then you note "hey, I thought you had 10 kids, I've only met 9" and he says "AHH, Pat is by the punch-bowl, go say hi". Well you see a man and a woman by the punch bowl. Who do you introduce yourself to? One is Pat, the other is a stranger. And you don't want to ask your "supposedly" good friend that you don't know the sex of one of his children...

So you think through the birth announcements you've received over the years (and promptly thrown away without opening them). And you wish you could go through each of those unopened letters to see if one of them was pink or all of them was blue. Now, you think "If I only had those letters now...". You would open them one at a time until you saw either 1 that was pink (AHH beautiful Patricia, problem solved) or ALL of them were blue (ok, Patrick it is, mother must be Irish...).

AH-HAH! well I might not have the letters with me, but while walking over to the punch bowl this little thought experiment has yielded something! Just like you wouldn't bet that it is very likely I could flip a coin 10 times and get all heads... I can also wager that it isn't likely I would open all 10 letters and see blue each time... (GO breakdown of the wave function) Because opening those letters all those years ago vs opening them now is the same... to me

So with some confidence you walk over and say hi to the (beautiful by the way) woman... "You must be Patricia" you have ~91% chance of being right (Is my math right I hesitate with percentages?)

As for the Days of the week... I'm not going to get into that, but again it is assuming that by saying that 1 was born on a Tuesday that you are similarly cutting down the lists of the combination/tree... but you can see how that would make a difference now in a similar way.

*gender assignment is not 50/50, it is like 50/49/1 (intersex) but for purposes of mathemagical fun... lets forget about the hermaphrodites and b vs g disparity :)

posted by DetonatedManiac at 8:59 PM on May 25, 2010 [1 favorite]

Milestogo said:

*yet you also say: 28 of them will be in 2 boy families and 28 of them will be in Boy-girl families.*

but fail to conclude from here that if you are a boy, you will have about a 50% chance of being in a two boy family. (no day-of-week mention). In our problem, we'd like an answer of 1/3 here. So, your chain of reasoning doesn't apply.

You are forgetting all of the girls. If you have 56 boys in two kid families you will also have 56 girls. This makes for 56 total families with 14 being two girls, 14 two boys, and 28 being boy girl. So you have a 25% chance of being in a two boy family and a 50% chance of being in a boy-girl family.

But WAIT: you have a fucking point. If you are a boy you are more likely to be in a boy-boy family because there are more boys in those types of families.

Ok,,, need to think...dammit, I thought I figured this out.

posted by spaceviking at 8:59 PM on May 25, 2010

but fail to conclude from here that if you are a boy, you will have about a 50% chance of being in a two boy family. (no day-of-week mention). In our problem, we'd like an answer of 1/3 here. So, your chain of reasoning doesn't apply.

You are forgetting all of the girls. If you have 56 boys in two kid families you will also have 56 girls. This makes for 56 total families with 14 being two girls, 14 two boys, and 28 being boy girl. So you have a 25% chance of being in a two boy family and a 50% chance of being in a boy-girl family.

But WAIT: you have a fucking point. If you are a boy you are more likely to be in a boy-boy family because there are more boys in those types of families.

Ok,,, need to think...dammit, I thought I figured this out.

posted by spaceviking at 8:59 PM on May 25, 2010

I was clearly wrong above. I had it backwards about which probabilities were twice as likely. Sorry about that.

The 1/3 answer with no additional information makes perfect sense. By revealing that one child is a boy, the possibility of two girls is removed, leaving three possibilities of which only one is another boy. That's pretty intuitive. Of people with two children who have at least one boy, the other child is a boy in 1/3 of cases.

However, by identifying the day the child was born, we move away from talking about the general population of two child families and towards a specific child, and the chances for a specific child to be a boy is 1/2. The more specific (less probable) the additional information is, the closer to a probability of 1/2 we get.

That makes sense to me, though it might still be the entirely wrong way to think about it.

posted by Nothing at 9:12 PM on May 25, 2010

The 1/3 answer with no additional information makes perfect sense. By revealing that one child is a boy, the possibility of two girls is removed, leaving three possibilities of which only one is another boy. That's pretty intuitive. Of people with two children who have at least one boy, the other child is a boy in 1/3 of cases.

However, by identifying the day the child was born, we move away from talking about the general population of two child families and towards a specific child, and the chances for a specific child to be a boy is 1/2. The more specific (less probable) the additional information is, the closer to a probability of 1/2 we get.

That makes sense to me, though it might still be the entirely wrong way to think about it.

posted by Nothing at 9:12 PM on May 25, 2010

Okay. I finally got it to make sense in my mind, thanks mainly to chortly's chart, and I think I might be able to explain it in a way that makes it make sense to some of the people having trouble with it.

Let's start with the basic problem, without the Tuesday. What's slipping some people up is that "one of them is a boy" can describe both BG and GB. This isn't about birth order, though birth order is an easy way to think about it. For people getting tripped up on birth order, try thinking of the order that they sit in in a family photo. The important thing is that*you don't know which child the speaker is talking about*. If you did, *then you'd be determining the gender of only one child*, and it would be 50/50.

When you bring the Tuesday in, you're adding a layer of specificity about the kid that the speaker is talking about. It's still not 50/50, because you still don't know for certain that you know which child is being discussed (in other words, it might be two boys both born on Tuesday).

Now, let's change the statement to, "One of them is a boy born on Christmas." Imagine a chart like chortly's in which each quadrant is 365x365. Now it's much more specificity: because there's such a slim chance that both children are boys born on Christmas, the chance that they're both boys is very, very close to 50/50. Someone correct me if I'm wrong, but I think it's 729/1459, or .499657. It's not 50/50 because there's still that 1/1459 chance that both children are boys born on Christmas.

The limit is 50/50; it'll never quite reach 50/50 unless*you know which child the speaker is talking about*. If the speaker says, "The older child is a boy," then you have a 50/50 chance that they're both boys.

posted by roll truck roll at 9:13 PM on May 25, 2010 [1 favorite]

Let's start with the basic problem, without the Tuesday. What's slipping some people up is that "one of them is a boy" can describe both BG and GB. This isn't about birth order, though birth order is an easy way to think about it. For people getting tripped up on birth order, try thinking of the order that they sit in in a family photo. The important thing is that

When you bring the Tuesday in, you're adding a layer of specificity about the kid that the speaker is talking about. It's still not 50/50, because you still don't know for certain that you know which child is being discussed (in other words, it might be two boys both born on Tuesday).

Now, let's change the statement to, "One of them is a boy born on Christmas." Imagine a chart like chortly's in which each quadrant is 365x365. Now it's much more specificity: because there's such a slim chance that both children are boys born on Christmas, the chance that they're both boys is very, very close to 50/50. Someone correct me if I'm wrong, but I think it's 729/1459, or .499657. It's not 50/50 because there's still that 1/1459 chance that both children are boys born on Christmas.

The limit is 50/50; it'll never quite reach 50/50 unless

posted by roll truck roll at 9:13 PM on May 25, 2010 [1 favorite]

Ok, I think I got it, and I haven't changed my mind again yet...

You have to think in terms of families. If you are a boy there are 42 possible families you can be in, 14 of them are boy-boy and 28 of them are boy-girl.

So perhaps the problem here is perspective. If you think in terms of the family then it is one way, but if you think in terms of the kid its another.

For example: I am a family with two kids and I have one boy -- what are the chances that have two boys? Ans: 1/3

BUT: I am a boy, what are the chances that I will be in a family with two boys? Ans: 1/2

In the second case it is more like saying: I'm a family with two kids and the first kid was a boy -- what is the chance the second will be also a boy?

Ug.

posted by spaceviking at 9:15 PM on May 25, 2010

You have to think in terms of families. If you are a boy there are 42 possible families you can be in, 14 of them are boy-boy and 28 of them are boy-girl.

So perhaps the problem here is perspective. If you think in terms of the family then it is one way, but if you think in terms of the kid its another.

For example: I am a family with two kids and I have one boy -- what are the chances that have two boys? Ans: 1/3

BUT: I am a boy, what are the chances that I will be in a family with two boys? Ans: 1/2

In the second case it is more like saying: I'm a family with two kids and the first kid was a boy -- what is the chance the second will be also a boy?

Ug.

posted by spaceviking at 9:15 PM on May 25, 2010

Now in the case of coins I think this is the way to think about it:

You flip two coins and hide them in a box. You ask your friend: "If one of these is heads and is minted in 1983 what is the chance that the other will also be heads?" Answ: 1/2

Because it is a really low probability that it will be heads and minted in 1983 (you don't know before you start of course) it is equally as likely for it to occur with two heads as with heads-tails even though heads-tails will happen more often.

posted by spaceviking at 9:23 PM on May 25, 2010

You flip two coins and hide them in a box. You ask your friend: "If one of these is heads and is minted in 1983 what is the chance that the other will also be heads?" Answ: 1/2

Because it is a really low probability that it will be heads and minted in 1983 (you don't know before you start of course) it is equally as likely for it to occur with two heads as with heads-tails even though heads-tails will happen more often.

posted by spaceviking at 9:23 PM on May 25, 2010

At the risk of being an ass... I think I buried the lead in my post. SO let me re-post this part of my post above since I think it puts a nice verbal image of what happens (NO MATH):

In the example below we are taking the question from the article:

S2: "one of my children is a boy, what is the probably that they both are boys" (A: 1/3)

and turning it into a more extreme example in order to illustrate the point

S2b: - "9 of my children are boys, what is the probability that all 10 are boys" (A: 1/11) chck this please

You visit your friend at a reunion after years of not seeing him. You never really kept up with his Christmas cards, correspondence etc of all the kids names who married who, what jobs they took, colleges, or even their sexes and ages, but you remember he had 10 kids (all adults now and so you cannot easily tell age, so you cannot guess who is oldest/youngest etc). He introduces you to his 9 grown sons sitting at a table. Then you note:

"hey, I thought you had 10 kids, I've only met 9"

and he says "Pat is by the punch-bowl, go say hi".

Well you see a man and a woman by the punch bowl. Who do you introduce yourself to? One is Pat, the other is a stranger. And you don't want to ask your "supposedly" good friend that you don't know the sex of one of his children...

So you think through the birth announcements you've received over the years (and promptly thrown away without opening them). And you wish you could go through each of those unopened letters to see if one of them was pink or all of them was blue. Now, you think "If I only had those letters now...". You would open them one at a time until you saw either 1 that was pink (AHH beautiful Patricia, problem solved) or ALL of them were blue (ok, Patrick it is, mother must be Irish...). Notice, you would only have to find 1 pink notice before stopping... if the first one was pink you wouldn't even have to look at the other 9... you KNOW they are all guys. But if Pat is Patrick then you will only KNOW that he is so after opening ALL of those unopened letters.

AH-HAH! well I might not have the letters with me, but while walking over to the punch bowl this little thought experiment has yielded something! Just like you wouldn't bet that it is very likely I could flip a coin 10 times and get all heads... I can also wager that it isn't likely I would open all 10 letters and see blue each time... (wave to the kitty) Because opening those letters all those years ago vs opening them now is the same... to me now. Each is it's own %50 chance.

So with some confidence you walk over and say hi to the (beautiful by the way) woman... "You must be Patricia, your father has written so much about you" you have ~91% chance of being right (Is my math right I hesitate with percentages?)

As for the Days of the week... I'm not going to get into that, but again it is assuming that by saying that 1 was born on a Tuesday that you are similarly cutting down the lists of the combination/tree... but you can see how that would make a difference now in a similar way.

posted by DetonatedManiac at 9:26 PM on May 25, 2010

In the example below we are taking the question from the article:

S2: "one of my children is a boy, what is the probably that they both are boys" (A: 1/3)

and turning it into a more extreme example in order to illustrate the point

S2b: - "9 of my children are boys, what is the probability that all 10 are boys" (A: 1/11) chck this please

You visit your friend at a reunion after years of not seeing him. You never really kept up with his Christmas cards, correspondence etc of all the kids names who married who, what jobs they took, colleges, or even their sexes and ages, but you remember he had 10 kids (all adults now and so you cannot easily tell age, so you cannot guess who is oldest/youngest etc). He introduces you to his 9 grown sons sitting at a table. Then you note:

"hey, I thought you had 10 kids, I've only met 9"

and he says "Pat is by the punch-bowl, go say hi".

Well you see a man and a woman by the punch bowl. Who do you introduce yourself to? One is Pat, the other is a stranger. And you don't want to ask your "supposedly" good friend that you don't know the sex of one of his children...

