# The Sleeping Beaty problem

July 21, 2004 3:15 AM Subscribe

"We plan to put Beauty to sleep by chemical means, and then we’ll flip a fair coin. If the coin lands Heads, we will awaken Beauty on Monday afternoon and interview her. If it lands Tails, we will awaken her Monday afternoon, interview her, put her back to sleep, and then awaken her again on Tuesday afternoon and interview her again. The (each?) interview is to consist of the one question : what is your credence now for the proposition that our coin landed Heads? When awakened (and during the interview) Beauty will not be able to tell which day it is, nor will she remember whether she has been awakened before. She knows about the above details of our experiment. What credence should she state in answer to our question?"

In light of the recent thread on the Monty Hall problem, here's a probability puzzle that's even more mind-bending: the Sleeping Beauty problem. Some people say the answer is 1/2. Some people say the answer is 1/3. Some people say there is no answer. Papers have been written which can't resolve this one.

In light of the recent thread on the Monty Hall problem, here's a probability puzzle that's even more mind-bending: the Sleeping Beauty problem. Some people say the answer is 1/2. Some people say the answer is 1/3. Some people say there is no answer. Papers have been written which can't resolve this one.

I won't explain my logic, by the way. It's just the way it is. Send $25 for a brochure.

posted by Jimbob at 5:12 AM on July 21, 2004

posted by Jimbob at 5:12 AM on July 21, 2004

A quick and dirty program (may be wrong) to model this problem shows that for 100,000 iterations of this experiment (150,000 throws), and randomly guessing heads or tails each time...

heads guesses will be correct 25,000 times,

tails guesses will will be correct 50,000 times.

if she guesses heads all the time, she'll get 50,000 right.

if she guesses tails all the time, she'll get 100,000 guesses right.

I don't know how to say that in posh statistic-ese, but I think that this means the answer is ...

if you tell me that the coin landed heads, then you're only going to be right 1/3 of the time and your credence (credibility?) is 1/3.

posted by seanyboy at 5:30 AM on July 21, 2004

heads guesses will be correct 25,000 times,

tails guesses will will be correct 50,000 times.

if she guesses heads all the time, she'll get 50,000 right.

if she guesses tails all the time, she'll get 100,000 guesses right.

I don't know how to say that in posh statistic-ese, but I think that this means the answer is ...

if you tell me that the coin landed heads, then you're only going to be right 1/3 of the time and your credence (credibility?) is 1/3.

posted by seanyboy at 5:30 AM on July 21, 2004

seanyboy: How do you generate your throws?

It probably doesn't matter, but if anyone trying this out for themselves want to be _completely_ sure, have a look at HotBits.

posted by spazzm at 5:36 AM on July 21, 2004

It probably doesn't matter, but if anyone trying this out for themselves want to be _completely_ sure, have a look at HotBits.

posted by spazzm at 5:36 AM on July 21, 2004

Your modelling is brilliant seanyboy (although your perfectly rounded numbers are a big disconcerting) - why didn't someone think of that before? I guess people enjoy the mental anguish.

posted by Jimbob at 6:12 AM on July 21, 2004

posted by Jimbob at 6:12 AM on July 21, 2004

the results were pseudo random, but good enough.

After thinking about it a bit more, the problem seems to be in how the question is asked. Because S.B. is asked the question twice as much when the result comes up tails, then the probability is that that she is asked the question when the result is heads drops. The probability that the result is heads stays at 50%

The people asking the question stretch the numbers forwards (six sided dice / lottery numbers) to show that the 1/3 probability calculation is incorrrect.

However, if I said to S.B. that I would only wake her up on tails (1 awakening on tails, zero on heads), then she would be 100% sure upon being asked the question that the probability of heads is 0%

posted by seanyboy at 6:17 AM on July 21, 2004

After thinking about it a bit more, the problem seems to be in how the question is asked. Because S.B. is asked the question twice as much when the result comes up tails, then the probability is that that she is asked the question when the result is heads drops. The probability that the result is heads stays at 50%

The people asking the question stretch the numbers forwards (six sided dice / lottery numbers) to show that the 1/3 probability calculation is incorrrect.

