I'll take the door with TEN goats please Monty
October 21, 2021 9:52 PM Subscribe
The Two Envelope Problem - a Mystifying Probability Paradox (YouTube, 28m23s) presents and analyzes a paradox that the kind of person who likes to argue about the Monty Hall Problem (previously) will probably enjoy.
Infinite diversity in infinite dimensions if the math doesn't work out just keep adding dimensions until it does.
posted by zengargoyle at 10:52 PM on October 21, 2021
posted by zengargoyle at 10:52 PM on October 21, 2021
If I see $5,000,000 in the envelope, I'm keeping it.
The point is that you need to decide what to do before you've seen what's in the envelope.
posted by flabdablet at 12:12 AM on October 22, 2021 [1 favorite]
The point is that you need to decide what to do before you've seen what's in the envelope.
posted by flabdablet at 12:12 AM on October 22, 2021 [1 favorite]
Not a million miles from this anecdote from Surely you're joking Mr Feynman
I once did an experiment in the cyclotron laboratory at Princeton that had some startling results. There was a problem in a hydrodynamics book that was being discussed by all the physics students. The problem is this: You have an S-shaped lawn sprinkler--an S-shaped pipe on a pivot--and the water squirts out at right angles to the axis and makes it spin in a certain direction. Everybody knows which way it goes around; it backs away from the outgoing water. Now the question is this: If you had a lake, or swimming pool--a big supply of water--and you put the sprinkler completely under water, and sucked the water in, instead of squirting it out, which way would it turn? Would it turn the same way as it does when you squirt water out into the air, or would it turn the other way?
The answer is perfectly clear at first sight. The trouble was, some guy would think it was perfectly clear one way, and another guy would think it was perfectly clear the other way. So everybody was discussing it. I remember at one particular seminar, or tea, somebody went nip to Prof John Wheeler and said, "Which way do you think it goes around?"
Wheeler said, "Yesterday, Feynman convinced me that it went backwards. Today, he's convinced me equally well that it goes around the other way. I don't know what he'll convince me of tomorrow!"
posted by BobTheScientist at 12:24 AM on October 22, 2021 [7 favorites]
I once did an experiment in the cyclotron laboratory at Princeton that had some startling results. There was a problem in a hydrodynamics book that was being discussed by all the physics students. The problem is this: You have an S-shaped lawn sprinkler--an S-shaped pipe on a pivot--and the water squirts out at right angles to the axis and makes it spin in a certain direction. Everybody knows which way it goes around; it backs away from the outgoing water. Now the question is this: If you had a lake, or swimming pool--a big supply of water--and you put the sprinkler completely under water, and sucked the water in, instead of squirting it out, which way would it turn? Would it turn the same way as it does when you squirt water out into the air, or would it turn the other way?
The answer is perfectly clear at first sight. The trouble was, some guy would think it was perfectly clear one way, and another guy would think it was perfectly clear the other way. So everybody was discussing it. I remember at one particular seminar, or tea, somebody went nip to Prof John Wheeler and said, "Which way do you think it goes around?"
Wheeler said, "Yesterday, Feynman convinced me that it went backwards. Today, he's convinced me equally well that it goes around the other way. I don't know what he'll convince me of tomorrow!"
posted by BobTheScientist at 12:24 AM on October 22, 2021 [7 favorites]
Does it still work if you put the sprinkler on a treadmill?
posted by flabdablet at 12:30 AM on October 22, 2021 [7 favorites]
posted by flabdablet at 12:30 AM on October 22, 2021 [7 favorites]
Bob, you forgot the part where they did the science! The result is neither direction: the sprinkler does not turn, since the water has the same density and imparts no net torque..
posted by autopilot at 1:09 AM on October 22, 2021 [13 favorites]
posted by autopilot at 1:09 AM on October 22, 2021 [13 favorites]
When considering whether to have another donut I usually reason thus:
posted by Tell Me No Lies at 1:19 AM on October 22, 2021 [15 favorites]
- Statistically speaking there are at least 100 people in the world who are currently making this exact same decision.
- At least 10 of them will decide on the same action as me.