So you think through the birth announcements you've received over the years (and promptly thrown away without opening them). And you wish you could go through each of those unopened letters to see if one of them was pink or all of them was blue. Now, you think "If I only had those letters now...". You would open them one at a time until you saw either 1 that was pink (AHH beautiful Patricia, problem solved) or ALL of them were blue (ok, Patrick it is, mother must be Irish...). Notice, you would only have to find 1 pink notice before stopping... if the first one was pink you wouldn't even have to look at the other 9... you KNOW they are all guys. But if Pat is Patrick then you will only KNOW that he is so after opening ALL of those unopened letters.

AH-HAH! well I might not have the letters with me, but while walking over to the punch bowl this little thought experiment has yielded something! Just like you wouldn't bet that it is very likely I could flip a coin 10 times and get all heads... I can also wager that it isn't likely I would open all 10 letters and see blue each time... (wave to the kitty) Because opening those letters all those years ago vs opening them now is the same... to me now. Each is it's own %50 chance.

So with some confidence you walk over and say hi to the (beautiful by the way) woman... "You must be Patricia, your father has written so much about you" you have ~91% chance of being right (Is my math right I hesitate with percentages?)

As for the Days of the week... I'm not going to get into that, but again it is assuming that by saying that 1 was born on a Tuesday that you are similarly cutting down the lists of the combination/tree... but you can see how that would make a difference now in a similar way.

posted by DetonatedManiac at 9:26 PM on May 25, 2010

How does the probability change if the boy has red hair? And recieved three sets of shoes as a birth present?

posted by vertriebskonzept at 9:32 PM on May 25, 2010

posted by vertriebskonzept at 9:32 PM on May 25, 2010

The born on tuesday thing seems to me to be a red herring. His statement did not exclude the other child being born on tuesday, he just mentioned that one was. If he has said one child was born on tuesday, the other not, or something to that effect you can whittled down the probability tree otherwise it's a deadend and besides the point. Gubo has made it very clear in his coin example, it would be folly to suggest date of one coin affects the head or tailness of another.

posted by ExitPursuedByBear at 9:35 PM on May 25, 2010

posted by ExitPursuedByBear at 9:35 PM on May 25, 2010

I'm going to go with empath's version of gubo's question, because it's a little easier to work with.

empath: "*I have flipped two coins and hid them under a box.*

I look under the box and tell you that one of the coins is heads and was minted in a year ending in 3.

What are the odds that both coins are heads?"

I'm going to start from the premise that if we ignore the year part, the answer is 1/3. That's already been hashed out so many times above, and it doesn't seem worth doing again.

Assuming that a coin has a 1/10 chance of being made in a year ending in 3, we imagine another chart like this one, but in which each quadrant is 10x10. The chances of both coins being heads is 19/39. The reason why it's not a straight 50/50 is because*there's still a 1/39 chance that both coins are showing heads and made in a year ending in 3*.

Let's change the question. Let's say you look at one of the coins under the box and tell me, "It's heads and**oh my gosh it's a 1920 Franklin McKrugerrand there's only one of these in the world!**" Now, I can safely say that the chances of both coins being heads is 50/50.

To put it in completely unmathy terms, the more information I have about the coin, the more I know which one you're talking about.

posted by roll truck roll at 9:43 PM on May 25, 2010 [3 favorites]

empath: "

I look under the box and tell you that one of the coins is heads and was minted in a year ending in 3.

What are the odds that both coins are heads?

I'm going to start from the premise that if we ignore the year part, the answer is 1/3. That's already been hashed out so many times above, and it doesn't seem worth doing again.

Assuming that a coin has a 1/10 chance of being made in a year ending in 3, we imagine another chart like this one, but in which each quadrant is 10x10. The chances of both coins being heads is 19/39. The reason why it's not a straight 50/50 is because

Let's change the question. Let's say you look at one of the coins under the box and tell me, "It's heads and

To put it in completely unmathy terms, the more information I have about the coin, the more I know which one you're talking about.

posted by roll truck roll at 9:43 PM on May 25, 2010 [3 favorites]

Okay, I've been reading the comments above, and as my way of thinking things through, I have a few scenarios here, which may or may not have faulty reasoning:

1. Someone is looking in a room and tells you there are two children in it. They tell you that one of the children is a girl, and then ask you to guess if the other child is a boy or a girl. Suppose this exercise is repeated many times, with new pairs of children each time but always with one of them being a girl. You always answer "girl." How often will you be right? The answer is 1/3, I think.

2. Now a slightly different scenario. Someone is looking in a room and tells you there are two children in it. They tell you that one of the children is a girl, and they give some other fact about that girl, height or weight or day of birth or whathaveyou, and then ask you to guess if the other child is a boy or girl. Suppose this exercise is repeated many times. New pairs of children are present each time but always with one of them being a girl. Furthermore, the fact given about the selected girl changes each time: sometimes the day of birth is given, sometimes the age, whathaveyou. You always answer that the other child is a girl. How often will you be right? It's still 1/3, right?

3. Now here comes the scenario I think the fpp question is aiming at. Someone is looking in a room and tells you there are two children in it. They tell you that one of the children is a girl, and that the day of birth of that girl is Tuesday, and then ask you to guess if the other child is a boy or girl. Suppose this exercise is repeated many times with different pairs of children, always with one of the children being a girl born on Tuesday. You always answer that the other child is a girl. How often will you be right? Here's where I can visualize the answer being 13/27. Am I getting there?

posted by gubo at 9:48 PM on May 25, 2010 [1 favorite]

1. Someone is looking in a room and tells you there are two children in it. They tell you that one of the children is a girl, and then ask you to guess if the other child is a boy or a girl. Suppose this exercise is repeated many times, with new pairs of children each time but always with one of them being a girl. You always answer "girl." How often will you be right? The answer is 1/3, I think.

2. Now a slightly different scenario. Someone is looking in a room and tells you there are two children in it. They tell you that one of the children is a girl, and they give some other fact about that girl, height or weight or day of birth or whathaveyou, and then ask you to guess if the other child is a boy or girl. Suppose this exercise is repeated many times. New pairs of children are present each time but always with one of them being a girl. Furthermore, the fact given about the selected girl changes each time: sometimes the day of birth is given, sometimes the age, whathaveyou. You always answer that the other child is a girl. How often will you be right? It's still 1/3, right?

3. Now here comes the scenario I think the fpp question is aiming at. Someone is looking in a room and tells you there are two children in it. They tell you that one of the children is a girl, and that the day of birth of that girl is Tuesday, and then ask you to guess if the other child is a boy or girl. Suppose this exercise is repeated many times with different pairs of children, always with one of the children being a girl born on Tuesday. You always answer that the other child is a girl. How often will you be right? Here's where I can visualize the answer being 13/27. Am I getting there?

posted by gubo at 9:48 PM on May 25, 2010 [1 favorite]

Okay, I generated a population of 1100 pairs of kids twice with random days of weeks, and after filtering out all the ones with no boys born on tuesday, I got 36% with 2 boys on the first run through and 32.5% on the second run through.

I am 100% positive the answer is 1/3, and this guy is completely full of shit.

If anyone else wants to program it and run it, I'd be happy to see the numbers. I just did it with an excel spreadsheet.

posted by empath at 9:55 PM on May 25, 2010 [1 favorite]

I am 100% positive the answer is 1/3, and this guy is completely full of shit.

If anyone else wants to program it and run it, I'd be happy to see the numbers. I just did it with an excel spreadsheet.

posted by empath at 9:55 PM on May 25, 2010 [1 favorite]

But you don't have any infornation about the coin unless the information tells you which coin is which.

For example, if you have 1 quarter and one dime, and the person tells you the dime is heads, then yes, it's 50% that the quarter is heads.

Someone else run a simulation, please, that's really the only way to resolve this.

posted by empath at 9:57 PM on May 25, 2010

empath, that's exactly the point of my made up "McKrugerrand." In that case, there's only one in the world, so yeah, 1/2.

empath: "*Someone else run a simulation, please, that's really the only way to resolve this.*"

Even I know the difference between simulations and proofs.

posted by roll truck roll at 9:59 PM on May 25, 2010

empath: "

Even I know the difference between simulations and proofs.

posted by roll truck roll at 9:59 PM on May 25, 2010

If your theory doesn't match results, your theory is wrong.

posted by empath at 10:06 PM on May 25, 2010

posted by empath at 10:06 PM on May 25, 2010

Well, here are the results. That's every possible combination of two children born on two days of the week. What's the difference between your model and this one?

posted by roll truck roll at 10:13 PM on May 25, 2010

posted by roll truck roll at 10:13 PM on May 25, 2010

When I'm confused, I just ask the basic Unix commands:

There are 196 possible configurations of two genders and seven weekdays for two children, which we assume are equally likely in the general population.

Of those, only 27 have at least one boy born on a Tuesday; we know we're talking to a parent in that group.

And of*these*, 13 have two boys.

posted by nicwolff at 10:26 PM on May 25, 2010 [11 favorites]

`> echo {b,g}{M,T,W,t,F,S,s}{b,g}{M,T,W,t,F,S,s} | fold -w 5 | wc -l`

**196**

There are 196 possible configurations of two genders and seven weekdays for two children, which we assume are equally likely in the general population.

`> echo {b,g}{M,T,W,t,F,S,s}{b,g}{M,T,W,t,F,S,s} | fold -w 5 | grep bT | wc -l`

**27**

Of those, only 27 have at least one boy born on a Tuesday; we know we're talking to a parent in that group.

`> echo {b,g}{M,T,W,t,F,S,s}{b,g}{M,T,W,t,F,S,s} | fold -w 5 | grep bT | grep b.b. | wc -l`

**13**

And of

posted by nicwolff at 10:26 PM on May 25, 2010 [11 favorites]

The problem when the person is giving the information as opposed to being asked the question is that when you have a mixed pair, the person could tell you about the boy or the girl, which reduces the probability where you have where there is a boy girl pair and you are being told about the boy.

There is a big difference whether you are observing whether there is at least one boy, and observing someone telling you there is at least one boy with no bidding to do so.

The first case has the four options outlined by Rhomboid earlier, the latter has six:

1. eldest boy youngest girl, told about a boy

2. eldest boy youngest girl, told about a girl

3. eldest girl, youngest boy, told about a girl

4. eldest girl, youngest boy, told about a boy

5. eldest girl youngest girl, told about a girl

6. eldest boy youngest boy, told about a boy

But they are not all of equal probability... numbers 5 and 6 each have a probability of 1/4 since there is a 1/4 chance of having two girls overall and a 25% chance of having two boys overall. Numbers 1 & 2 must add up to a total of 1/4, same with 3 & 4 as having a boy first and then a girl or a girl first and then a boy each have a 1/4 probability as well.

So if we assume that there is an equal chance of having a boy or girl, and also an equal chance of the parent talking about a boy as a girl when there is a choice the options and probabilities are:

1. eldest boy youngest girl, told about a boy (1/8)

4. eldest girl youngest boy, told about a boy (1/8)

6. eldest boy youngest boy, told about a boy (1/4)

So the boy/boy pair has half the share of the remaining probabilities. That means a 50% chance that the second child is also a boy.

If the parent had a psychology that strongly favored talking about a boy, the probability would indeed be 1/3 that the second child was a boy. However, if the parent was inclined to talk about a girl over a boy, the probability of the second child being a boy would be 100%--if that parent had any girls at all, they would have talked about one.

posted by Zalzidrax at 10:28 PM on May 25, 2010

There is a big difference whether you are observing whether there is at least one boy, and observing someone telling you there is at least one boy with no bidding to do so.

The first case has the four options outlined by Rhomboid earlier, the latter has six:

1. eldest boy youngest girl, told about a boy

2. eldest boy youngest girl, told about a girl

3. eldest girl, youngest boy, told about a girl

4. eldest girl, youngest boy, told about a boy

5. eldest girl youngest girl, told about a girl

6. eldest boy youngest boy, told about a boy

But they are not all of equal probability... numbers 5 and 6 each have a probability of 1/4 since there is a 1/4 chance of having two girls overall and a 25% chance of having two boys overall. Numbers 1 & 2 must add up to a total of 1/4, same with 3 & 4 as having a boy first and then a girl or a girl first and then a boy each have a 1/4 probability as well.

So if we assume that there is an equal chance of having a boy or girl, and also an equal chance of the parent talking about a boy as a girl when there is a choice the options and probabilities are:

1. eldest boy youngest girl, told about a boy (1/8)

4. eldest girl youngest boy, told about a boy (1/8)

6. eldest boy youngest boy, told about a boy (1/4)

So the boy/boy pair has half the share of the remaining probabilities. That means a 50% chance that the second child is also a boy.

If the parent had a psychology that strongly favored talking about a boy, the probability would indeed be 1/3 that the second child was a boy. However, if the parent was inclined to talk about a girl over a boy, the probability of the second child being a boy would be 100%--if that parent had any girls at all, they would have talked about one.

posted by Zalzidrax at 10:28 PM on May 25, 2010

A crazy uncle of mine used to pose the following brainteaser to me when I was 7 or 8:

"My Aunt Sue, she died last night, she died last night. Did she die?"