However, if I said to S.B. that I would only wake her up on tails (1 awakening on tails, zero on heads), then she would be 100% sure upon being asked the question that the probability of heads is 0%

posted by seanyboy at 6:17 AM on July 21, 2004

I rounded the numbers up myself... The truth was a bit dirtier.

posted by seanyboy at 6:20 AM on July 21, 2004

posted by seanyboy at 6:20 AM on July 21, 2004

It is obviously a third (I guess this makes me a "thirder").

There are two possible results for the coin, heads and tails. They have equal probability. Half the time (when it's heads), sleeping beauty gets woken up once. The other half (when it's tails), she gets woken up twice.

If I'm sleeping beauty, I know that there are three possibilites:

1. the throw came up heads, and this is my only interview,

2. the throw came up tails, and this is my first (Monday) interview,

3. or the throw came up tails, and this is my second (Tuesday) interview.

It is only in the first of these possibilites that the throw was heads -- so the probability is 1/3. I don't understand how you'd justify the "halfer" position.

posted by reklaw at 6:34 AM on July 21, 2004

There are two possible results for the coin, heads and tails. They have equal probability. Half the time (when it's heads), sleeping beauty gets woken up once. The other half (when it's tails), she gets woken up twice.

If I'm sleeping beauty, I know that there are three possibilites:

1. the throw came up heads, and this is my only interview,

2. the throw came up tails, and this is my first (Monday) interview,

3. or the throw came up tails, and this is my second (Tuesday) interview.

It is only in the first of these possibilites that the throw was heads -- so the probability is 1/3. I don't understand how you'd justify the "halfer" position.

posted by reklaw at 6:34 AM on July 21, 2004

Does this have anything to do with the Art Carney situation?

posted by spilon at 6:34 AM on July 21, 2004

posted by spilon at 6:34 AM on July 21, 2004

If the actual problem is stated exactly as above (and in the link), then the semantics of "what is your credence now for the proposition that our coin landed Heads?" and "What credence should she state in answer to our question?" do not make a lot of sense, because neither the tester's actions nor Beauty's future are determined by Beauty's answers.

She could just as easily (and correctly) answer, "Fuck you, asshole. Get me some coffee."

Also, one very important detail is missing: Does Beauty know the rules of the game?

posted by mischief at 6:50 AM on July 21, 2004

She could just as easily (and correctly) answer, "Fuck you, asshole. Get me some coffee."

Also, one very important detail is missing: Does Beauty know the rules of the game?

posted by mischief at 6:50 AM on July 21, 2004

reklaw: the flaw in your reasoning is that you are assuming all three possibilities are equally likely. They're not. It's just as valid to say that 1/2 the time the coin will be heads, 1/4 time the coin will be tails and this is the first interview, and 1/4 the time the coin will be tails and this is the second interview. It all depends where your frame of reference is.

posted by salmacis at 6:52 AM on July 21, 2004

posted by salmacis at 6:52 AM on July 21, 2004

*Also, one very important detail is missing: Does Beauty know the rules of the game?*

Yes: "She knows about the above details of our experiment." However, I do think "Fuck you asshole" is the better answer!

posted by salmacis at 6:53 AM on July 21, 2004

*I don't understand how you'd justify the "halfer" position.*

halfer position.

I throw a coin. If it comes up heads, I'll only ask you what the coin was once. If it comes up tails, I'll keep waking you up every year for the next 100 years to ask you the same question. What credence do you have upon being woken up that the answer is heads?

There is a fifty percent chance that the coin was heads. If S.B is woken up, then she can assume that it's the first time she's been woken (as no information passes between her and the other awakenings, and the other awakenings don't affect the answer). Therefore, the chance that the coin is heads is 50% not 1%

or... How does asking you the question more than once reduce the chance of an item being heads or tails?

I'm still a thirder. I just think the question is misleading.

posted by seanyboy at 6:56 AM on July 21, 2004

Hmm...

I think I get this. The reason that my asking the question more than once when it is tails changes the probability is that

posted by reklaw at 7:05 AM on July 21, 2004

*How does asking you the question more than once reduce the chance of an item being heads or tails?*I think I get this. The reason that my asking the question more than once when it is tails changes the probability is that

*sleeping beauty knows I'm going to do that*, and so she can adjust her probability estimate accordingly. If she had no idea that I was going to wake her up more times if it came up tails, then the halfer position would be correct.posted by reklaw at 7:05 AM on July 21, 2004

When she gets woken up, she doesn't know if it's the first time or the second time. But she knows it's twice as likely that she will be woken up if it's tails. So you gotta go with 1/3.