- Who am I to deny these 10 people their donuts?
posted by Tell Me No Lies at 1:19 AM on October 22, 2021 [15 favorites]
Navier-Stokes Equations - Numberphile - YouTube fluid mechanics is a bitch, there are equations thought correct but the working out is way tooooooo hard so there are approximations. Water going around a strict right angle corner reaches infinite velocity WTF? BTW, FLP Vol. I Table of Contents the Feynman lectures on physics are online.
posted by zengargoyle at 3:05 AM on October 22, 2021 [1 favorite]
posted by zengargoyle at 3:05 AM on October 22, 2021 [1 favorite]
That was a really interesting video and I learned something weird about infinite series and ordering, but...
...why in the set-up to the "better defined" problem is the 100 vs 1,000 pair any less likely than the 1 vs 10 pair? That isn't part of the original problem, is it? Surely any value in the envelope is equally likely? I'm not a mathematician (in fact I'm pretty abysmal at most maths) but it took me out of the video - it seemed as though the narrator was using the envelope problem as a jumping off point, rather than actually talking about the original paradox in the end.
posted by chappell, ambrose at 3:52 AM on October 22, 2021
...why in the set-up to the "better defined" problem is the 100 vs 1,000 pair any less likely than the 1 vs 10 pair? That isn't part of the original problem, is it? Surely any value in the envelope is equally likely? I'm not a mathematician (in fact I'm pretty abysmal at most maths) but it took me out of the video - it seemed as though the narrator was using the envelope problem as a jumping off point, rather than actually talking about the original paradox in the end.
posted by chappell, ambrose at 3:52 AM on October 22, 2021
...why in the set-up to the "better defined" problem is the 100 vs 1,000 pair any less likely than the 1 vs 10 pair? That isn't part of the original problem, is it?
Yeah, I was wondering about that too. In the real world, there would be an upper limit, which is that the top envelope can't contain more than 10/11 of all the money in the world, and the bottom envelope 1/11 of all the money in the world. The chance that this envelope benefactor would have all the money in the world would be vanishingly small, so the likely amount is much lower. Let's say it's the richest person in the world, currently Jeff Bezos, giving away all of his money, i.e. his estimated worth of $177 billion. He puts $160.9 billion in the top envelope and $16.09 billion in the bottom (and keeps $10m for himself, so he can buy himself a hamburger now and then). Still seems pretty unlikely that he'd go that far, when there are perfectly good phallic rockets to build. The chance that this envelope benefactor is someone with less money, and/or puts far less in the envelopes because they're tight, has to be higher. So we're looking at a decrease from higher chances that the envelopes contain relatively lower amounts, to smaller chances that they contain higher amounts. Even if the rate of decrease wasn't from 1/4 to 1/8 to 1/16 etc, but went down more rapidly or more slowly, and the amounts weren't $1 and $10 etc, but $2.35 and $23.50 and other random pairings increasing in size, you'd still end up with the same overall conclusion that you're approaching a probability of zero as the amounts in the envelopes go up; the calculations would just be more involved.
posted by rory at 4:23 AM on October 22, 2021 [2 favorites]
Yeah, I was wondering about that too. In the real world, there would be an upper limit, which is that the top envelope can't contain more than 10/11 of all the money in the world, and the bottom envelope 1/11 of all the money in the world. The chance that this envelope benefactor would have all the money in the world would be vanishingly small, so the likely amount is much lower. Let's say it's the richest person in the world, currently Jeff Bezos, giving away all of his money, i.e. his estimated worth of $177 billion. He puts $160.9 billion in the top envelope and $16.09 billion in the bottom (and keeps $10m for himself, so he can buy himself a hamburger now and then). Still seems pretty unlikely that he'd go that far, when there are perfectly good phallic rockets to build. The chance that this envelope benefactor is someone with less money, and/or puts far less in the envelopes because they're tight, has to be higher. So we're looking at a decrease from higher chances that the envelopes contain relatively lower amounts, to smaller chances that they contain higher amounts. Even if the rate of decrease wasn't from 1/4 to 1/8 to 1/16 etc, but went down more rapidly or more slowly, and the amounts weren't $1 and $10 etc, but $2.35 and $23.50 and other random pairings increasing in size, you'd still end up with the same overall conclusion that you're approaching a probability of zero as the amounts in the envelopes go up; the calculations would just be more involved.
posted by rory at 4:23 AM on October 22, 2021 [2 favorites]
why in the set-up to the "better defined" problem is the 100 vs 1,000 pair any less likely than the 1 vs 10 pair? That isn't part of the original problem, is it? Surely any value in the envelope is equally likely?
You could certainly set up a version of the problem in which any value between limits you'd need to define is indeed equally likely, or another version with any other probability distribution you like, and if you did, I expect you'd find that the analysis runs along very similar lines.