My answer was almost always wrong, no matter which one I picked, and he took great pleasure in never telling me*why*. While he was also the one who taught me the classic one-sentence mysteries such as George and Martha lying dead on the floor with broken glass and water around them, or the man who never would have died if he'd seen the sawdust, he never let me get anywhere near the secret behind his Aunt Sue one. So I came to the conclusion that he was merely messing with my head just to see what answers I could come up with and the explanations I could provide.

I feel this brainteaser is presented as kind of the same thing, only there's no Aunt Sue.

posted by Spatch at 10:46 PM on May 25, 2010

"My Aunt Sue, she died last night, she died last night. Did she die?"

My answer was almost always wrong, no matter which one I picked, and he took great pleasure in never telling me

I feel this brainteaser is presented as kind of the same thing, only there's no Aunt Sue.

posted by Spatch at 10:46 PM on May 25, 2010

Sorry but I have to stick my oar in as there are far too many people accepting that the answer in the simple case is 1/3.

In 100 cases of 2 child families there will be

BB 25 BG 25 GB 25 GG 25

But the number of boys in each family would be

BB 50

BG 25

GB 25

GG 0

So there are 50 boys with brothers and 50 with sisters hence it's 50-50.

posted by jamespake at 11:04 PM on May 25, 2010

In 100 cases of 2 child families there will be

BB 25 BG 25 GB 25 GG 25

But the number of boys in each family would be

BB 50

BG 25

GB 25

GG 0

So there are 50 boys with brothers and 50 with sisters hence it's 50-50.

posted by jamespake at 11:04 PM on May 25, 2010

Delightfully entertaining indeed.

posted by captainsohler at 11:08 PM on May 25, 2010

posted by captainsohler at 11:08 PM on May 25, 2010

But the boys are not evenly distributed over the four possible family configurations, and the question is about family configurations (i.e. families as a whole) not about individual boys. The answer really is 1/3.

posted by Rhomboid at 11:22 PM on May 25, 2010

I am 100% positive the answer is 1/3, and this guy is completely full of shit.

If anyone else wants to program it and run it, I'd be happy to see the numbers. I just did it with an excel spreadsheet.

(OK, I can't resist a programming task.)

1100 is way too small of a sample size for this. From an initial sample size of 10000 pairs of kids, I'm getting values hovering around 48%. (13/27 = 0.481481481) I did this in PHP, source is here. (I threw it together very quickly, forgive the bad style. Hopefully the variables and logic are clear.)

On preview: jamespake, how many of those families satisfy the condition "one or both children are boys?" Of that subset, how many are both boys?

posted by kmz at 11:22 PM on May 25, 2010 [1 favorite]

Empath, never use Excel for anything when Perl is available:

So, how often, out of those among one hundred million families that have a boy born on Tuesday, does he have a brother? 13/27 = .481481481… and we're close enough to that. (In one minute and two seconds, using 1 core of a new 2.66 GHz MacBook Pro, if anyone cares.)

posted by nicwolff at 11:32 PM on May 25, 2010 [1 favorite]

`> cat | perl -l`

for (1..100000000) {

$boy1 = rand() < 1/2; $tues1 = rand() < 1/7;

$boy2 = rand() < 1/2; $tues2 = rand() < 1/7;

if ( $boy1 && $tues1 or $boy2 && $tues2 ) {

$boy_tues++;

if ( $boy1 && $boy2 ) { $boy_boy++ }

}

}

print $boy_boy / $boy_tues;

**0.481796652224822**

So, how often, out of those among one hundred million families that have a boy born on Tuesday, does he have a brother? 13/27 = .481481481… and we're close enough to that. (In one minute and two seconds, using 1 core of a new 2.66 GHz MacBook Pro, if anyone cares.)

posted by nicwolff at 11:32 PM on May 25, 2010 [1 favorite]

jamespake, you're sampling the boys; we're sampling the families. *Of course* ½ of boys with one sibling have a brother. But ⅓ of *families* with two kids and at least one boy have two boys.

For that matter, ½ of boys born on Tuesday with one sibling have a brother. But in*families* with two kids, one a boy born on Tuesday, 13/27 have two boys.

posted by nicwolff at 11:46 PM on May 25, 2010 [1 favorite]

For that matter, ½ of boys born on Tuesday with one sibling have a brother. But in

posted by nicwolff at 11:46 PM on May 25, 2010 [1 favorite]

OK, that's why I shouldn't code at 1AM. My code is crap compared to yours, nicwolff.

posted by kmz at 11:49 PM on May 25, 2010

posted by kmz at 11:49 PM on May 25, 2010

As long as you have 47 GB of free RAM to hold 7 x 10^{8} PHP hash entries at 68 bytes each your approach would work too, kmz ;)

posted by nicwolff at 12:32 AM on May 26, 2010

posted by nicwolff at 12:32 AM on May 26, 2010

The question is ambiguous as to how the boy has been sampled. If we have a list of all the two child families that contain at least one boy then we can either pick a family at random in which case the odds of getting 2 boys are 1 in 3 or we can pick a boy at random in which case the odds are 1 in 2. This is because although there are half as many 2 boys configurations, they contain exactly twice as many boys. I'd say the 50-50 case is more natural - for example if you meet a boy at random who tells you that he has one sibling then you would be wrong to assume that there is a 2/3 chance of it being a sister.

posted by jamespake at 12:42 AM on May 26, 2010

posted by jamespake at 12:42 AM on May 26, 2010

Yeah, I just rewrote it in python doing 10000 at a time and i'm steadily getting 33% without knowing about tuesday and 48% with.

posted by empath at 1:15 AM on May 26, 2010

posted by empath at 1:15 AM on May 26, 2010

Yeah, these are the numbers I'm getting, I did it with arrays, too, because I'm a terrible programmer:

families with at least one boy:7467 families with at least one boy born on tuesday:1361 families with two boys:2481 families with two boys, at least one of them born on tuesday:640 percentage of 2 boy families if at least one boy:33.226195259140219 percentage of 2 boy families if at least one boy born on tuesday:47.02424687729610833.2261952591

posted by empath at 1:24 AM on May 26, 2010

families with at least one boy:7467 families with at least one boy born on tuesday:1361 families with two boys:2481 families with two boys, at least one of them born on tuesday:640 percentage of 2 boy families if at least one boy:33.226195259140219 percentage of 2 boy families if at least one boy born on tuesday:47.02424687729610833.2261952591

posted by empath at 1:24 AM on May 26, 2010

I'm still hung up on the syntax. Sorry for returning so late to the thread.

I'm against trying to recontextualize the question. I can see how the riddle works, but it is phrased to ambiguously for me to parse it as such.

Take each sentence in order to create the formula.

"I have 2 children." This defines the given set as a 2 individuals, each with a possibility of being either a boy or a girl.

"One is a boy born on a Tuesday." Ok, so that eliminates the variable of one of the children being a boy or a girl.

"What is the probability I have two boys?" Ok, here's where most of us lose it, since we're not mathematicians. Well, since we know you have one boy, then the only variable we are left to concern ourselves with is the gender of the other child. Since the gender of one of the children is known, we begin calculating it as a simple equation. The unknown child's gender is defined already as a 50/50 probability, since it can only be a boy or a girl. Thus, the reasoning is that since you already know that one is a boy, it is removed from the set, and the only variable that needs to be defined is the gender of one child, not two.

Or we can be really silly and say that the probability is really strange because of the variable of hermaphrodites, though the rarity of this mutation in the population reduces the variable to within .000000000000e1 of 50/50. Which is still a pretty neat demonstration of the principal that the original solution provided by the article tries to make. And it still completely ignores the Tuesday part of the riddle.

I will say, I understand what most people are trying to say about the solution involving creating a set of 2, with 4 possible combination (BB, BG, GB, GG, etc) and then the added thing with the Tuesday variable.

This also explains why trying to "get" this type of math riddle is hard for many people, since the interpretation is ambiguous. If someone can try and explain why you can't eliminate the child whose gender has been revealed as a variable, I'd be happy to hear it.

posted by daq at 1:30 AM on May 26, 2010

I'm against trying to recontextualize the question. I can see how the riddle works, but it is phrased to ambiguously for me to parse it as such.

Take each sentence in order to create the formula.

"I have 2 children." This defines the given set as a 2 individuals, each with a possibility of being either a boy or a girl.

"One is a boy born on a Tuesday." Ok, so that eliminates the variable of one of the children being a boy or a girl.

"What is the probability I have two boys?" Ok, here's where most of us lose it, since we're not mathematicians. Well, since we know you have one boy, then the only variable we are left to concern ourselves with is the gender of the other child. Since the gender of one of the children is known, we begin calculating it as a simple equation. The unknown child's gender is defined already as a 50/50 probability, since it can only be a boy or a girl. Thus, the reasoning is that since you already know that one is a boy, it is removed from the set, and the only variable that needs to be defined is the gender of one child, not two.

Or we can be really silly and say that the probability is really strange because of the variable of hermaphrodites, though the rarity of this mutation in the population reduces the variable to within .000000000000e1 of 50/50. Which is still a pretty neat demonstration of the principal that the original solution provided by the article tries to make. And it still completely ignores the Tuesday part of the riddle.

I will say, I understand what most people are trying to say about the solution involving creating a set of 2, with 4 possible combination (BB, BG, GB, GG, etc) and then the added thing with the Tuesday variable.

This also explains why trying to "get" this type of math riddle is hard for many people, since the interpretation is ambiguous. If someone can try and explain why you can't eliminate the child whose gender has been revealed as a variable, I'd be happy to hear it.

posted by daq at 1:30 AM on May 26, 2010

I'm supposed to be working right now, so sorry if this has been said before but... I disagree with the answer 1/3, because I disagree with the question.

The question states*"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"* so, the one child is 100%. We are talking about the probability of the second child being a boy.

In which case, the chance is >50% (although not by much), because the chance of XX/XY on a lone or non-identical child is 50:50, to which we need to add the chance of an identical twin.

The 1/3 makes no sense at all, given the statement that there is already one child of known gender.

It's like those times someone flips a coin ten times and gets heads every time - the chance of the next flip being heads is 50%...

posted by sodium lights the horizon at 1:49 AM on May 26, 2010

The question states

In which case, the chance is >50% (although not by much), because the chance of XX/XY on a lone or non-identical child is 50:50, to which we need to add the chance of an identical twin.

The 1/3 makes no sense at all, given the statement that there is already one child of known gender.

It's like those times someone flips a coin ten times and gets heads every time - the chance of the next flip being heads is 50%...

posted by sodium lights the horizon at 1:49 AM on May 26, 2010

Prepare to have your mind blown:

Consider a game where I flip (in secret) two coins. Your objective is to guess how many heads I have flipped.

If you know that I have at least one head, then you've narrowed down my possibilities to HH, HT, and TH, so if you guess 1H1T you'll be right 2/3 of the time.

Likewise, if you know that I have at least one tail, by the same argument you'll be right 2/3 of the time if you guess 1H1T.

Now, I flip the coins, pick one (at random) and announce that coin. How frequently can you correctly guess the composition of my flip?

According to the above, if I announce Heads (so you know I have at least one head), you should pick 1H1T, and succeed 2/3 of the time.

Similarly, if I announce Tails, you know I have at least one Tail, and so you should (again!) pick 1H1T, and succeed 2/3 of the time.

This leads to the conclusion that*regardless of what I announce*, you should guess 1H1T, and that you'll be right 2/3 of the time.

But this is clearly a contradiction of the assumption that I am flipping independent fair coins, where I should only get 1H1T 1/2 of the time, not 2/3.

posted by Pyry at 1:50 AM on May 26, 2010 [2 favorites]

Consider a game where I flip (in secret) two coins. Your objective is to guess how many heads I have flipped.

If you know that I have at least one head, then you've narrowed down my possibilities to HH, HT, and TH, so if you guess 1H1T you'll be right 2/3 of the time.

Likewise, if you know that I have at least one tail, by the same argument you'll be right 2/3 of the time if you guess 1H1T.

Now, I flip the coins, pick one (at random) and announce that coin. How frequently can you correctly guess the composition of my flip?

According to the above, if I announce Heads (so you know I have at least one head), you should pick 1H1T, and succeed 2/3 of the time.

Similarly, if I announce Tails, you know I have at least one Tail, and so you should (again!) pick 1H1T, and succeed 2/3 of the time.

This leads to the conclusion that

But this is clearly a contradiction of the assumption that I am flipping independent fair coins, where I should only get 1H1T 1/2 of the time, not 2/3.

posted by Pyry at 1:50 AM on May 26, 2010 [2 favorites]

daq, because they haven't mentioned a specific child. You know one is a boy. Either of the two (the younger or the older or both) could be a boy, so you have 3 possibilites left: They are both boys, the younger is a girl, or the older a girl.