As with the monty hall thing, empirical modeling will shed light on the answer.

posted by PrinceValium at 7:12 AM on July 21, 2004

As with the monty hall thing, empirical modeling will shed light on the answer.

posted by PrinceValium at 7:12 AM on July 21, 2004

The only problem in the question comes from the ambiguity of the question itself.

Look, suppose she'll be woken up 99 times for a tails, and 1 time for a heads (makes things clearer). Each time she's woken up, she has no memory of anything. On any given waking, she has a 99% chance of being currently in a "tails" loop rather than a "heads" loop. On the other hand, the actual chance of the coin landing heads or tails is still, and always, 50/50.

Since the wording of the question is "what is your credence now for the proposition that our coin landed Heads?", I'd have to say 50/50 - regardless of how long she spends in the "tails" loop, there was still a 50/50 chance of the coin landing on heads to start with.

If the question was, "Are you now in a tails loop or in a heads loop?", she should obviously pick tails, since she's got a 99% chance of being correct (or 2/3 in the original problem). But since the question is about the original event, not what comes afterward, that's 50/50 no matter how many times you ask her. Tricking her by asking her lots of times doesn't change the original event!

(Suppose you set up a problem where you flip a coin, then ask a bunch of kids whether it was heads or tails. If they guess right, you stop. If they guess wrong, you ask them again, with the caveat that they ARE REQUIRED to give the SAME wrong answer that they just gave. You ask them five hundred times. So, after we repeat this trial many times, overall your kids guess WRONGLY about 500 times more often than they guess correctly. Are they bad guessers? The coin isn't fair? Neither, you just set up the problem to produce this skewed result. Asking Beauty multiple times in the "tails" loop without letting her benefit from the knowledge due to amnesia is the same as asking the kids repeatedly without letting them change their answer/benefit from their knowledge.)

People argue about this question because it's worded badly, not because it's inherently a great problem. The "thirders" are interpreting the QUESTION to be asking a different thing than the halfers are, which means it's a sucky problem, not a difficult one.

Extend this to the Monty Hall problem: suppose the contestant picks a door, Monty opens one of the other two doors to show a goat, then you take the contestant into another room and flash him with the "Men in Black" light flasher. He forgets everything. Now you walk him back to the stage. He sees Monty, two closed doors, and an open door with a goat. Which door should he pick? In the original Monty Hall problem the contestant gets 2/3 odds by switching, because he's taking advantage of Monty's knowledge of where the prize is. In this case, the contestant has lost Monty's knowledge (flash!), and can't take advantage of it (he forgot which door he originally picked), and thus he can't get 2/3 odds - he has to settle for 1/2 odds.

If you ask him, what is your credence for the proposition that the prize is behind door 1, he has to say 50%. If he could use Monty's knowledge, sure, he could get up to 66%. But he doesn't have Monty's knowledge. And neither does Sleeping Beauty.

posted by jellicle at 7:20 AM on July 21, 2004

Look, suppose she'll be woken up 99 times for a tails, and 1 time for a heads (makes things clearer). Each time she's woken up, she has no memory of anything. On any given waking, she has a 99% chance of being currently in a "tails" loop rather than a "heads" loop. On the other hand, the actual chance of the coin landing heads or tails is still, and always, 50/50.

Since the wording of the question is "what is your credence now for the proposition that our coin landed Heads?", I'd have to say 50/50 - regardless of how long she spends in the "tails" loop, there was still a 50/50 chance of the coin landing on heads to start with.

If the question was, "Are you now in a tails loop or in a heads loop?", she should obviously pick tails, since she's got a 99% chance of being correct (or 2/3 in the original problem). But since the question is about the original event, not what comes afterward, that's 50/50 no matter how many times you ask her. Tricking her by asking her lots of times doesn't change the original event!