The point of putting more limits on the original problem is that the original problem is so vaguely posed as to resist analysis. Also note that the first part of the analysis of the restricted problem is a demonstration that even with the added restrictions, the paradox is still firmly evident.
posted by flabdablet at 4:32 AM on October 22, 2021 [3 favorites]
You could certainly set up a version of the problem in which any value between limits you'd need to define is indeed equally likely, or another version with any other probability distribution you like, and if you did, I expect you'd find that the analysis runs along very similar lines.
The point of putting more limits on the original problem is that the original problem is so vaguely posed as to resist analysis. Also note that the first part of the analysis of the restricted problem is a demonstration that even with the added restrictions, the paradox is still firmly evident.
posted by flabdablet at 4:32 AM on October 22, 2021 [3 favorites]
...why in the set-up to the "better defined" problem is the 100 vs 1,000 pair any less likely than the 1 vs 10 pair? That isn't part of the original problem, is it? Surely any value in the envelope is equally likely? I'm not a mathematician (in fact I'm pretty abysmal at most maths) but it took me out of the video - it seemed as though the narrator was using the envelope problem as a jumping off point, rather than actually talking about the original paradox in the end.
The assertion that you gain no information by opening the envelope is false (sort of).
If you assume any amount of money is equally likely, and has no upper bound then the amount of money you expect to get when you open the envelope is an infinite amount of money. Not having found an infinite amount of money opening the envelope, you can conclude that the case is infinitely unlikely.
Mind you, you do not have to open the envelope in the first place to deduce that it does not contain an infinite amount of money, presumably. So you have to assume another distribution of some sort.
posted by Zalzidrax at 4:38 AM on October 22, 2021 [1 favorite]
The assertion that you gain no information by opening the envelope is false (sort of).
If you assume any amount of money is equally likely, and has no upper bound then the amount of money you expect to get when you open the envelope is an infinite amount of money. Not having found an infinite amount of money opening the envelope, you can conclude that the case is infinitely unlikely.
Mind you, you do not have to open the envelope in the first place to deduce that it does not contain an infinite amount of money, presumably. So you have to assume another distribution of some sort.
posted by Zalzidrax at 4:38 AM on October 22, 2021 [1 favorite]
If you assume any amount of money is equally likely, and has no upper bound then the amount of money you expect to get when you open the envelope is an infinite amount of money.
Oh goody, something to argue about!
No it isn't, because regardless of whether there's an upper bound or not, the amount in any particular envelope is a member of the set of real numbers, and all members of that set are finite.
posted by flabdablet at 4:43 AM on October 22, 2021 [3 favorites]
Oh goody, something to argue about!
No it isn't, because regardless of whether there's an upper bound or not, the amount in any particular envelope is a member of the set of real numbers, and all members of that set are finite.
posted by flabdablet at 4:43 AM on October 22, 2021 [3 favorites]
In the real world, there would be an upper limit, which is that the top envelope can't contain more than 10/11 of all the money in the world
In the real world the envelope would contain a cheque, and you can write any amount you want on a cheque. There are not a googol of dollars in circulation in the world but you could write that amount on a cheque if you wanted to.
Whether or not you could cash the cheque you got from your envelope is beyond the scope of the problem as presented.
posted by flabdablet at 4:46 AM on October 22, 2021 [2 favorites]
In the real world the envelope would contain a cheque, and you can write any amount you want on a cheque. There are not a googol of dollars in circulation in the world but you could write that amount on a cheque if you wanted to.
Whether or not you could cash the cheque you got from your envelope is beyond the scope of the problem as presented.
posted by flabdablet at 4:46 AM on October 22, 2021 [2 favorites]
Water going around a strict right angle corner reaches infinite velocity
If you turn off a water tap instantaneously it will create infinite water pressure and explode the pipes.
posted by StickyCarpet at 4:49 AM on October 22, 2021 [2 favorites]
If you turn off a water tap instantaneously it will create infinite water pressure and explode the pipes.
posted by StickyCarpet at 4:49 AM on October 22, 2021 [2 favorites]
In the real world, there would be an upper limit, which is that the top envelope can't contain more than 10/11 of all the money in the world, and the bottom envelope 1/11 of all the money in the world. The chance that this envelope benefactor would have all the money in the world would be vanishingly small, so the likely amount is much lower.