Let's say you don't know that there is at least one boy.

You then find out the older child is a girl.

The younger child still has a 50/50 chance of being boy.

Okay, start again now:

You are first told that at least one child is a boy.

Then lets say you find out that the older child is a girl.

What are the odds then that the younger child is a boy?

Now do you see why there is a difference?

posted by empath at 1:52 AM on May 26, 2010

Let's say you don't know that there is at least one boy.

You then find out the older child is a girl.

The younger child still has a 50/50 chance of being boy.

Okay, start again now:

You are first told that at least one child is a boy.

Then lets say you find out that the older child is a girl.

What are the odds then that the younger child is a boy?

Now do you see why there is a difference?

posted by empath at 1:52 AM on May 26, 2010

Man, I used to be so much better at this sort of stuff when I taught the LSAT.

Anyway, it seems to me that this is all about how you organize information.

posted by Saxon Kane at 2:01 AM on May 26, 2010

Anyway, it seems to me that this is all about how you organize information.

posted by Saxon Kane at 2:01 AM on May 26, 2010

Pyry wins the thread.

posted by The Tensor at 2:16 AM on May 26, 2010

posted by The Tensor at 2:16 AM on May 26, 2010

If your options are 1H1T, 2H or 2T, then this is obviously correct. 1H1T is twice as likely as the other two options.

posted by empath at 2:23 AM on May 26, 2010

Here's my attempt at explaining. I'll start by answering the question about the coin that comes down "Heads, minted in 1993" or what have you, and progress from there. That *does* change the odds, as you will see ...

Two coins are flipped. The first one comes up heads. What are the chances that the other is heads? There are two cases: H/H, H/T. 1 in 2. In only one of the two cases is the second coin heads.

You’ve specified the first one, and the second one is just random chance.

The same two coins are flipped. Now all you know is that of them is heads – you don’t know which. What are the chances that the other is heads? There are three possible cases: H/H, H/T, T/H. Only in one case of those cases are they both heads. 1 in 3.

You haven’t specified which is which, so the odds changed. This is not a verbal trick. If you flipped the coins a million times, and separated out all the instances where the first coin was heads, you would find that both coins were heads half the time. If you flipped the coins a million times, and separated out all the instances where at least one coin, any coin, was heads, you would find both coins were heads a third of the time.

I'll explain why this is this case in a bit.

But first. OK. Now. Let’s see what happens when add a bit of interesting information.

The two coins we’re flipping were both minted in the 1990’s. One of them comes up heads/1993 coin. What are the chances that the other is heads?

Stay with me, this isn't as complex as it looks:

- There are 10 cases where H1993 is the first coin and the second is tails:

(You don’t need to wade through it, but it’s: H1993 T1990, H1993 T1991, H1993 T1992, H1993 T1993, H1993 T1994, H1993 T1995, H1993 T1996, H1993 T1997, H1993 T1998, H1993 T1999)

- There are 10 cases where H1993 is the second coin and the first one is tails:

(T1990 H1993, T1991 H1993, T1992 H1993, T1993 H1993, T1994 H1993, T1995 H1993, T1996 H1993, T1997 H1993, T1998 H1993, T1999 H1993)

- There are 10 cases where H1993 is the first coins and the second one is heads:

(H1993 H1990, H1993 H1991, H1993 H1992, H1993 H1993, H1993 H1994, H1993 H1995, H1993 H1996, H1993 H1997, H1993 H1998, H1993 H1999)

- There are only 9 more cases where H1993 is the second coin and the first one is heads (since we just already counted the case where both coins were H1993):

(H1990 H1993, H1991 H1993, H1992 H1993, H1994 H1993, H1995 H1993, H1996 H1993, H1997 H1993, H1998 H1993, H1999 H1993)

The last 19 answers are both H/H situations. There are 39 total cases. So the answer is 19 in 39 -- MUCH closer to 1 in 2 than case where you only know that one is heads, but don't know the date!

Again, this is NOT a verbal trick – IF YOU FLIP RANDOM COINS MINTED IN THE 1990’S A MILLION TIMES AND SEPARATE OUT ALL CASES WHERE ONE IS HEADS/1993, THEN 19 OUT OF 39 TIMES, BOTH WILL BE HEADS.

Why is this?

You’ve come very close to, but not quite reached, a “the first coin is heads, what’s the second coin?” scenario. One coin is close to being specified. It’s “the 1993” coin. That’s close to being “the first” coin. If you knew only one of the coins was minted in 1993, it might as well BE “the first” coin. Then you’d be saying “the 1993 coins comes up heads, what’s the other?” 1 in 2 chance.

The only reason it isn’t *quite* this is because THAT OTHER COIN MIGHT BE A 1993 COIN, TOO. Remember when we shaved that last set of numbers from 10 to 9? That’s what happened there. A tiny bit of uncertainty kept one coin from being “the 1993” coin. Either coin *could* be 1993, after all, so you can’t be absolutely sure that one is “the 1993” coin.

Now, in most of the cases, it is in fact “the 1993” coin. But there are a few cases where it isn’t, so instead of being 1 in 2, it’s 19 in 39 – close but not quite.

Now, let’s take the case that was actually asked about. You flip two coins, not necessarily minted in the 1990’s. One – you don’t know which – turns out to be a 1993 heads. What are the chances the other is heads?

Really, really close to 1 in 2. If there are, say, a hundred years where coins were minted, then the odds of both coins being 1993 are really small. You’re really close to one being “the 1993” coin.

OK.

This same logic applies to families. The “Tuesday” boy isn’t as specific as the “firstborn” boy, but changes the odds. The “December 28” boy would change the odds more. The “December 28, 1987” boy would change the odds even more.

The explanation I promised. Here’s a way of thinking about it which might not hurt your head.

Take all families with two children. How many have two boys? About 1 in 4.

Now just take all families with two children and at least one boy. How many have two boys? About 1 in 3.

Why? While keeping the end you’re looking for the same (2 boys) you’ve reduced your set. You’ve shaved off all the families with two girls. Within the set that remains, the number of families with two boys grows proportionally.

Now. Take all families with two children and at least one boy born on December 28, 1987. How many have two boys?

You’re still looking for two boys, but now you’re looking at ONLY those families with one boy born on December 28, 1987. You’ve gotten rid of most of your families. Of the ones that remain, the odds that the one who wasn’t born on December 28, 1987 is pretty close to one in two.

But not quite. Here's why:

Take all families with two children and at least one boy born on a Tuesday. How many have two boys?

You’ve still shaved away a lot of families. You’re only looking at about 1/14 of your original number of families. You’ve changed the odds a lot.

But how have you changed the odds? Well, for most of those families, the “Tuesday” child is only going to be one of the children. In that case, if the “Tuesday” child is male, the odds of the second one being male is 1 in 2.

But ... in some of those families, both children were born on Tuesday. There is no “Tuesday” child, so the distinction has no meaning. For those families, the chances of both children being male are still 1 in 3!

But that’s not a lot of families, so when you tally all the families together, it still gets pretty close to 1 in 2.

13 in 27, as it happens.

posted by kyrademon at 2:25 AM on May 26, 2010 [3 favorites]

Two coins are flipped. The first one comes up heads. What are the chances that the other is heads? There are two cases: H/H, H/T. 1 in 2. In only one of the two cases is the second coin heads.

You’ve specified the first one, and the second one is just random chance.

The same two coins are flipped. Now all you know is that of them is heads – you don’t know which. What are the chances that the other is heads? There are three possible cases: H/H, H/T, T/H. Only in one case of those cases are they both heads. 1 in 3.

You haven’t specified which is which, so the odds changed. This is not a verbal trick. If you flipped the coins a million times, and separated out all the instances where the first coin was heads, you would find that both coins were heads half the time. If you flipped the coins a million times, and separated out all the instances where at least one coin, any coin, was heads, you would find both coins were heads a third of the time.

I'll explain why this is this case in a bit.

But first. OK. Now. Let’s see what happens when add a bit of interesting information.

The two coins we’re flipping were both minted in the 1990’s. One of them comes up heads/1993 coin. What are the chances that the other is heads?

Stay with me, this isn't as complex as it looks:

- There are 10 cases where H1993 is the first coin and the second is tails:

(You don’t need to wade through it, but it’s: H1993 T1990, H1993 T1991, H1993 T1992, H1993 T1993, H1993 T1994, H1993 T1995, H1993 T1996, H1993 T1997, H1993 T1998, H1993 T1999)

- There are 10 cases where H1993 is the second coin and the first one is tails:

(T1990 H1993, T1991 H1993, T1992 H1993, T1993 H1993, T1994 H1993, T1995 H1993, T1996 H1993, T1997 H1993, T1998 H1993, T1999 H1993)

- There are 10 cases where H1993 is the first coins and the second one is heads:

(H1993 H1990, H1993 H1991, H1993 H1992, H1993 H1993, H1993 H1994, H1993 H1995, H1993 H1996, H1993 H1997, H1993 H1998, H1993 H1999)

- There are only 9 more cases where H1993 is the second coin and the first one is heads (since we just already counted the case where both coins were H1993):

(H1990 H1993, H1991 H1993, H1992 H1993, H1994 H1993, H1995 H1993, H1996 H1993, H1997 H1993, H1998 H1993, H1999 H1993)

The last 19 answers are both H/H situations. There are 39 total cases. So the answer is 19 in 39 -- MUCH closer to 1 in 2 than case where you only know that one is heads, but don't know the date!

Again, this is NOT a verbal trick – IF YOU FLIP RANDOM COINS MINTED IN THE 1990’S A MILLION TIMES AND SEPARATE OUT ALL CASES WHERE ONE IS HEADS/1993, THEN 19 OUT OF 39 TIMES, BOTH WILL BE HEADS.

Why is this?

You’ve come very close to, but not quite reached, a “the first coin is heads, what’s the second coin?” scenario. One coin is close to being specified. It’s “the 1993” coin. That’s close to being “the first” coin. If you knew only one of the coins was minted in 1993, it might as well BE “the first” coin. Then you’d be saying “the 1993 coins comes up heads, what’s the other?” 1 in 2 chance.

The only reason it isn’t *quite* this is because THAT OTHER COIN MIGHT BE A 1993 COIN, TOO. Remember when we shaved that last set of numbers from 10 to 9? That’s what happened there. A tiny bit of uncertainty kept one coin from being “the 1993” coin. Either coin *could* be 1993, after all, so you can’t be absolutely sure that one is “the 1993” coin.

Now, in most of the cases, it is in fact “the 1993” coin. But there are a few cases where it isn’t, so instead of being 1 in 2, it’s 19 in 39 – close but not quite.

Now, let’s take the case that was actually asked about. You flip two coins, not necessarily minted in the 1990’s. One – you don’t know which – turns out to be a 1993 heads. What are the chances the other is heads?

Really, really close to 1 in 2. If there are, say, a hundred years where coins were minted, then the odds of both coins being 1993 are really small. You’re really close to one being “the 1993” coin.

OK.

This same logic applies to families. The “Tuesday” boy isn’t as specific as the “firstborn” boy, but changes the odds. The “December 28” boy would change the odds more. The “December 28, 1987” boy would change the odds even more.

The explanation I promised. Here’s a way of thinking about it which might not hurt your head.

Take all families with two children. How many have two boys? About 1 in 4.

Now just take all families with two children and at least one boy. How many have two boys? About 1 in 3.

Why? While keeping the end you’re looking for the same (2 boys) you’ve reduced your set. You’ve shaved off all the families with two girls. Within the set that remains, the number of families with two boys grows proportionally.

Now. Take all families with two children and at least one boy born on December 28, 1987. How many have two boys?

You’re still looking for two boys, but now you’re looking at ONLY those families with one boy born on December 28, 1987. You’ve gotten rid of most of your families. Of the ones that remain, the odds that the one who wasn’t born on December 28, 1987 is pretty close to one in two.

But not quite. Here's why:

Take all families with two children and at least one boy born on a Tuesday. How many have two boys?

You’ve still shaved away a lot of families. You’re only looking at about 1/14 of your original number of families. You’ve changed the odds a lot.

But how have you changed the odds? Well, for most of those families, the “Tuesday” child is only going to be one of the children. In that case, if the “Tuesday” child is male, the odds of the second one being male is 1 in 2.

But ... in some of those families, both children were born on Tuesday. There is no “Tuesday” child, so the distinction has no meaning. For those families, the chances of both children being male are still 1 in 3!

But that’s not a lot of families, so when you tally all the families together, it still gets pretty close to 1 in 2.

