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Topic: HARD: ENVELOPE GAMBLE (Read 9876 times) |
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S. Owen
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HARD: ENVELOPE GAMBLE
« on: Jul 28th, 2002, 7:38pm » |
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There is a flaw in the analysis presented in the problem. After you pick, there is a 0.5 chance that you picked the envelope with X and that switching gets you 2X. There is a 0.5 chance that you picked the envelope with 2X though (*not* X), and switching leaves you with X.
It's not valid to use the same X for both cases; that's the catch.
Your expected gain/loss from switching is:
0.5(2X-X) + 0.5(X-2X) = 0
...which is what you'd expect intuitively! So there is no expected gain for switching, and no loss. You can switch if you want.
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| « Last Edit: Jul 28th, 2002, 7:40pm by S. Owen » |
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Alex Harris
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Re: HARD: ENVELOPE GAMBLE
« Reply #1 on: Jul 29th, 2002, 2:24pm » |
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Actually I think its a bit more fundamental than that. The reasoning stated in the problem is just as valid as the reasoning of using the same X for both computations as you did. The problem is that we're implicitly drawing uniform random numbers from an unbounded distribution. The "payoff" for this game doesn't converge absolutely. By choosing different ways of analyzing it we're arranging the infinte sum in different ways and thus its converging to different answers, neither of which is really valid. Been a long while since statistics but I think thats right =P
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #2 on: Jul 29th, 2002, 2:41pm » |
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How does drawing random numbers comes into this... the 2X sum is equally likely to be in either envelope, that's about it.
What's indefinite about the analysis? The expected value of switching is definitely 0.5X + 0.5(-X) = 0, not 0.25X. The payoff is definitely 1.5X whether you switch or not.
The riddle is just about finding the flaw in the reasoning given. Are you saying that you can analyze this in a way that shows the expected value of switching is not 0?
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #3 on: Jul 29th, 2002, 10:35pm » |
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If X is drawn uniformly from the real numbers, or the positive powers of 2, or some other unbounded infinite set then the analysis listed in the problem is no more faulty than the analysis you use in your solution determing a payoff of 0. You've come up with alternative reasoning which gets a different answer, but you didn't actually justify why the problems reasoning is wrong other than the fact that it doesn't get the same answer as yours does.
You're attempting to define the payoff in terms of X, but the distribution of X is very naughty and is not amenable to typical statistical computations. For example, what is the mean value of X? Infinite/undefined. So when you say something like the payoff from switching is .5X - .5X = 0 what you're effectively saying is infinity - infinity = 0 but in truth infinty - infinity is not well-defined here.
If we put in a well behaved distribution from which to draw X (say a uniform random number from 1 -100), then the analysis stated in the problem breaks down because for a given X it is not equally likely that X/2 and 2X are in the other envelope and we do indeed get the expected payoff from switching equal to 0.
In short, the reasoning stated in the problem is "wrong" in the sense that it assumes something about the distribution of X (i.e. that given one envelope with some particular x in it, the odds of the other envelop having x/2 or 2x are even). This assumption, if true, causes the payoff of the game to not well-defined.
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S. Owen
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Value of X is immaterial
« Reply #4 on: Jul 30th, 2002, 5:56am » |
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The value of X is some constant and doesn't matter - for simplicity say that the envelopes contain $1 and $2. There is only one trial here, so there is no issue of the "distribution of X."
The statement that one has X and the other has 2X is simply a stated fact of the scenario. Nobody is randomly selecting values according to some distribution to go into these envelopes. Right?
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #5 on: Jul 30th, 2002, 9:49am » |
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Ok. For some reason I was solving a generalization of the problem  . Once we define a fixed pair (x,2x) then we can analyze just as you say, and the problem's logic breaks as I described: whether the opposite envelope is double your envelope is completely dependent on whats in your envelope. If we play the generalized game where we worry about how the game-setter chosses x, then he could be drawing it from a distribution that makes the reasoning of the problem correct, but if so he has to draw from a ugly distribution that makes the payoff outcome undefined (and is also rather hard to draw numbers from).