I feel like there are quite a large number of practical objections that you could make to the set up (but are there infinitely many?) which would help to resolve the paradox but would kind of miss the point.
On preview, what flabdablet said.
I'm also interested in the lower bound. Is money infinitely sub-dividable, or is there a sort of Planck length at some point? You're presumably just as likely to get a very small fraction of 1.00 Ariary as you are to get a very large multiple.
posted by chappell, ambrose at 4:52 AM on October 22, 2021
I feel like there are quite a large number of practical objections that you could make to the set up (but are there infinitely many?) which would help to resolve the paradox but would kind of miss the point.
On preview, what flabdablet said.
I'm also interested in the lower bound. Is money infinitely sub-dividable, or is there a sort of Planck length at some point? You're presumably just as likely to get a very small fraction of 1.00 Ariary as you are to get a very large multiple.
posted by chappell, ambrose at 4:52 AM on October 22, 2021
If you turn off a water tap instantaneously it will create infinite water pressure and explode the pipes.
That is pretty much the exact operating principle of the hydraulic ram pump.
posted by flabdablet at 4:52 AM on October 22, 2021 [2 favorites]
That is pretty much the exact operating principle of the hydraulic ram pump.
posted by flabdablet at 4:52 AM on October 22, 2021 [2 favorites]
No it isn't, because regardless of whether there's an upper bound or not, the amount in any particular envelope is a member of the set of real numbers, and all members of that set are finite.
That's why I said you don't need to open the envelope to conclude that the situation is infinitely unlikely. Half of infinity is still infinity, so your expectation value is infinity. Any amount of money you get is going to be finite, even given the ridiculous premise, so you get an amount infinitely less than the expectation value. And you know this without even opening the envelope.
Now admittedly I am fudging proper statistics by saying that finding a finite amount of money means you can conclude that such a distribution is infinitely unlikely, but you're going to keep ending up with infinities and undefined numbers until you give up and decide that you are trying to describe a situation so ridiculous that you cannot do useful math on it.
posted by Zalzidrax at 4:58 AM on October 22, 2021
That's why I said you don't need to open the envelope to conclude that the situation is infinitely unlikely. Half of infinity is still infinity, so your expectation value is infinity. Any amount of money you get is going to be finite, even given the ridiculous premise, so you get an amount infinitely less than the expectation value. And you know this without even opening the envelope.
Now admittedly I am fudging proper statistics by saying that finding a finite amount of money means you can conclude that such a distribution is infinitely unlikely, but you're going to keep ending up with infinities and undefined numbers until you give up and decide that you are trying to describe a situation so ridiculous that you cannot do useful math on it.
posted by Zalzidrax at 4:58 AM on October 22, 2021
Is money infinitely sub-dividable, or is there a sort of Planck length at some point?
Depends on the payment method. If you're doing transfers between bank accounts nominated in dollars, those are generally rounded to the nearest cent. The smallest denomination in which Australian cash is currently issued is five cents, though one-cent pieces can still occasionally be found in the backs of long-undisturbed drawers. US shares can trade at per-share valuations on 1/16 cent boundaries if I recall correctly; finest grain I've seen on an Australian share valuation is 1/20 cent. Bitcoin can be traded in quanta of millisatoshi (10-11 BTC). Stellar Lumens can be subdivided into stroops (10-7 XLM).
posted by flabdablet at 5:32 AM on October 22, 2021 [1 favorite]
Depends on the payment method. If you're doing transfers between bank accounts nominated in dollars, those are generally rounded to the nearest cent. The smallest denomination in which Australian cash is currently issued is five cents, though one-cent pieces can still occasionally be found in the backs of long-undisturbed drawers. US shares can trade at per-share valuations on 1/16 cent boundaries if I recall correctly; finest grain I've seen on an Australian share valuation is 1/20 cent. Bitcoin can be traded in quanta of millisatoshi (10-11 BTC). Stellar Lumens can be subdivided into stroops (10-7 XLM).
posted by flabdablet at 5:32 AM on October 22, 2021 [1 favorite]
I feel like there are quite a large number of practical objections that you could make to the set up (but are there infinitely many?) which would help to resolve the paradox but would kind of miss the point.
Well, sure, if you want to stick to pure mathematics. If you want your two-envelope problem to be considered alongside the Monty Hall problem, though, it ought to be practical for a game show.
posted by rory at 7:43 AM on October 22, 2021 [1 favorite]
Well, sure, if you want to stick to pure mathematics. If you want your two-envelope problem to be considered alongside the Monty Hall problem, though, it ought to be practical for a game show.
posted by rory at 7:43 AM on October 22, 2021 [1 favorite]
This was a really great video, thanks.