13 in 27, as it happens.

posted by kyrademon at 2:25 AM on May 26, 2010 [3 favorites]

Pyry - it's wrong to treat HH as one option. There are four options HT, HT, the first H in HH or the second H in HH. It's the same problem with the two brothers - the boy in the question could be either of the two.

posted by jamespake at 2:39 AM on May 26, 2010

posted by jamespake at 2:39 AM on May 26, 2010

There is a difference between "what proportion of families with at least one boy have two boys" and "a family has announced that it has at least one boy, what is the probability it has two". The difference depends on what *alternatives* the announcing family had with regards to the announcement.

For example, suppose a family has to announce that it either has at least one boy, or that it has at least one girl. If the family has one boy and one girl, it can announce either (assume that they uniformly randomly pick one). Then the question, "what is the probability that a family that has announced it has at least one boy has two boys?" has an answer different from 1/3.

That is, the probability tree looks like this:

(1/4) BB -> announce Boy (1) -> BBaB (1/4)

(1/4) GG -> announce Girl (1) -> GGaG (1/4)

(1/4) BG -> announce Boy (1/2) -> BGaB (1/8)

(1/4) BG -> announce Girl (1/2) -> BGaG (1/8)

(1/4) GB -> announce Boy (1/2) -> GBaB (1/8)

(1/4) GB -> announce Girl (1/2) -> GBaG (1/8)

Then it's clear that P(BB | announce B) = 1/2 and not 1/3.

If the family has to announce that it either has at least one boy, or that it has no boys, then the result is the expected 1/3:

(1/4) BB -> announce Boy (1) -> BBaB (1/4)

(1/4) GG -> announce -Boy (1) -> GGa-B (1/4)

(1/4) BG -> announce Boy (1) -> BGaB (1/4)

(1/4) GB -> announce Boy (1) -> GBaB (1/4)

P(BB | announce B) = (1/4) / (1/4 + 1/4 + 1/4) = 1/3.

posted by Pyry at 2:48 AM on May 26, 2010 [1 favorite]

For example, suppose a family has to announce that it either has at least one boy, or that it has at least one girl. If the family has one boy and one girl, it can announce either (assume that they uniformly randomly pick one). Then the question, "what is the probability that a family that has announced it has at least one boy has two boys?" has an answer different from 1/3.

That is, the probability tree looks like this:

(1/4) BB -> announce Boy (1) -> BBaB (1/4)

(1/4) GG -> announce Girl (1) -> GGaG (1/4)

(1/4) BG -> announce Boy (1/2) -> BGaB (1/8)

(1/4) BG -> announce Girl (1/2) -> BGaG (1/8)

(1/4) GB -> announce Boy (1/2) -> GBaB (1/8)

(1/4) GB -> announce Girl (1/2) -> GBaG (1/8)

Then it's clear that P(BB | announce B) = 1/2 and not 1/3.

If the family has to announce that it either has at least one boy, or that it has no boys, then the result is the expected 1/3:

(1/4) BB -> announce Boy (1) -> BBaB (1/4)

(1/4) GG -> announce -Boy (1) -> GGa-B (1/4)

(1/4) BG -> announce Boy (1) -> BGaB (1/4)

(1/4) GB -> announce Boy (1) -> GBaB (1/4)

P(BB | announce B) = (1/4) / (1/4 + 1/4 + 1/4) = 1/3.

posted by Pyry at 2:48 AM on May 26, 2010 [1 favorite]

Okay, here's another way of looking at it - if you meet someone who just tells you that they have two children then the odds of them having two boys is 25% and the odds of them having one boy are 50%. But if they say "Their names are Peter and..." then the chance that the other is a boy is 50% - Peter could have a younger brother, an older brother, a young sister or an older sister i.e. There are 4 options not 3 - Pb bP Pg gP.

posted by jamespake at 2:54 AM on May 26, 2010

posted by jamespake at 2:54 AM on May 26, 2010

This puzzle always pissed me off. It's not really a math puzzle at all; the actual problem--"Given that *at least one* of two children is a boy, what is the probability that both children are boys?"--isn't very interesting if stated clearly. The only real puzzle here is figuring out how to convince people that the deliberately coy problem statement is unambiguous enough that they should've been able to figure out what the hell you were asking them.

I mean, terse problem statements are nice, but if unpacking the statement is the whole goddamn puzzle, that's pretty sad. The Monty Hall problem is a much better puzzle than the two-boy one, despite having almost the same crux.

posted by equalpants at 3:18 AM on May 26, 2010 [5 favorites]

I mean, terse problem statements are nice, but if unpacking the statement is the whole goddamn puzzle, that's pretty sad. The Monty Hall problem is a much better puzzle than the two-boy one, despite having almost the same crux.

posted by equalpants at 3:18 AM on May 26, 2010 [5 favorites]

Just to be clear, the answer to this riddle hinges on what you think the possible knowledge states are:

If the three possible states are "we have no male children", "we have at least one male child, but none born on a Tuesday" and "we have at least one male child born on a Tuesday", then the answer is the given 13/27.

If the states are "we have no male children", and "we have a male child born on a Monday", "we have a male child born on Tuesday", "we have... Wednesday", and so on, then the answer is 1/3.

If the states are "we have a [gender] born on [weekday]" then the answer is 1/2.

posted by Pyry at 3:49 AM on May 26, 2010 [1 favorite]

If the three possible states are "we have no male children", "we have at least one male child, but none born on a Tuesday" and "we have at least one male child born on a Tuesday", then the answer is the given 13/27.

If the states are "we have no male children", and "we have a male child born on a Monday", "we have a male child born on Tuesday", "we have... Wednesday", and so on, then the answer is 1/3.

If the states are "we have a [gender] born on [weekday]" then the answer is 1/2.

posted by Pyry at 3:49 AM on May 26, 2010 [1 favorite]

There's a little confusion going on among here about something, I think.

Every time a child is born, there is a (roughly speaking) 50% chance that it will be a boy or a girl.

So, naturally, the tendency is to assume, that no matter what other outside information is given, the chance of any child being a boy or a girl is 50%.

But the thing is, precisely *because* the chance of any individual child being born a boy or a girl is 50%, that number changes when you look at the probability across different, limited sets of families, or talk about how many families overall end up having four boys in a row, or what have you.

This is pretty obvious, when you think about it. Sure, there's a 50% chance of having a boy each time, and THAT MEANS that your chance of having four boys in a row is less than 50%. Likewise, when you disregard all families that have only girls, the percentages of the remaining families shift around.

But it means you can cut up the sets in tricky, nonobvious ways. For example, you'd probably think that the distribution of 2-child families with at least one boy born on a Tuesday looks about the same as 2-child families with at least one boy.

It's not, quite.

Out of 27 million 2-child families with at least one boy, about 9 million will have two boys.

Out of 27 million 2-child families with at least one boy born on a Tuesday, about 13 million will have two boys.

That seem weird? Think about it this way.

Start with 147 million 2-child families with at least one boy. About a third, 49 million, will have two boys.

Now get rid of all the Sunday-Sunday families. That's about 3 million families gone, a third of which, 1 million, will have had two boys.

Do the same for all the Sunday-Mondays, Monday-Wednesdays, and so on. Everything that doesn't have a Tuesday in it. Now we've gotten rid of a good chunk, we're down to 39 million families, each with at least one child, boy or girl, born on a Tuesday. About a third, 13 million, will be 2-boy families.

Here's where it gets interesting.

Get rid of the families with A girl born on Tuesday and a boy born on Sunday. We're not interested in them. We specified the boy was born on Tuesday. 2 million families off the list -- none of them 2 boy families.

Keep going. Girl Tuesday and Boy Monday. Girl Tuesday and Boy Wednesday. And so on. We just got rid of 12 million more families. Not a single one was a two-boy family.

So suddenly, we're down to ... 27 million families, 13 million of which are 2-boy families!

Basically, by specifying the case that at least one boy was born on a Tuesday, we took the case of any-child-born-on-a-Tuesday (39M families, 13M 2-boy families), and got rid of a whole bunch of boy-girl families! No wonder the odds look different.

Sure, it's a little tricksy, but it's essentially the same as looking at the difference between the odds of having two boys in a row if you look at all families (25%), only families with at least one boy (33%), or only families where the first child was a boy (50%).

This is *because* the chance of any individual child being born a boy or girl is 50%, not in spite of it. It's moving the statistical frame around.

posted by kyrademon at 4:08 AM on May 26, 2010 [2 favorites]

Every time a child is born, there is a (roughly speaking) 50% chance that it will be a boy or a girl.

So, naturally, the tendency is to assume, that no matter what other outside information is given, the chance of any child being a boy or a girl is 50%.

But the thing is, precisely *because* the chance of any individual child being born a boy or a girl is 50%, that number changes when you look at the probability across different, limited sets of families, or talk about how many families overall end up having four boys in a row, or what have you.

This is pretty obvious, when you think about it. Sure, there's a 50% chance of having a boy each time, and THAT MEANS that your chance of having four boys in a row is less than 50%. Likewise, when you disregard all families that have only girls, the percentages of the remaining families shift around.

But it means you can cut up the sets in tricky, nonobvious ways. For example, you'd probably think that the distribution of 2-child families with at least one boy born on a Tuesday looks about the same as 2-child families with at least one boy.

It's not, quite.

Out of 27 million 2-child families with at least one boy, about 9 million will have two boys.

Out of 27 million 2-child families with at least one boy born on a Tuesday, about 13 million will have two boys.

That seem weird? Think about it this way.

Start with 147 million 2-child families with at least one boy. About a third, 49 million, will have two boys.

Now get rid of all the Sunday-Sunday families. That's about 3 million families gone, a third of which, 1 million, will have had two boys.

Do the same for all the Sunday-Mondays, Monday-Wednesdays, and so on. Everything that doesn't have a Tuesday in it. Now we've gotten rid of a good chunk, we're down to 39 million families, each with at least one child, boy or girl, born on a Tuesday. About a third, 13 million, will be 2-boy families.

Here's where it gets interesting.

Get rid of the families with A girl born on Tuesday and a boy born on Sunday. We're not interested in them. We specified the boy was born on Tuesday. 2 million families off the list -- none of them 2 boy families.

Keep going. Girl Tuesday and Boy Monday. Girl Tuesday and Boy Wednesday. And so on. We just got rid of 12 million more families. Not a single one was a two-boy family.

So suddenly, we're down to ... 27 million families, 13 million of which are 2-boy families!

Basically, by specifying the case that at least one boy was born on a Tuesday, we took the case of any-child-born-on-a-Tuesday (39M families, 13M 2-boy families), and got rid of a whole bunch of boy-girl families! No wonder the odds look different.

Sure, it's a little tricksy, but it's essentially the same as looking at the difference between the odds of having two boys in a row if you look at all families (25%), only families with at least one boy (33%), or only families where the first child was a boy (50%).

This is *because* the chance of any individual child being born a boy or girl is 50%, not in spite of it. It's moving the statistical frame around.

posted by kyrademon at 4:08 AM on May 26, 2010 [2 favorites]

So, this is how mathematicians masturbate? I much prefer the old-fashioned method...

posted by Thorzdad at 4:40 AM on May 26, 2010

posted by Thorzdad at 4:40 AM on May 26, 2010

Oh hello there, math -- a.k.a. why I went to law school.

posted by the littlest brussels sprout at 5:21 AM on May 26, 2010

posted by the littlest brussels sprout at 5:21 AM on May 26, 2010

"Oh hello there, math -- a.k.a. why I went to law school.

posted by the littlest brussels sprout at 5:21 AM on May 26 "

My brother! (or sister) - I calculate that at least 50% of us were in law school cause we read the requirements and math wasn't one, so we sighed with relief and applied to take the LSAT.

LOL!

Like others who have commented, I see how the numbers work, but it is definitely counter-intuitive. To me, a 50-50 chance looks like the only reality.

But I guess that leaves out hermaphrodites and I understand that they occur more frequently than one might think. ;)

posted by Tena at 6:25 AM on May 26, 2010

posted by the littlest brussels sprout at 5:21 AM on May 26 "

My brother! (or sister) - I calculate that at least 50% of us were in law school cause we read the requirements and math wasn't one, so we sighed with relief and applied to take the LSAT.

LOL!

Like others who have commented, I see how the numbers work, but it is definitely counter-intuitive. To me, a 50-50 chance looks like the only reality.

But I guess that leaves out hermaphrodites and I understand that they occur more frequently than one might think. ;)

posted by Tena at 6:25 AM on May 26, 2010

I'm curious whether anybody here would have assumed the odds are 1/2 (rather than 1/3) if the detail about Tuesday was not provided. Because if the day of the week was unstated, the odds are 1/3.

The funny thing is that the people insisting the odds are 1/2 (rather than 1/3) are actually closer to the correct 13/27 answer, although their reasoning is entirely incorrect.