(Suppose you set up a problem where you flip a coin, then ask a bunch of kids whether it was heads or tails. If they guess right, you stop. If they guess wrong, you ask them again, with the caveat that they ARE REQUIRED to give the SAME wrong answer that they just gave. You ask them five hundred times. So, after we repeat this trial many times, overall your kids guess WRONGLY about 500 times more often than they guess correctly. Are they bad guessers? The coin isn't fair? Neither, you just set up the problem to produce this skewed result. Asking Beauty multiple times in the "tails" loop without letting her benefit from the knowledge due to amnesia is the same as asking the kids repeatedly without letting them change their answer/benefit from their knowledge.)

People argue about this question because it's worded badly, not because it's inherently a great problem. The "thirders" are interpreting the QUESTION to be asking a different thing than the halfers are, which means it's a sucky problem, not a difficult one.

Extend this to the Monty Hall problem: suppose the contestant picks a door, Monty opens one of the other two doors to show a goat, then you take the contestant into another room and flash him with the "Men in Black" light flasher. He forgets everything. Now you walk him back to the stage. He sees Monty, two closed doors, and an open door with a goat. Which door should he pick? In the original Monty Hall problem the contestant gets 2/3 odds by switching, because he's taking advantage of Monty's knowledge of where the prize is. In this case, the contestant has lost Monty's knowledge (flash!), and can't take advantage of it (he forgot which door he originally picked), and thus he can't get 2/3 odds - he has to settle for 1/2 odds.

If you ask him, what is your credence for the proposition that the prize is behind door 1, he has to say 50%. If he could use Monty's knowledge, sure, he could get up to 66%. But he doesn't have Monty's knowledge. And neither does Sleeping Beauty.

posted by jellicle at 7:20 AM on July 21, 2004

Can she look at her hands and say "Four years!"? :)

posted by aeschenkarnos at 7:27 AM on July 21, 2004

posted by aeschenkarnos at 7:27 AM on July 21, 2004

*I don't understand how you'd justify the "halfer" position.*

I've been resisting commenting because my brain just doesn't handle this kind of thing well, but my gut says the whole rigamarole about mickey finns and whether it's monday or tuesday is moot. Right now, to SB, the coin is either heads or tails and both are a 50% shot.

Then again, I'm clearly missing something because why else would mathematically smart people be arguing the question?

posted by CunningLinguist at 7:31 AM on July 21, 2004

The chance that the coin is heads = 50%

The chance of the coin being heads each time she is asked = 33%

The difference is her Point of View.

(* and I refer you back to my previous comment*)

Even though the probability of heads throughout the experiment is 50%

posted by seanyboy at 7:47 AM on July 21, 2004

The chance of the coin being heads each time she is asked = 33%

The difference is her Point of View.

(* and I refer you back to my previous comment*)

*However, if I said to S.B. that I would only wake her up on tails (1 awakening on tails, zero on heads), then she would be 100% sure upon being asked the question that the probability of heads is 0%*Even though the probability of heads throughout the experiment is 50%

posted by seanyboy at 7:47 AM on July 21, 2004

Third. Or something close to third. Because we're not talking about a single coin toss, we're talking about a series of them. Granted, it's a series of either one or two tosses, but it's not a single toss.

I'm not particularly mathematically inclined, but it seems like the answer would not be 50/50 because there's a coin toss and a half.

posted by majick at 7:49 AM on July 21, 2004

I'm not particularly mathematically inclined, but it seems like the answer would not be 50/50 because there's a coin toss and a half.

posted by majick at 7:49 AM on July 21, 2004

Also sprach Wittgenstein:

"I am sitting with a philosopher in the garden; he says again and again, 'I know that that is a tree,' pointing to a tree that is near us. Someone else arrives and hears this, and I tell him, 'This fellow isn't insane. We are only doing philosophy.'"

posted by Sidhedevil at 8:20 AM on July 21, 2004 [2 favorites]

"I am sitting with a philosopher in the garden; he says again and again, 'I know that that is a tree,' pointing to a tree that is near us. Someone else arrives and hears this, and I tell him, 'This fellow isn't insane. We are only doing philosophy.'"

posted by Sidhedevil at 8:20 AM on July 21, 2004 [2 favorites]

Here's a good explanation of the "the question is ambiguous" position, i.e., "either 1/2 or 1/3 depending on what exactly you mean by 'your credence for the proposition.'"

posted by DevilsAdvocate at 8:34 AM on July 21, 2004

posted by DevilsAdvocate at 8:34 AM on July 21, 2004

Here, let me simplify this:

Call the following a coin flipping session: Flip a coin. If it's tails, you get to flip again. If it's heads, stop the session.