Also I assume the reason they arranged things in decreasing probability is because you can't generally define a uniform probability distribution over infinite values. The probabilities also need to satisfy some convergence properties (as in, being a series that sums to 1, which isn't true for repetitions of 1/c for any c). I think they elided that point just because early on in the video they were avoiding the headier maths.
posted by Alex404 at 8:42 AM on October 22, 2021 [3 favorites]
Also I assume the reason they arranged things in decreasing probability is because you can't generally define a uniform probability distribution over infinite values. The probabilities also need to satisfy some convergence properties (as in, being a series that sums to 1, which isn't true for repetitions of 1/c for any c). I think they elided that point just because early on in the video they were avoiding the headier maths.
posted by Alex404 at 8:42 AM on October 22, 2021 [3 favorites]
you get an amount infinitely less than the expectation value
This is known as the Great Disappointment.
posted by nickmark at 9:02 AM on October 22, 2021
This is known as the Great Disappointment.
posted by nickmark at 9:02 AM on October 22, 2021
In the case where the distribution of envelope contents is an unbounded uniform over the positive reals (0,+∞), I think the same argument holds. The expected value of this distribution is undefined (in fact, even the probability density function is undefined everywhere), therefore you can't reason about it. By modifying the original paradox to use a geometric distribution for envelope contents, that distribution has an expected value, but the expected value of profit from switching remains undefined. So I think it helps clarify where exactly the problem with reasoning lies.
This was a really nice video and very clearly explained!
posted by biogeo at 9:55 AM on October 22, 2021 [4 favorites]
This was a really nice video and very clearly explained!
posted by biogeo at 9:55 AM on October 22, 2021 [4 favorites]
Anyone got a link to a text description of the problem?
One time I did the Monty Hall problem live at a booth set up outside the math building as part of a university-wide fair day. It was a complete trip. Most people didn't get it at first, that's the point.
And I knew there would be some skepticism to my quick explanation of the right answer.
But what I did not expect was school kids to understand my explanation when their parents did not, or for a few people who tried it several times in a row as empirical evidence, and still refused to believe.
posted by SaltySalticid at 10:15 AM on October 22, 2021
One time I did the Monty Hall problem live at a booth set up outside the math building as part of a university-wide fair day. It was a complete trip. Most people didn't get it at first, that's the point.
And I knew there would be some skepticism to my quick explanation of the right answer.
But what I did not expect was school kids to understand my explanation when their parents did not, or for a few people who tried it several times in a row as empirical evidence, and still refused to believe.
posted by SaltySalticid at 10:15 AM on October 22, 2021
Anyone got a link to a text description of the problem?
Wikipedia's got you covered along with many interesting variations on the problem.
posted by sjswitzer at 10:20 AM on October 22, 2021 [1 favorite]
Wikipedia's got you covered along with many interesting variations on the problem.
posted by sjswitzer at 10:20 AM on October 22, 2021 [1 favorite]
It seems weird that they need to actually specify a distribution, as the argument doesn't actually depend on the amount of money in the envelopes; just call it x and 2x... You could draw x from a distribution, or it could be some fixed value.
Now the envelope I picked has either x or 2x. So, the value of switching is 0.5 * x + 0.5 * 2x = 1.5 * x. And the value of /not/ switching is 0.5 * x + 0.5 * 2x = 1.5 * x. The same, regardless of what distribution x has been drawn from.
posted by kaibutsu at 10:24 AM on October 22, 2021
Now the envelope I picked has either x or 2x. So, the value of switching is 0.5 * x + 0.5 * 2x = 1.5 * x. And the value of /not/ switching is 0.5 * x + 0.5 * 2x = 1.5 * x. The same, regardless of what distribution x has been drawn from.
posted by kaibutsu at 10:24 AM on October 22, 2021
Oh thanks, I somehow didn't register that the proper name was given in the title!
posted by SaltySalticid at 10:26 AM on October 22, 2021
posted by SaltySalticid at 10:26 AM on October 22, 2021
kaibutsu, the problem with that is that if x is a fixed value that you know, the problem disappears. If x=$10 and I open the envelope and see $10, I know for sure that the other envelope contains $20, and vice versa. And if you don't know x, then x is effectively a random variable (if you're a Bayesian anyway) and the problem remains the same as the originally specified envelope paradox.
posted by biogeo at 12:13 PM on October 22, 2021
posted by biogeo at 12:13 PM on October 22, 2021
Seems to me that kaibutsu's analysis still holds.