Incidentally, three different commenters have provided graphic illustrations of all possible boy,girl,day combinations - you can work it out yourself by counting up the colored squares. If you continue to insist that the odds are 1/2, you may be interested in Gene Ray's theories as well.

posted by ardgedee at 6:35 AM on May 26, 2010

The funny thing is that the people insisting the odds are 1/2 (rather than 1/3) are actually closer to the correct 13/27 answer, although their reasoning is entirely incorrect.

Incidentally, three different commenters have provided graphic illustrations of all possible boy,girl,day combinations - you can work it out yourself by counting up the colored squares. If you continue to insist that the odds are 1/2, you may be interested in Gene Ray's theories as well.

posted by ardgedee at 6:35 AM on May 26, 2010

Tena, if you do in fact find that the only reality is "whatever my intuition indicates, despite seeing how the numbers work", then a really excellent career in US politics is awaiting you if you should tire of whatever you're doing now.

posted by Wolfdog at 7:14 AM on May 26, 2010

posted by Wolfdog at 7:14 AM on May 26, 2010

The problem is: How does Tuesday change the odds when it is arbitrary? As someone pointed out above, every child is born on some day of the week. So why can't we say the odds are always 13/27 without the knowledge of which day of the week? I.e., someone poses the question without telling us the day of the week, we do the calculation the tuesday way and get the 13/27 answer **and then** we ask the day of the week the child was born: if its not tuesday, we don't have to go back and change the calculation, since the numbers are the same for any day of the week.

Or what if the questioner says at the end "I lied, the boy was actually born on Friday", again, no change in the calculation.

The answer is in what question you really asking, the 1/2, 1/3 and the 13/27 are all true answers to only very subtly different questions. 13/27 doesn't seem to make sense, unless you pose the question "what is the fraction of 2 children families, who have at least one boy born on a tuesday" We are suspicious of an answer that doesn't have a convenient way of adding up to 100%, but, even though there are 7 days, there are overlaps in the sets for each day.

posted by 445supermag at 7:16 AM on May 26, 2010 [3 favorites]

Or what if the questioner says at the end "I lied, the boy was actually born on Friday", again, no change in the calculation.

The answer is in what question you really asking, the 1/2, 1/3 and the 13/27 are all true answers to only very subtly different questions. 13/27 doesn't seem to make sense, unless you pose the question "what is the fraction of 2 children families, who have at least one boy born on a tuesday" We are suspicious of an answer that doesn't have a convenient way of adding up to 100%, but, even though there are 7 days, there are overlaps in the sets for each day.

posted by 445supermag at 7:16 AM on May 26, 2010 [3 favorites]

I won ten bucks at a party once with exactly this trick. There was a small bet on each pair of flips, and the guy just wouldn't give it up.

In fact, this would be a great game for a casino. You break even on people who know the trick, and make a pile on the people who don't. You could even push the payout a bit to make it look really attractive to the common rube...

posted by kaibutsu at 7:23 AM on May 26, 2010 [1 favorite]

The probability is 1/3: GB, BG, BB -- I think everyone agrees with that.

So here's my issue -- which always confuses me and which has a simple yes or no answer that I'm sure a smart MeFite can provide. It's a semantic question, not a mathematical one:

Is there a difference between the following two questions:

1. You have two kids and one is a boy -- what is the chance that both are boys?

2. You have two kids and one is a boy -- what is the chance that the other is a boy?

I read those two questions differently. The first (to me) is obviously 1/3, the second is obviously 1/2, because they seem to be asking for different things. I freely admit that my mind doesn't work well on these problems, so maybe I'm missing something.

posted by The Bellman at 7:36 AM on May 26, 2010

Privacy Law

Regulation of Environment

Regulation of Telecommunications

Health Care Law

Tax Law

Law school is why I did an undergrad in math.

I kid because I love.

posted by Lemurrhea at 7:37 AM on May 26, 2010

Bellman: The difference is with that word "other". It is ill-defined. Clearly, in a boy-girl family, the girl is the "other" child. However, in a family with two boys, John and Jack, which one is the "other"? This ambiguity is deeply tied to the reasons why it is 1/3 and not 1/2.

The more precise way to ask the question is the first.

posted by milestogo at 7:41 AM on May 26, 2010

The more precise way to ask the question is the first.

posted by milestogo at 7:41 AM on May 26, 2010

Bellman, I would read the two questions the same. There's no difference in information requested or given. You know that one is a boy, and you want to know if the other is a boy. If the other is a boy, it's necessarily true that both are boys.

posted by Lemurrhea at 7:41 AM on May 26, 2010

posted by Lemurrhea at 7:41 AM on May 26, 2010

THERE ARE FOUR LIGHTS

posted by Uther Bentrazor at 7:42 AM on May 26, 2010

posted by Uther Bentrazor at 7:42 AM on May 26, 2010

Why are the days of the week equally weighted? In North America (if not most of the first world) the weekend days see much fewer births.

Here are some stats I found for the US in 2005:

posted by ODiV at 8:15 AM on May 26, 2010

Here are some stats I found for the US in 2005:

Sunday 7,374 Monday 11,704 Tuesday 13,169 Wednesday 13,038 Thursday 13,013 Friday 12,664 Saturday 8,459

posted by ODiV at 8:15 AM on May 26, 2010

The boy/girl distribution ain't exactly 50% either. Oh, well.

posted by kyrademon at 8:20 AM on May 26, 2010

posted by kyrademon at 8:20 AM on May 26, 2010

Lemurrhea: *Bellman, I would read the two questions the same. There's no difference in information requested or given. You know that one is a boy, and you want to know if the other is a boy. If the other is a boy, it's necessarily true that both are boys.*

I understand that reading, but I read it differently. I read the second as asking for the separate, (non-conditional?) probability that the "other" child is a boy, which is (approximately) 50% -- the probability that any given child is any given gender. So that's where I always get confused -- semantics not math because of course the math is absolutely clear once I understand what's being asked.

I think milestogo answered it for me -- if you're seeking the 1/3 answer, the more precise way to ask it is the first and the "riddle" obscures that by using the second.

posted by The Bellman at 9:08 AM on May 26, 2010

I understand that reading, but I read it differently. I read the second as asking for the separate, (non-conditional?) probability that the "other" child is a boy, which is (approximately) 50% -- the probability that any given child is any given gender. So that's where I always get confused -- semantics not math because of course the math is absolutely clear once I understand what's being asked.

I think milestogo answered it for me -- if you're seeking the 1/3 answer, the more precise way to ask it is the first and the "riddle" obscures that by using the second.

posted by The Bellman at 9:08 AM on May 26, 2010

Because the question doesn't really have anything at all to do with the probabilities of particular gender and births. I'm fairly certain that if your first child is one sex you are more likely to have another child of that sex. I know if you have two children of one sex you are much more likely (I've seen 80%) to have a third child of that same sex. Biology/psychology is messy, but not what this question is about.

posted by OmieWise at 9:25 AM on May 26, 2010

posted by

Eponysterical!

posted by phatkitten at 9:31 AM on May 26, 2010

Yeah, we're assuming perfectly spherical children in a vacuum...

posted by kmz at 9:41 AM on May 26, 2010

I'm not sure how serious I am about this.

If you find Photo A on the street, there's a 1/2 chance that both children are boys. If you find Photo B on the street, there's a 1/3 chance that both children are boys.

posted by roll truck roll at 9:55 AM on May 26, 2010

If you find Photo A on the street, there's a 1/2 chance that both children are boys. If you find Photo B on the street, there's a 1/3 chance that both children are boys.

posted by roll truck roll at 9:55 AM on May 26, 2010

Isn't it the case that both Einstein and Paul Erdos backed the wrong answer to the Monty Hall problem?

posted by StickyCarpet at 10:02 AM on May 26, 2010

Why is it assumed that the only allowable sexes are male and female? You could be born with both sexual characteristics or neither.

Now what are the odds?

posted by Kilovolt at 10:14 AM on May 26, 2010

Now what are the odds?

posted by Kilovolt at 10:14 AM on May 26, 2010

The decision matrix that Upton O'Good laid out should really be a tree once you have more than two options. The simple tree assumes that each option has equal probability. If you want to have multiple options per decision node then you just annotate each edge with the probability and when you go to sum up the odds for each leaf in the tree you do it with the product of the probabilities of all the edges between the leaf and the root.

I doubt it changes the overall odds much given that such births are extremely rare.

posted by GuyZero at 10:19 AM on May 26, 2010

> *Isn't it the case that both Einstein and Paul Erdos backed the wrong answer to the Monty Hall problem? *

God doesn't play dice with the menfolk of the universe on Tuesdays.

posted by ardgedee at 10:26 AM on May 26, 2010

God doesn't play dice with the menfolk of the universe on Tuesdays.

posted by ardgedee at 10:26 AM on May 26, 2010

I dunno, but it is entirely possible. The thing that makes the Monty Hall problem unintuitive, and the reason people complain about the phrasing of this problem, is that in each there is a person in the equation who is actively trying to mess with you, something not encountered in physics (we hope).

In the Monty Hall problem, the person is only ever showing you an empty door. In this case, apparently this guy would only ever tell you about a boy if he had at least one. Both of these alter the probability counter to what our intuition would tell us.

Of course with the Monty Hall problem the door switching strategy has no downside, statistically, since even if the guy showing the door was doing it fairly and there were a possibility of showing the prize door, learning that one room is empty reduces it to a 50/50 split between the room you picked and the remaining room you didn't.

posted by Zalzidrax at 10:51 AM on May 26, 2010

Well the 50-50 split for boy and girl is understandable because it's close enough. What I don't get how people see "Tuesday" and then start automatically working out the probabilities using seven equally weighted days. That's like hearing "I was shot once in either my head, my leg, my torso, my foot, or my hand. What is the probability I was shot in my hand?" and then deciding it's 1/7.

posted by ODiV at 10:52 AM on May 26, 2010

It is a semantics problem, not a mathematics problem. All similar problems are as well, because if they were sufficiently specified, then the maths are straightforward and obvious.

Kyrademon's explanation is wonderful. However, I think a case can be made that the answer is 14/28 not 13/28 because you SHOULD double count Tuesday/Tuesday. Chortly's graphic grid was also wonderful, but it could be said that it is hiding information, because if you graph the segments independently and tally, you get a different answer.

posted by discountfortunecookie at 10:57 AM on May 26, 2010

Kyrademon's explanation is wonderful. However, I think a case can be made that the answer is 14/28 not 13/28 because you SHOULD double count Tuesday/Tuesday. Chortly's graphic grid was also wonderful, but it could be said that it is hiding information, because if you graph the segments independently and tally, you get a different answer.

posted by discountfortunecookie at 10:57 AM on May 26, 2010

Says who? If you work it out as above you get a number that has no relevance to anything the person was asking about.

Are we looking for the probability that this guy has two male children or not?

posted by ODiV at 11:02 AM on May 26, 2010

Tuesday is mentioned for a reason, the way I see it, and it has nothing to do with mathematics. He says, *I have one boy, born on a tuesday* to differentiate from his *other*

boy, who was born on a different day. If his other child was a girl, he would have no need to mention the day of the boy's birth. I'm sticking by this no matter what numbers (which only confuse me) are thrown at me.

posted by Hobgoblin at 11:42 AM on May 26, 2010

boy, who was born on a different day. If his other child was a girl, he would have no need to mention the day of the boy's birth. I'm sticking by this no matter what numbers (which only confuse me) are thrown at me.

posted by Hobgoblin at 11:42 AM on May 26, 2010

discountfortunecookie: "*Chortly's graphic grid was also wonderful, but it could be said that it is hiding information, because if you graph the segments independently and tally, you get a different answer.*"

What does this mean?

posted by roll truck roll at 11:51 AM on May 26, 2010

What does this mean?

posted by roll truck roll at 11:51 AM on May 26, 2010

This thread is funny, given that based on some Mefi comment I bought and just read The Drunkard's Walk, which uses this example to show how **the human mind is not optimized for probability and statistics** and why most of the time **our gut feelings are wrong.**

Most people in this thread who don't get it are either going completely with their gut or having a conflict. If probability and statistics came as naturally to us as recognizing faces, people would not need to go to school for years and years to get a degree in the subject.

BTW, the book may be interesting for those who did not love their probability and statistics classes in college. I loved mine and already was familiar with most of the book, but it gave me an excuse to freshen up my Bayes.

posted by dirty lies at 12:01 PM on May 26, 2010

Most people in this thread who don't get it are either going completely with their gut or having a conflict. If probability and statistics came as naturally to us as recognizing faces, people would not need to go to school for years and years to get a degree in the subject.