You'll note that, after a hundred sessions, you have far more tails than heads. Duh -- each "tails" gave you another toss. That doesn't mean tails was more likely to come up, or more importantly, that you had any ability to predict the future by saying heads or tails. You might want to bet on tails, because you'll have fewer opportunities to bet on heads, but the odds remain the same.

In political terms, if there's a 50/50 split in the electorate, voter purges make all the difference. In this question, there's a "heads" purge -- one heads win eliminates the potential for a second heads win.

posted by effugas at 8:51 AM on July 21, 2004

Call the following a coin flipping session: Flip a coin. If it's tails, you get to flip again. If it's heads, stop the session.

You'll note that, after a hundred sessions, you have far more tails than heads. Duh -- each "tails" gave you another toss. That doesn't mean tails was more likely to come up, or more importantly, that you had any ability to predict the future by saying heads or tails. You might want to bet on tails, because you'll have fewer opportunities to bet on heads, but the odds remain the same.

In political terms, if there's a 50/50 split in the electorate, voter purges make all the difference. In this question, there's a "heads" purge -- one heads win eliminates the potential for a second heads win.

posted by effugas at 8:51 AM on July 21, 2004

*Call the following a coin flipping session: Flip a coin. If it's tails, you get to flip again. If it's heads, stop the session.*

You'll note that, after a hundred sessions, you have far more tails than heads. Duh -- each "tails" gave you another toss.

You'll note that, after a hundred sessions, you have far more tails than heads. Duh -- each "tails" gave you another toss.

No, you won't. Each tails may give you a subsequent toss, but that subsequent toss is equally likely to be heads or tails.

Theoretical results from a run of 100 sessions, rounded off, would be:

50 sessions H

25 sessions TH

13 sessions TTH

6 sessions TTTH

3 sessions TTTTH

2 sessions TTTTTH

1 session TTTTTTH

Total: 100 heads, 97 tails. If I didn't round off and didn't stop at 6 tails, it would have come out exactly to 100 heads, 100 tails.

But don't take my word for it--code up the simulation yourself if you don't believe me.

I don't see that this has any relevance to the SB problem, though.

posted by DevilsAdvocate at 9:08 AM on July 21, 2004

*Call the following a coin flipping session: Flip a coin. If it's tails, you get to flip again. If it's heads, stop the session.*

You'll note that, after a hundred sessions, you have far more tails than heads. Duh -- each "tails" gave you another toss.

You'll note that, after a hundred sessions, you have far more tails than heads. Duh -- each "tails" gave you another toss.

And as this example shows, "common sense" probability can be very misleading! Actually, you'll have exactly the same number of heads and tails, on average. Think of it like this:

50 sessions: 0T 1H

25 sessions: 1T 1H

13 sessions: 2T 1H

6 sessions: 3T 1H

3 sessions: 4T 1H

1 session: 5T 1H

posted by salmacis at 9:16 AM on July 21, 2004

Halfer.

The probability tree on salmacis's "1/2" link provides enlightenment.

posted by saintsguy at 10:56 AM on July 21, 2004

The probability tree on salmacis's "1/2" link provides enlightenment.

posted by saintsguy at 10:56 AM on July 21, 2004

Halfer.

To me, it seems as though, and seanyboy touched on it with his

Whether this is the first or second time she wakes up to the question, to her, it should be irrelevent in her answer... she has no way of knowing anyway.

posted by Witty at 11:57 AM on July 21, 2004

To me, it seems as though, and seanyboy touched on it with his

*point of view*reponse, her answer to the question "is it heads or tails"**should**always be 50/50... and the fact that she might be tempted to answer tails more often would only be because she's influenced by the terms of the experiment. But like others have said, the actual probability of the coin toss is certainly 50/50 and she should ultimately answer the question under those odds.Whether this is the first or second time she wakes up to the question, to her, it should be irrelevent in her answer... she has no way of knowing anyway.

posted by Witty at 11:57 AM on July 21, 2004

Say you have two Sleeping Beauties, in two different rooms. You flip a coin. If it comes up heads, you randomly choose one of the Sleeping Beauties, wake her up, and ask her what the probability is that the coin came up heads. If it comes up tails, you wake both Sleeping Beauties, and ask them both the same question.