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #6 on: Jul 30th, 2002, 10:40am » |
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on Jul 30th, 2002, 9:49am, AlexH wrote:| If we play the generalized game where we worry about how the game-setter chosses x, then he could be drawing it from a distribution that makes the reasoning of the problem correct, but if so he has to draw from a ugly distribution that makes the payoff outcome undefined (and is also rather hard to draw numbers from). |
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Wouldn't the generalization be that we play a bunch of times, and each time, X is chosen according to some distribution, and he puts X and 2X into envelopes?
Instead of: X and Y are chosen from some distribution in such a way that one is twice the other - is that the right interpretation of what you're saying? I don't know how you would do that.
Regardless, either way, in this generalization, the expected value of switching envelopes remains 0. I don't see how it can be shown to be anything else.
The expected value of the game may be problematic, but the value of switching is not.
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #7 on: Jul 30th, 2002, 11:27am » |
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I'm fine with your method of drawing some X from a distribution and letting the second envelope be 2X. Lets let that distribution have the property that for any particular x the probabilities of drawing x/2, x, and 2x are all equal (uniform on the positive real numbers is one such distribution).
If we're drawing numbers from this distribution, then what do you claim is wrong with the reasoning in the problem? You can come up with alternative, equally valid, reasoning that says that the answer should be different, but I'm asking you to point to where the original reasoning goes wrong.
The reason two "correct" (i.e. correct except for sweeping the infinities under the rug) reasonings yield different answers is that the answer just isn't well defined.
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #8 on: Jul 30th, 2002, 11:48am » |
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on Jul 30th, 2002, 11:27am, AlexH wrote:
If we're drawing numbers from this distribution, then what do you claim is wrong with the reasoning in the problem? You can come up with alternative, equally valid, reasoning that says that the answer should be different, but I'm asking you to point to where the original reasoning goes wrong.
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Sorry, I still don't see what is not well-defined about any of this.
The original problem "computes" the expected value of switching as follows:
"You think to yourself that if your envelope has x dollars there is a 50% chance the other one has x/2 dollars and a 50% chance it has 2x dollars. The expected return, you compute, is .5[.5x + 2x]=1.25x which seems like a favorable gamble."
I claim that this is not valid, and should proceed like this:
"You think to yourself that there is a 50% chance that your envelope has x dollars and the other one has 2x dollars, and a 50% chance that your envelope has 2x dollars and the other one has x dollars. The expected return, you compute, is .5[x + 2x]=1.5x which is the same as the expected value without switching (1.5x)."
Or to be more succint: "50% of the time, switching loses me X dollars. 50% of the time it gains me X dollars. The expected value of switching is .5(-x) + .5(x) = 0."
The best I can do to explain intuitively what is wrong with the problem's analysis is to say that it is subtly comparing apples to oranges... meaning that yes, there is a 50% chance that the other envelope has twice as much (2x) as yours (x), and a 50% chance that it has half as much, but half of twice as much as we were talking about in the first case (2x)!
Now, I don't believe you are arguing that the expected value of switching is *not* 0. That is definitely wrong. But the problem is clearly concluding otherwise, that the expected value is positive. So, let me ask, what is right about the problem's analysis then!
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| « Last Edit: Jul 30th, 2002, 11:49am by S. Owen » |
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #9 on: Jul 30th, 2002, 12:20pm » |
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Remember that we've chosen this special distribution. Let's look at the distribution conditional on knowing that our envelope contains the value X.
This reduces us to 2 possiblities, either the pair of numbers is X, 2X and we drew low, or the pair is X/2, X and we drew high. The pair X/2, X has the same probability of occuring as does the pair X, 2X (according to the rules of our distribution) and the fact that we got X tells us nothing about which case we're in because in both cases we have a 50/50 chance of selecting X as our envelope. So the case X,2X has the same odds as the case X,X/2. As in the problem statement this gives us expected payoff of X/4.