One of the envelopes contains an amount of money X, the other contains a larger amount Y, and you don't know which contains which. But you don't need to know what the actual numbers are in order to write down an expression for the expected return from picking an envelope: p(envelope contains X) = p(envelope contains Y) = 1/2 for both envelopes, absent other information, so the expected return for picking either envelope is (X + Y) / 2.
However, opening just one envelope doesn't give you enough information to evaluate that expression, whose value remains undefined unless you open both. So you can't tell whether the one you did open contains more or less than the expected return because the expected return is undefined, which means that opening just one envelope cannot help you decide whether or not you should switch.
Even if the problem setup guarantees that X and Y have some defined arithmetic relationship, this doesn't help you; the expected return was (X + Y) / 2 before you looked inside any envelope, and you cannot tell by opening just one envelope whether you're looking at a value for X or a value for Y except in the special cases where the number you're looking at has the smallest possible value (in which case it is X, and you know for sure that (X + Y) / 2 is bigger so you should switch) or the largest possible value (in which case it is Y, and you know for sure that (X + Y) / 2 is smaller so you shouldn't).
So either opening an envelope doesn't help you at all, or it might (with equal probability) reveal that you should switch or that you shouldn't. Therefore, opening an envelope does not yield consistent advice to switch if the expected-return calculation is done correctly, and there is no paradox.
Even the seeming asymmetry induced by removing any upper bound on Y doesn't help, because you now have an infinite number of possible values, which makes the probability of opening an envelope and seeing any given value strictly zero. So reducing the probability of seeing a maximum possible value to zero by explicitly defining one away has the side effect of reducing the probability of seeing the minimum value to zero as well, which means that the probabilities of seeing minimum and maximum values remain equal, so there is still no switch-preference bias in the information you can glean from opening a single envelope.
posted by flabdablet at 2:25 PM on October 22, 2021
One of the envelopes contains an amount of money X, the other contains a larger amount Y, and you don't know which contains which. But you don't need to know what the actual numbers are in order to write down an expression for the expected return from picking an envelope: p(envelope contains X) = p(envelope contains Y) = 1/2 for both envelopes, absent other information, so the expected return for picking either envelope is (X + Y) / 2.
However, opening just one envelope doesn't give you enough information to evaluate that expression, whose value remains undefined unless you open both. So you can't tell whether the one you did open contains more or less than the expected return because the expected return is undefined, which means that opening just one envelope cannot help you decide whether or not you should switch.
Even if the problem setup guarantees that X and Y have some defined arithmetic relationship, this doesn't help you; the expected return was (X + Y) / 2 before you looked inside any envelope, and you cannot tell by opening just one envelope whether you're looking at a value for X or a value for Y except in the special cases where the number you're looking at has the smallest possible value (in which case it is X, and you know for sure that (X + Y) / 2 is bigger so you should switch) or the largest possible value (in which case it is Y, and you know for sure that (X + Y) / 2 is smaller so you shouldn't).
So either opening an envelope doesn't help you at all, or it might (with equal probability) reveal that you should switch or that you shouldn't. Therefore, opening an envelope does not yield consistent advice to switch if the expected-return calculation is done correctly, and there is no paradox.
Even the seeming asymmetry induced by removing any upper bound on Y doesn't help, because you now have an infinite number of possible values, which makes the probability of opening an envelope and seeing any given value strictly zero. So reducing the probability of seeing a maximum possible value to zero by explicitly defining one away has the side effect of reducing the probability of seeing the minimum value to zero as well, which means that the probabilities of seeing minimum and maximum values remain equal, so there is still no switch-preference bias in the information you can glean from opening a single envelope.
posted by flabdablet at 2:25 PM on October 22, 2021
And if the values are drawn from a probability distribution deliberately skewed so as to give the minimum value a nonzero probability of appearing while the maximum value remains unbounded, then that's the case analyzed in the video.
posted by flabdablet at 2:37 PM on October 22, 2021
posted by flabdablet at 2:37 PM on October 22, 2021
They can't be all zero, they have to be normalized which the video does say very briefly.