This is, in essense, the same as the original problem. And the answer is the same: one half.

posted by Silune at 2:43 PM on July 21, 2004

This is, in essense, the same as the original problem. And the answer is the same: one half.

posted by Silune at 2:43 PM on July 21, 2004

Or is it? Say you have one hundred Sleeping Beauties, in one hundred rooms. If the coin is heads, you wake one at random, and if the coin is tails you wake all of them. Obviously, any individual Sleeping Beauty is more likely to be woken up if the coin comes up tails, and they all know that...

posted by Silune at 3:05 PM on July 21, 2004

posted by Silune at 3:05 PM on July 21, 2004

True. But that's not the same as only one Sleeping Beauty. One SB knows she's going to be woken up no matter what. So her decision is always the same.

posted by Witty at 3:11 PM on July 21, 2004

posted by Witty at 3:11 PM on July 21, 2004

The question is stilted, but I think it's asking Beauty how confident she should be that it came up heads. In other words, given that she was woken up, how likely is it that the flip was heads. That's a Baysian problem -- the prob of X given Y is the prob of X && Y divided by the prob of Y. Thus, the prob of heads given she was woken up is the prob of heads AND being woken up (1/3) divided by the prob of being woken up (1). So the answer is 1/3.

That's my 2 cents worth. The total value of my mathematical education.

posted by dness2 at 3:17 PM on July 21, 2004

That's my 2 cents worth. The total value of my mathematical education.

posted by dness2 at 3:17 PM on July 21, 2004

Silune, that answer would be 1/200 divided by 101/200, so the probability would be 1/101 (that heads came up, given they were woken up).

posted by dness2 at 3:20 PM on July 21, 2004

posted by dness2 at 3:20 PM on July 21, 2004

I have to wonder about some nerd who dreams up scenarios of drugging women for his nefarious experiments.

posted by mischief at 3:21 PM on July 21, 2004

posted by mischief at 3:21 PM on July 21, 2004

Freshman "philosophy": Given that we exist, the conditional probability that circumstances permit our existence is 1.

The same principle applies fractionally as well. If Beauty's credence is predicated on the event of her waking, it should skew toward ci rcumstances that result in more awakenings. I don't buy the argument that Beauty gains no information by waking up! Her waking tends to confirm the fruitfulness of awakenings.

However, if Beauty's credence is

<!--o

posted by aws17576 at 4:26 PM on July 21, 2004

The same principle applies fractionally as well. If Beauty's credence is predicated on the event of her waking, it should skew toward ci rcumstances that result in more awakenings. I don't buy the argument that Beauty gains no information by waking up! Her waking tends to confirm the fruitfulness of awakenings.

However, if Beauty's credence is

**not**predicated on the event of her wak ing, then, as others have pointed out, the whole premise of the "thirder" view evaporates. The reason I incline to the thirder premise, I think, is expressed by this fragment from a conversation about the problem:

**My friend**: In the case of t ails, is it assumed that Beauty will answer the same way twice?**Me**: Who knows! You're Beauty, and I'm asking you a goddam question.<!--o

posted by aws17576 at 4:26 PM on July 21, 2004

One more way of looking at it:

There are three possible situations she can be in when she wakes up:

"Heads Monday"

"Tails Monday"

"Tails Tuesday"

However, these situations are not all equally likely.

It is actually twice as likely that she will wake up on "Heads Monday" than she will at "Tails Monday". Same goes for "Tails Tuesday".

Sorry but you can't just state this as it is essentially what the question is asking us to determine - as above, I would argue that this is 1/2.

posted by saintsguy at 4:37 PM on July 21, 2004

There are three possible situations she can be in when she wakes up:

"Heads Monday"

"Tails Monday"

"Tails Tuesday"

However, these situations are not all equally likely.

It is actually twice as likely that she will wake up on "Heads Monday" than she will at "Tails Monday". Same goes for "Tails Tuesday".