BTW, the book may be interesting for those who did not love their probability and statistics classes in college. I loved mine and already was familiar with most of the book, but it gave me an excuse to freshen up my Bayes.

posted by dirty lies at 12:01 PM on May 26, 2010

Brilliant! He combined the monty hall trick with the shared birthday trick. I love it!

posted by furtive at 12:08 PM on May 26, 2010

posted by furtive at 12:08 PM on May 26, 2010

If anyone is still confused or for some ridiculous reason actually thinks the answer to this riddle **isn't** 13/27, I have proved it with brute force for your edification.

posted by turaho at 1:39 PM on May 26, 2010 [3 favorites]

posted by turaho at 1:39 PM on May 26, 2010 [3 favorites]

Sigh. That's not the same question though. The whole point of this entire thread is that the probabilities change depending on how much specific information is used to refer to the child. If you just call it "one of them is a boy" that is the most vague as it could mean either the older child or the younger child, and in that case the probability is 1/3. When you specify he was born on a Tuesday, that's another bit of specificity that raises the probability closer to 1/2. When you actually refer to him by name, that removes all doubt about which one you are referring to, as Peter can only be a given one of the two children, so the probability is 50%. Go back and read what I wrote before:

Or another way of stating this: when the problem says "I have two kids, one of which is a boy, what the the probability that the other is a boy", it is not the same thing as asking "my younger child is a boy, what are the chances that my older child is a boy?"posted by Rhomboid at 2:14 PM on May 26, 2010

I've also drawn out the solution to the riddle that empath posted earlier and wouldn't you know, it is even closer to 1/2 than the Tuesday Boy riddle.

posted by turaho at 2:23 PM on May 26, 2010 [1 favorite]

posted by turaho at 2:23 PM on May 26, 2010 [1 favorite]

Is the weekend drop in birthrate not applicable for some reason?

posted by ODiV at 2:42 PM on May 26, 2010

It is not applicable because that's not what the riddle is about. It's usually understood that riddles are self-contained and have all the information stated in them required to solve them. For example a real penny thrown by real humans is only fair to two digits but not more [PDF] but when a problem says flip a coin, everyone knows that means an exact 50/50 coin, not a .505 coin or a .498 coin.

posted by Rhomboid at 2:59 PM on May 26, 2010

Here's where I'm confused. A coin turning up heads and a child being a boy are assumed to be 50% likely because that's reasonably close to what it is in reality. Why, when presented with a number of possibilities that you know nothing about, would you assume that they're equally likely?

"One of my sons found a penny on the ground. My other son also found a coin on the ground. What are the chances he found a nickel?"

posted by ODiV at 3:07 PM on May 26, 2010

"One of my sons found a penny on the ground. My other son also found a coin on the ground. What are the chances he found a nickel?"

posted by ODiV at 3:07 PM on May 26, 2010

For the same reason physics problems all seem exist in a frictionless environment.

posted by turaho at 3:13 PM on May 26, 2010

I still don't get how you can just take a number of possibilities and just assume that they're equally weighted because it's an easy way to categorize something. It's like trying to figure out during what hour I am most likely to eat breakfast and then concluding that there's a 1 in 24 chance I will eat breakfast in any given hour.

posted by ODiV at 3:28 PM on May 26, 2010

posted by ODiV at 3:28 PM on May 26, 2010

Well ... for one thing, ODiv, the birth rates worldwide may very well be closer to an even distribution across 7 days than the stats you posted. The U.S. rates appear to most likely be an anomaly caused largely be C-sections and induced delivery. It's not unreasonable to assume a fairly even distribution of birth by day of week.

But yeah, I do take your point. You do need to know real-world distributions to solve these kinds of problems accurately once you start making them real-world problems and not just abstract math. For example, in your coin problem:

Assuming that the only coins likely to be found are pennies, nickels, dimes, and quarters, then with a completely even distribution –

The chance of one of two random coins being a nickel is 7 in 16.

The chance of one of the coins being a nickel if the other is a penny is 2 in 7.

But an even distribution seems hardly likely. Let's assume you did research and found that on average, 40% of dropped coins are quarters, 30% are dimes, 20% are nickels, and 10% are pennies –

The chance of one of two random coins being a nickel is 9 in 25.

The chance of one of the coins being a nickel if the other is a penny is 4 in 19.

It does change the numbers. If you want the answer in the real world, you are entirely correct that you need the parameters, and the answer to the 2-boy problem is probably somewhat different than the simplified math implies.

posted by kyrademon at 3:31 PM on May 26, 2010

But yeah, I do take your point. You do need to know real-world distributions to solve these kinds of problems accurately once you start making them real-world problems and not just abstract math. For example, in your coin problem:

Assuming that the only coins likely to be found are pennies, nickels, dimes, and quarters, then with a completely even distribution –

The chance of one of two random coins being a nickel is 7 in 16.

The chance of one of the coins being a nickel if the other is a penny is 2 in 7.

But an even distribution seems hardly likely. Let's assume you did research and found that on average, 40% of dropped coins are quarters, 30% are dimes, 20% are nickels, and 10% are pennies –

The chance of one of two random coins being a nickel is 9 in 25.

The chance of one of the coins being a nickel if the other is a penny is 4 in 19.

It does change the numbers. If you want the answer in the real world, you are entirely correct that you need the parameters, and the answer to the 2-boy problem is probably somewhat different than the simplified math implies.

posted by kyrademon at 3:31 PM on May 26, 2010

milestogo: Pretty much everyone in this thread?

Here's turaho's math where he shows the 196 possibilities as all equally likely.

posted by ODiV at 3:32 PM on May 26, 2010

Here's turaho's math where he shows the 196 possibilities as all equally likely.

posted by ODiV at 3:32 PM on May 26, 2010

Oh, I didn't realize you meant the Tuesday thing. That struck me as a given uniform distribution, unlike breakfast.

posted by milestogo at 3:34 PM on May 26, 2010

posted by milestogo at 3:34 PM on May 26, 2010

Both are pretty obviously not uniform distribution in reality, which is my point.

posted by ODiV at 3:37 PM on May 26, 2010

posted by ODiV at 3:37 PM on May 26, 2010

Would it help your understanding of the riddle if it had explicitly said, "Assume boys and girls are equally likely to be born and that the probability of being born on any day is 1/7"? Because implicitly, that is what it is saying.

posted by Rhomboid at 3:44 PM on May 26, 2010

posted by Rhomboid at 3:44 PM on May 26, 2010

Would it help your understanding of this if in reality girls were 20% more likely to be born than boys, but when solving the riddle everyone just assumed an equal distribution?

posted by ODiV at 3:48 PM on May 26, 2010

posted by ODiV at 3:48 PM on May 26, 2010

It's a thought experiment. It has nothing to do with reality. Nobody is *actually* predicting anything.

posted by Rhomboid at 3:50 PM on May 26, 2010

posted by Rhomboid at 3:50 PM on May 26, 2010

And just to be clear, if you really wanted to you could work this problem out with non-equal weights given to each day. It would be much harder, and you probably could not longer do it in your head, and you'd get an answer that was pretty close to but slightly less than .50, just as you do with equal weights. Except you haven't gained anything but make the problem unnecessarily more complicated by adding extra work that doesn't do anything towards proving the point that specifying a birth-weekday changes the answer from 1/3 to something close to 1/2.

posted by Rhomboid at 3:55 PM on May 26, 2010 [1 favorite]

posted by Rhomboid at 3:55 PM on May 26, 2010 [1 favorite]

So if the riddle specified months, would the number of days in a month be similarly irrelevant? Which answer would be the correct answer to the riddle, the one that took into account the fact that months have different lengths or the one that didn't? They'd be very close, granted, but which one would be right on a test?

posted by ODiV at 4:01 PM on May 26, 2010

posted by ODiV at 4:01 PM on May 26, 2010

The word for what you're doing right now is "spoil sport."

posted by roll truck roll at 4:18 PM on May 26, 2010

posted by roll truck roll at 4:18 PM on May 26, 2010

If the riddle specified months then it wouldn't be as good a riddle, because there is no ambiguity as to the fact that months have different lengths -- everyone knows that. However there is no inherent biological reason why birth weekdays would not be uniform, so you'd have to be familiar with the obscure details of hospital management to know that surgical procedures tend to be scheduled on weekdays so that surgeons can have their weekends off [or whatever] in order to even wonder if it's not a uniform distribution. I think it's pretty much obvious to everyone that the question is not about obscure public health statistics and that the author wants us to use 1/7, because again that's how riddles work: everything you need to answer is contained in the text.

Now as to a question on a test, that's a very different matter. Unless it's their first day on the job, the professor knows to state all these assumptions so that there is no room for uncertainty. If this was a question on a statistics test, it would certainly say "Assume boys and girls are equally likely to be born and that the probability of being born on any day is 1/7". But this isn't a formal test, it's a riddle, and riddles omit these kind of things because, again, the unstated rule of riddles is that you don't need to bring in outside information.

posted by Rhomboid at 4:19 PM on May 26, 2010

Now as to a question on a test, that's a very different matter. Unless it's their first day on the job, the professor knows to state all these assumptions so that there is no room for uncertainty. If this was a question on a statistics test, it would certainly say "Assume boys and girls are equally likely to be born and that the probability of being born on any day is 1/7". But this isn't a formal test, it's a riddle, and riddles omit these kind of things because, again, the unstated rule of riddles is that you don't need to bring in outside information.

posted by Rhomboid at 4:19 PM on May 26, 2010

Are you being willfully obtuse, or have you never seen a riddle before, or a physics problem? The real world has nothing whatsoever to do with this, which is why people have been able to transpose it, in this thread, to coins or whatever. Further, as has been pointed out that if you want to talk about the real world,

I suspect you just know a fact that you consider neat and cannot bear that it doesn't have any relevance here.

posted by OmieWise at 4:34 PM on May 26, 2010 [1 favorite]

I understand how you guys get the answer and I understand that's what the asker is driving at. I guess I just don't understand how this is a good riddle.

If someone presented this to me as a riddle, I would find it as unanswerable as "I have a son who is on the varsity high school volleyball team. If there are 12 teams in the league, what is the probability of his team winning the league round robin?" There's vital information that is missing, in my opinion.

I have no issue with them tacking "Assume boys and girls are equally likely to be born and that the probability of being born on any day is 1/7" on to make it answerable. I have an issue with tacking it on myself though, because births/days of the week is not something I would automatically assume to be equal. If most people have no problem assuming that themselves, then I guess that's where I fail.

*Are you being willfully obtuse, or have you never seen a riddle before, or a physics problem?*

Physics problems are usually pretty clear and have fairly well defined assumptions and practices. I've seen riddles before too (obviously), but they usually involve things I have no problems with making assumptions about, I guess.

*The word for what you're doing right now is "spoil sport."*

Sorry, I'm spoiling it for you. Can you skip my comments?

posted by ODiV at 4:58 PM on May 26, 2010

If someone presented this to me as a riddle, I would find it as unanswerable as "I have a son who is on the varsity high school volleyball team. If there are 12 teams in the league, what is the probability of his team winning the league round robin?" There's vital information that is missing, in my opinion.

I have no issue with them tacking "Assume boys and girls are equally likely to be born and that the probability of being born on any day is 1/7" on to make it answerable. I have an issue with tacking it on myself though, because births/days of the week is not something I would automatically assume to be equal. If most people have no problem assuming that themselves, then I guess that's where I fail.

Physics problems are usually pretty clear and have fairly well defined assumptions and practices. I've seen riddles before too (obviously), but they usually involve things I have no problems with making assumptions about, I guess.

Sorry, I'm spoiling it for you. Can you skip my comments?

posted by ODiV at 4:58 PM on May 26, 2010

(The following comment assumes that there are 7 days in a week, 365 days in a year, and equal numbers of odd and even numbered days. I know that’s not exactly accurate, but it makes the math so much less messy, and the problem is hard enough to understand as it is, without muddying the waters further. That being said…)

Well, the original article was right, I was smiling about this one all day. But I also puzzled out a general formula for the problem, and I’m surprised no one had mentioned it yet. I’m not sure it helps to make the problem less counter-intuitive, but seeing the numbers work out is somehow comforting.

Where n = the number of different days the boy could have been born on,

And x = the probability that both children are boys:

x = 2n-1 / 4n-1

Case 1: Two children, one is a boy born on a Tuesday (n=7, or 1 of 7 days):

x = (2 * 7) -1 / (4 * 7) -1

x = 14-1 / 28-1

x = 13/27 ( = 48%)

Case 2: Two children, one is a boy born on an odd day (n=2, or 1 of 2 days):

x = (2 * 2) -1 / (4 * 2) -1

x = 4-1 / 8-1

x = 3/7 ( = 43%)

Case 3: Two children, one is a boy born on March 21 (n=365, or 1 of 365 days):

x = (2 * 365) -1 / (4 * 365) -1

x = 730-1 / 1460-1

x = 729/1459 ( = 49.9657%)

And trickily, this one:

Case 4: Two children, one is a boy (n=1, or 1 of 1 days):

x = (2 * 1) -1 / (4 * 1) -1

x = 2-1 / 4-1

x = 1/3 ( = .33%)

Note that as mentioned in the original article, when the trait is more rare than 1 in 7, the probability approaches 1/2, and when the trait is less rare than 1 in 7, the probability approaches 1/3.