posted by polymodus at 5:54 PM on October 22, 2021
posted by polymodus at 5:54 PM on October 22, 2021
I thought the point is that the variable x cannot be a proper random variable. So the paradox is misusing x in the erroneous step where it claims that E[switching] = half 2X + half (X/2), because those two terms are outcomes from two entirely different rounds: either the game set up by dealer with (X, X/2) in the envelopes or the game set up with (2X, X). My understanding is that such a move would not be allowed in standard probability where games are clearly defined and played separately. What the video does with the long chart is lay that out using concrete values. Wikipedia also mentions this conflation explanation, but what's interesting about both the video and Wikipedia is that they cover and discuss a lot other issues/objections/explanations that seem to arise, some of which I don't understand but look very interesting.
In comparison, a human being who isn't trying to formalize the example mathematically can just reason using symmetry and say it's absurd because at the end of the day it's a simple coin flip exercise with a clear answer, it doesn't matter which envelope you pick, so why is this even hard? But for defining expectation, outcome, event, etc. and showing formally where the mistake is that's super interesting in terms of understanding probability theory better.
Interestingly (though maybe I missed it?) Wikipedia's page doesn't discuss what if the set of numbers was relatively small and finite, like say 1 xx up to 1 million. That could be simulated as an exercise and could be super useful showing once and for all what's going on.
posted by polymodus at 9:51 PM on October 22, 2021
In comparison, a human being who isn't trying to formalize the example mathematically can just reason using symmetry and say it's absurd because at the end of the day it's a simple coin flip exercise with a clear answer, it doesn't matter which envelope you pick, so why is this even hard? But for defining expectation, outcome, event, etc. and showing formally where the mistake is that's super interesting in terms of understanding probability theory better.
Interestingly (though maybe I missed it?) Wikipedia's page doesn't discuss what if the set of numbers was relatively small and finite, like say 1 xx up to 1 million. That could be simulated as an exercise and could be super useful showing once and for all what's going on.
posted by polymodus at 9:51 PM on October 22, 2021
Oh, just found this (short!) article on stanford.plato entitled, "Tacitly Infinite Decision Problems: Two Envelopes".
posted by polymodus at 10:05 PM on October 22, 2021 [1 favorite]
posted by polymodus at 10:05 PM on October 22, 2021 [1 favorite]
at the end of the day it's a simple coin flip exercise with a clear answer, it doesn't matter which envelope you pick, so why is this even hard?
Quite so, and that is clearly the only applicable decision procedure. The interesting part of this exercise, as with any paradox, is not the actual making of the decision but the identification of what is wrong with a chain of reasoning that yields a demonstrably incorrect strategy - especially after having been softened up by stuff like Monty Hall, which superficially appears to be using a similar kind of somewhat convoluted reasoning but does lead to a correct though intuitively implausible strategy.
Paradoxes are about the ways minds work, not envelopes and doors and goats.
posted by flabdablet at 1:43 AM on October 23, 2021 [5 favorites]
Quite so, and that is clearly the only applicable decision procedure. The interesting part of this exercise, as with any paradox, is not the actual making of the decision but the identification of what is wrong with a chain of reasoning that yields a demonstrably incorrect strategy - especially after having been softened up by stuff like Monty Hall, which superficially appears to be using a similar kind of somewhat convoluted reasoning but does lead to a correct though intuitively implausible strategy.
Paradoxes are about the ways minds work, not envelopes and doors and goats.
posted by flabdablet at 1:43 AM on October 23, 2021 [5 favorites]
One thing that this doesn't seem to cover is why the conditional expected value of profit by switching seems to always be positive (or always negative) if you inspect one of the envelopes. That's weird, and weird in a way that's different from the unconditional expected value not being defined. Inspecting your envelope vs inspecting the other envelope should not be able to alter the expected profit from switching!
posted by BungaDunga at 10:03 AM on October 23, 2021 [1 favorite]
posted by BungaDunga at 10:03 AM on October 23, 2021 [1 favorite]
Wikipedia's page doesn't discuss what if the set of numbers was relatively small and finite, like say 1 xx up to 1 million
Since percentages are easier to think about, let's say you're given that the numbers are both between 1 and 100 (inclusive) and that one is 10x the other and the distribution is uniform.
You don't know what's in your envelope, but these are the cases:
If your number is 9 or less, switching would win... 9% - switch
If your number is a multiple of ten greater than ten, you hold... 9% - hold
If your number is 10 it could go either way, equal chance... 1% - doesn't matter
If your number is greater than 10 but not a multiple of ten, well that can't happen since it isn't 10x anything and 10x your number is too big.