*On average*, for every four repetitions of this experiment, there will be two occurrences of "Heads Monday", and 1 each of "Tails Monday" and "Tails Tuesday".*...the prob of heads AND being woken up (1/3)...*Sorry but you can't just state this as it is essentially what the question is asking us to determine - as above, I would argue that this is 1/2.

posted by saintsguy at 4:37 PM on July 21, 2004

i was going to link to the old metafilter discussion about the doomsday paradox, but the link's broken for me.

anyway here's a few good articles about it, which feels to me like a natural extension of the SBP. The first link is to Nick Bostrum's site, author of the famous "are we living in a computer simulation?" The doomsday paradox is an application of the anthropic principle, whence derives also the famous fermi paradox, essential to setiologists.

(and if anybody can remember which sci-fi book features diek's "urns" version of the doomsday problem, i would be grateful.)

and as far as discussion of the SBP goes, things like this post directly above do show how sure of themselves people are.

HAHAHAHA no. try again

posted by mitchel at 4:59 PM on July 21, 2004

anyway here's a few good articles about it, which feels to me like a natural extension of the SBP. The first link is to Nick Bostrum's site, author of the famous "are we living in a computer simulation?" The doomsday paradox is an application of the anthropic principle, whence derives also the famous fermi paradox, essential to setiologists.

(and if anybody can remember which sci-fi book features diek's "urns" version of the doomsday problem, i would be grateful.)

and as far as discussion of the SBP goes, things like this post directly above do show how sure of themselves people are.

*It is actually twice as likely that she will wake up on "Heads Monday" than she will at "Tails Monday".*HAHAHAHA no. try again

posted by mitchel at 4:59 PM on July 21, 2004

*It is actually twice as likely that she will wake up on "Heads Monday" than she will at "Tails Monday".*

HAHAHAHA no. try again

HAHAHAHA no. try again

You are right, of course. My bad. To rephrase ...

**If she gets just one shot at guessing**then it is twice as likely to be a "Heads Monday" than it is a "Tails Monday".

That, I think, is the route of the problem - whether we are looking at just one guess, or whether we are looking at many repetitions of the entire experiment, and she is considered to win every time she gets it right.

Oh well, whatever the answer is, great post ... really makes you think!

posted by saintsguy at 12:57 AM on July 22, 2004

The fact that it COULD be Tuesday doesn't mean that it is or has any more chance of being so... from Sleeping Beauty's point of view. In fact, it only WILL be a Tuesday 50% of the time. Since she will have no recollection of waking up on a Monday when she is woken up on Tuesday, then she should have no more reason to believe that it is Tuesday than it is Monday. If it's a Tuesday, then Monday never happened as far as she's concerned. There's a 50/50 chance that it is heads or tails and therefore there's only a 50/50 chance of it being Monday or Tuesday. Hopefully that makes sense.

Halfer... still (if I were Sleeping Beauty).

(on preview)

If it's tails, it will always be Tuesday eventually... which will be 50% of the time. So, to me, the terms of the experiment should not influence how she guesses.

posted by Witty at 9:25 AM on July 22, 2004

Halfer... still (if I were Sleeping Beauty).

(on preview)

If it's tails, it will always be Tuesday eventually... which will be 50% of the time. So, to me, the terms of the experiment should not influence how she guesses.

posted by Witty at 9:25 AM on July 22, 2004

What's amazing about this is that over the course of reading this thread I have changed my mind a number of times, each time thinking, well

posted by mdn at 10:54 AM on July 22, 2004

*of course*it has to be (x or y). I started on and have (for now) ended on being a thirder, but it is an interesting twist to complicate matters temporally, so that you only get around to the second interview at all because of a certain result on monday. But I think that's just a monty-hall type of intuitive hijacking - what it comes down to is that a H ends the session and T automatically adds an additional T - so it's not a split between H/T, it's a split between H/TT...posted by mdn at 10:54 AM on July 22, 2004

I hate to break in like this, but after reading through these links and discussions, I really need to know: Where does one study these things? Surely it's a branch of mathematics, but what courses at what level of education deals with these things. This thread made my head work in very unusual ways for me... I really got a kick out of these problems.

posted by trivirgata at 10:26 PM on July 22, 2004

posted by trivirgata at 10:26 PM on July 22, 2004

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But I say 2/5.

posted by Jimbob at 5:08 AM on July 21, 2004