Wow. Very neat.

posted by LEGO Damashii at 5:27 PM on May 26, 2010 [2 favorites]

Well, the original article was right, I was smiling about this one all day. But I also puzzled out a general formula for the problem, and I’m surprised no one had mentioned it yet. I’m not sure it helps to make the problem less counter-intuitive, but seeing the numbers work out is somehow comforting.

Where n = the number of different days the boy could have been born on,

And x = the probability that both children are boys:

x = 2n-1 / 4n-1

Case 1: Two children, one is a boy born on a Tuesday (n=7, or 1 of 7 days):

x = (2 * 7) -1 / (4 * 7) -1

x = 14-1 / 28-1

x = 13/27 ( = 48%)

Case 2: Two children, one is a boy born on an odd day (n=2, or 1 of 2 days):

x = (2 * 2) -1 / (4 * 2) -1

x = 4-1 / 8-1

x = 3/7 ( = 43%)

Case 3: Two children, one is a boy born on March 21 (n=365, or 1 of 365 days):

x = (2 * 365) -1 / (4 * 365) -1

x = 730-1 / 1460-1

x = 729/1459 ( = 49.9657%)

And trickily, this one:

Case 4: Two children, one is a boy (n=1, or 1 of 1 days):

x = (2 * 1) -1 / (4 * 1) -1

x = 2-1 / 4-1

x = 1/3 ( = .33%)

Note that as mentioned in the original article, when the trait is more rare than 1 in 7, the probability approaches 1/2, and when the trait is less rare than 1 in 7, the probability approaches 1/3.

Wow. Very neat.

posted by LEGO Damashii at 5:27 PM on May 26, 2010 [2 favorites]

For my sanity to summarize...

First, for the pedanterists: assuming 1/2 probability for girl or boy and 1/7 probability for a birth day on any given day and neglecting twins altogether...

All three answers (1/2,1/3,13/27) are correct given three distinct questions or scenarios probing different statistical samples.

Questions:

1) I have two children, one is a boy (or girl...no bearing), what is the chance the other is a boy? A: 1/2.

2) I have two children, one is a boy, what is the chance both are boys? A: 1/3.

3) I have two children, one is a boy born on tuesday, what is the chance the other is a boy? A: 13/27

These correspond to statistical samples:

1) out of all families with two children, selecting one of the children at random, what is the probability it is a boy? A: 1/2.

2) out of all families with two children one of whom is a boy, what is the probability both are boys? A: 1/3

3) out of all families with two children one of whom is a boy born on tuesday (or any day), what is the probability that both are boys? A: 13/27.

These samples can be created by the scenarios introduced in this blog post noted upthread; quoting:

1) "A father of two children is picked at random. He is instructed to choose a child by flipping a coin. Then he has to provide information about the chosen child in the following format: “I have a son/daughter born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?"

2) "A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son. If he has one son, he is instructed to provide information on his son’s day of birth. If he has two sons, he has to choose one at random. His statement will be, “I have a son born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?"

3) "A father of two children is picked at random. If he doesn’t have a son who is born on Tuesday, he is sent home and another is picked at random until one who has a son that was born on Tuesday is found. He is instructed to tell you, “I have a son born on Tuesday.” What is the probability that the second child is also a son?"

The maths are shown for each case there as well, but the numbers are well established by now.

In the end, to cite a number, one must ask the age old question: what's the scenario?

posted by sloe at 5:37 PM on May 26, 2010

First, for the pedanterists: assuming 1/2 probability for girl or boy and 1/7 probability for a birth day on any given day and neglecting twins altogether...

All three answers (1/2,1/3,13/27) are correct given three distinct questions or scenarios probing different statistical samples.

Questions:

1) I have two children, one is a boy (or girl...no bearing), what is the chance the other is a boy? A: 1/2.

2) I have two children, one is a boy, what is the chance both are boys? A: 1/3.

3) I have two children, one is a boy born on tuesday, what is the chance the other is a boy? A: 13/27

These correspond to statistical samples:

1) out of all families with two children, selecting one of the children at random, what is the probability it is a boy? A: 1/2.

2) out of all families with two children one of whom is a boy, what is the probability both are boys? A: 1/3

3) out of all families with two children one of whom is a boy born on tuesday (or any day), what is the probability that both are boys? A: 13/27.

These samples can be created by the scenarios introduced in this blog post noted upthread; quoting:

1) "A father of two children is picked at random. He is instructed to choose a child by flipping a coin. Then he has to provide information about the chosen child in the following format: “I have a son/daughter born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?"

2) "A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son. If he has one son, he is instructed to provide information on his son’s day of birth. If he has two sons, he has to choose one at random. His statement will be, “I have a son born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?"

3) "A father of two children is picked at random. If he doesn’t have a son who is born on Tuesday, he is sent home and another is picked at random until one who has a son that was born on Tuesday is found. He is instructed to tell you, “I have a son born on Tuesday.” What is the probability that the second child is also a son?"

The maths are shown for each case there as well, but the numbers are well established by now.

In the end, to cite a number, one must ask the age old question: what's the scenario?

posted by sloe at 5:37 PM on May 26, 2010

Wow. Reading back that was a pointless derail, annoying and needlessly argumentative. Sorry for that. Please ignore.

posted by ODiV at 6:12 PM on May 26, 2010 [1 favorite]

posted by ODiV at 6:12 PM on May 26, 2010 [1 favorite]

This is why I love Metafliter. When this kind of shit is going on, the rest of the world is temporarily safe from math geeks!

But as a non math geek, this is really interesting...

posted by salishsea at 10:13 PM on May 26, 2010

But as a non math geek, this is really interesting...

posted by salishsea at 10:13 PM on May 26, 2010

I am not math savvy but I enjoy semantics. I still say that everyone is wrong so far. Ok, I never said it but I thought it. The question states 'what is the probability I have two boys?' Having two boys is the crux. Well then how many different types of children can you have? I count more than two. Boy, girl, unknown, both sexes in one child, neither sex in one child. However you figure it the odds are going to be less than 1/2 or less than 1/3.

You must figure into the equation all outcomes of having another child. Boy , girl, other.

Plus you would have to figure into the equation where you live and what are the chances of having a certain type of sex (or non sex) born.

There is no right answer. It is like asking what colour is Thursday?

posted by Kilovolt at 7:12 AM on May 27, 2010

You must figure into the equation all outcomes of having another child. Boy , girl, other.

Plus you would have to figure into the equation where you live and what are the chances of having a certain type of sex (or non sex) born.

There is no right answer. It is like asking what colour is Thursday?

posted by Kilovolt at 7:12 AM on May 27, 2010

Urg.

The whole point of using boys on a Tuesday as that it’s something people are familiar with and intuitive about. It amplifies the counter-intuivity.

If, like Kilovolt, myself, and others, you have a hard time allowing assumptions that simplify the math of the problem, you can work it out yourself with coins and dice, and get essentially the same results.

You have two fair, six-sided dice, and two fair, double-sided coins.

If you flip both the coins (and they land normally) and one comes up heads, there is a 1 in 3 chance that the other is also heads (and not 1 in 2, as you might think). Do it a twenty times and you’ll see the pattern emerge.

If you then assign one die to each coin, and roll one die with one coin and one die with the other coin, and one of the sets comes up with a 3 on the die, and heads on the coin, the chances that the other coin is also heads will be 11/23.

Repeat the procedure, but remember to only count the times that one of the two dice shows a 3 (or which ever number you chose, but it has to stay the same) and the coin from that set shows heads.

Track the results. The math works out.

Everyone knows that a coin has a 1 in 2 chance of landing on heads. So how can it be 1 in 3, or 11 in 23? That’s crazy!

See, it's the math that's the amazing part, not the minutiae. Don't stare so hard at the trees. They’re cool in their own way, but right now we’re experiencing the forest.

posted by LEGO Damashii at 8:41 AM on May 27, 2010 [2 favorites]

The whole point of using boys on a Tuesday as that it’s something people are familiar with and intuitive about. It amplifies the counter-intuivity.

If, like Kilovolt, myself, and others, you have a hard time allowing assumptions that simplify the math of the problem, you can work it out yourself with coins and dice, and get essentially the same results.

You have two fair, six-sided dice, and two fair, double-sided coins.

If you flip both the coins (and they land normally) and one comes up heads, there is a 1 in 3 chance that the other is also heads (and not 1 in 2, as you might think). Do it a twenty times and you’ll see the pattern emerge.

If you then assign one die to each coin, and roll one die with one coin and one die with the other coin, and one of the sets comes up with a 3 on the die, and heads on the coin, the chances that the other coin is also heads will be 11/23.

Repeat the procedure, but remember to only count the times that one of the two dice shows a 3 (or which ever number you chose, but it has to stay the same) and the coin from that set shows heads.

Track the results. The math works out.

Everyone knows that a coin has a 1 in 2 chance of landing on heads. So how can it be 1 in 3, or 11 in 23? That’s crazy!

See, it's the math that's the amazing part, not the minutiae. Don't stare so hard at the trees. They’re cool in their own way, but right now we’re experiencing the forest.

posted by LEGO Damashii at 8:41 AM on May 27, 2010 [2 favorites]

Again: it's 50%. Any explanations otherwise are trying to go beyond what the original problem said.

Beanplatin' motherfuckers...

posted by grubi at 8:58 AM on May 27, 2010 [1 favorite]

Beanplatin' motherfuckers...

posted by grubi at 8:58 AM on May 27, 2010 [1 favorite]

Apparently Martin Gardner acknowledged the ambiguity in the question.

posted by jamespake at 9:00 AM on May 27, 2010 [1 favorite]

posted by jamespake at 9:00 AM on May 27, 2010 [1 favorite]

It's a riddle, a thought experiment; it's designed to induce beanplating. The answer is most assuredly not 50%, and arriving at that conclusion does not require going beyond what is stated.

posted by Rhomboid at 2:19 PM on May 27, 2010

posted by Rhomboid at 2:19 PM on May 27, 2010

I've been away a few days, but this problem went with me. Two nights ago I was lying in bed and figured out pretty much what LEGO D says above. It took me a while to wrap my head around it, but now I'm with the 13/27 crowd.

Man, I love beans!

posted by MtDewd at 5:17 AM on May 31, 2010

Man, I love beans!

posted by MtDewd at 5:17 AM on May 31, 2010

There has been common assumptions according to jamespake and Wikipedia: "this problem is about probability and not about obstetrics or demography. The problem would be the same if it was phrased using a gold coin and a silver coin.

I agree that the problem is about probability but it is not the same as using a gold coin and a silver coin. A coin has two sides only to land on. Having a child has more than boy or girl to have for a result.

This forest is unending.

posted by Kilovolt at 9:16 PM on May 31, 2010

I agree that the problem is about probability but it is not the same as using a gold coin and a silver coin. A coin has two sides only to land on. Having a child has more than boy or girl to have for a result.

This forest is unending.

posted by Kilovolt at 9:16 PM on May 31, 2010

Having a child has more than boy or girl to have for a result.

Are you assuming hermaphrodites, then? Because generally XX and XY are the two choices we get as parents.

If we start getting into XXY's and hermaphrodites, this question is going to have a million possibilities, instead of just two: the other child is a boy, or the other child is a girl.

Which makes it a 50/50 probability, no matter how you all care to beanplate it.

posted by misha at 7:07 AM on June 1, 2010

Are you assuming hermaphrodites, then? Because generally XX and XY are the two choices we get as parents.

If we start getting into XXY's and hermaphrodites, this question is going to have a million possibilities, instead of just two: the other child is a boy, or the other child is a girl.

Which makes it a 50/50 probability, no matter how you all care to beanplate it.

posted by misha at 7:07 AM on June 1, 2010

So I dunno, misha, you could scroll up and read the discussion. I'm not sure if people are up for having it again.

posted by roll truck roll at 9:40 AM on June 1, 2010

posted by roll truck roll at 9:40 AM on June 1, 2010

Sorry, roll-truck-roll, but the Engineer got home over the weekend and supported my position of the 50/50 solution, which left me feeling all math-confident and super-powerful for once, so I had to come in and start things back up again.

I'm only half kidding.

posted by misha at 1:01 PM on June 1, 2010

I'm only half kidding.

posted by misha at 1:01 PM on June 1, 2010

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