So there are 9 ways where switching wins, 9 ways holding wins and one way it's a toss-up.
But this doesn't take into account maximizing your gain. So among the equally probable possible values, what is the expected takings?
If you hold, you just get one of the 19 possible values in the envelope, considered in order:
hold: (1+2+...+10 + 20+30+...+100)/19 = (55 + 450)/19 = 505/19 ~= 26.58
If you switch, you get (note corresponding positions):
switch: (10+20+...+100 + 2+3+...+9 + 1), which is just the same numbers rearranged, or 505/19
So either way you stand to make 505/19, approx 26.58.
A way to look at this is that if your draw is very low, the winnings are higher but there are fewer ways for the number to be low. Similarly, if your draw is very high, your losings are high but there are also fewer ways of having a very high number and in the end it's all a wash.
posted by sjswitzer at 11:04 AM on October 23, 2021
Since percentages are easier to think about, let's say you're given that the numbers are both between 1 and 100 (inclusive) and that one is 10x the other and the distribution is uniform.
You don't know what's in your envelope, but these are the cases:
If your number is 9 or less, switching would win... 9% - switch
If your number is a multiple of ten greater than ten, you hold... 9% - hold
If your number is 10 it could go either way, equal chance... 1% - doesn't matter
If your number is greater than 10 but not a multiple of ten, well that can't happen since it isn't 10x anything and 10x your number is too big.
So there are 9 ways where switching wins, 9 ways holding wins and one way it's a toss-up.
But this doesn't take into account maximizing your gain. So among the equally probable possible values, what is the expected takings?
If you hold, you just get one of the 19 possible values in the envelope, considered in order:
hold: (1+2+...+10 + 20+30+...+100)/19 = (55 + 450)/19 = 505/19 ~= 26.58
If you switch, you get (note corresponding positions):
switch: (10+20+...+100 + 2+3+...+9 + 1), which is just the same numbers rearranged, or 505/19
So either way you stand to make 505/19, approx 26.58.
A way to look at this is that if your draw is very low, the winnings are higher but there are fewer ways for the number to be low. Similarly, if your draw is very high, your losings are high but there are also fewer ways of having a very high number and in the end it's all a wash.
posted by sjswitzer at 11:04 AM on October 23, 2021
(oops, there are a few errors in the above but instead of fixing them I'll just leave it as an exercise :) )
posted by sjswitzer at 12:28 PM on October 23, 2021
posted by sjswitzer at 12:28 PM on October 23, 2021
Just wanting to clear up some misunderstandings, here - the fact that the possible values have an unbounded range doesn't mean that the expected value of a given envelope is necessarily undefined, and certainly doesn't mean that you should expect to see an infinite amount of money in the envelope. As an easy way to see this, consider the procedure of "I started with $1 and flipped a coin until I got tails, adding $1 to the envelope for each heads I got": I think you can believe both that this process has a nonzero probability of resulting in any given positive integer, and that you'd expect on average to see a fairly small amount of money in the envelope; you wouldn't accuse me of lying or failing to generate a properly random variable if the envelope contained less than $20, for example. The distribution used in the video is exactly that of "I wrote $1 on a check, added a 0 to the end of it for each heads I flipped, then wrote another check for ten times that much and put it in the other envelope." The paradox arises from such a seemingly straightforward situation running into a weird corner case of statistical math behind the scenes.
posted by NMcCoy at 2:59 PM on October 23, 2021 [1 favorite]
posted by NMcCoy at 2:59 PM on October 23, 2021 [1 favorite]
It's a bit akin to the old classic "here's an algebraic proof that 1=2" that involves an obfuscated division by zero somewhere in the process, only most people know that you can't divide by zero, whereas in this case the division-by-zero equivalent is both harder to spot and not as well known of an undefined operation in the first place.
posted by NMcCoy at 3:06 PM on October 23, 2021 [3 favorites]
posted by NMcCoy at 3:06 PM on October 23, 2021 [3 favorites]
Oh, yeah, I'm fond of this "proof" of 1=0:
0=0+0+0+… =(1−1)+(1−1)+(1−1)+… =1+(−1+1)+(−1+1)… =1posted by BungaDunga at 9:19 AM on October 28, 2021 [1 favorite]
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posted by polymodus at 10:30 PM on October 21, 2021 [3 favorites]