If you can't point to the flaw in the reasoning that gets the X/4 when you "know" that its really 0 that should indicate that something weird is going on.
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #10 on: Jul 30th, 2002, 1:02pm » |
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on Jul 30th, 2002, 12:20pm, AlexH wrote:
This reduces us to 2 possiblities, either the pair of numbers is X, 2X and we drew low, or the pair is X/2, X and we drew high. |
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This is the heart of the problem with the reasoning: the possibilites are X,2X and 2X,X - not X,2X and X/2,X.
The game you are analyzing (and the problem is analyzing) would go like this:
You pick one of two envelopes. It has X dollars in it. I put, with equal probability, either 2X or X/2 dollars in the other. Should you switch?
The expected value of switching in *this* game is indeed 0.25X.
Our game is:
You pick one of two envelopes. There is a 50% chance it has X dollars in it (in which case there is a 100% chance that the other has 2X dollars). There is a 50% chance that it has 2X dollars in it (in which case there is a 100% chance that the other has X dollars). Should you switch?
These are crucially different situations... in our game, once you pick an envelope, there is in fact not a 50/50 chance that the other has half or double. In fact, the contents of the other is determined; it's either 2X or X, depending on whether you got X or 2X the first time.
on Jul 30th, 2002, 12:20pm, AlexH wrote:
If you can't point to the flaw in the reasoning that gets the X/4 when you "know" that its really 0 that should indicate that something weird is going on.
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Well, I think I've pointed out the flaw as clearly as is possible... not sure how much more I can say.
I understand that it's not enough to say "the argument is wrong because it doesn't produce the answer that I think is right," but in this case one can prove unequivocally that the expected outcome of switching is 0. All sorts of contradictions follow if it is not... it's like the proof that 2=1 problem.
So again, do you agree or disagree that the expected value of switching is 0? As in, should we get into this, or are you playing devil's advocate?
Any body else have an opinion?
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #11 on: Jul 30th, 2002, 2:27pm » |
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Ok one last try. I'm not just playing devils advocate, I really do believe what I'm saying, but I am definitely not saying that 1/4X is ever the correct answer for the payoff.
Let A be the envelope you have and B be the other envelope. We chose some value Y from our distribution which has the property that for any value Z, p(Y=Z/2) = p(Y=Z) and we put the values Y and 2Y randomly into the envelopes A and B. If our distribution is well-behaved (in particular I want the probability densities to be well defined) then everything I'm about to do is legitimate to the best of my knowledge.
I'll label the number in envelope A to be X. X is either Y or 2Y but we haven't determined which. For any particular value V:
p(A=V,B=V/2) = .5 * p(Y=V/2)
p(A=V,B=2V) = .5 * p(Y=V)
However our choice of distribution tells us that
p(Y=V) = p(Y=V/2)
so:
p(A=V,B=V/2) = (A=V,B=2V).
This is true for any V, so lets pick V=X.
p(B= 2X|A=X) = p(A=X,B=2X) / p(A=X)
= p(A=X,B=X/2) / p(A=X)
= p(B=X/2|A=X)
Since p(B=2X|A=X) = p(B=X/2|A=X) and we know A=X we now have:
p(B=2X) = p(B=X/2) = .5
The ONLY thing that I'm aware of that makes this reasoning fail is the fact that our distribution is not well behaved enough to apply probability densities. If there existed a well behaved distribution which still had the property that every pair of envelopes x/2,x was as likely as the pair x,2x then this reasoning would be perfectly correct in that domain and the payoff really would be X/4. Of course there is no such distribution and this is why the problem's logic must fail.
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #12 on: Jul 30th, 2002, 3:16pm » |
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Agreed, once more!
on Jul 30th, 2002, 2:27pm, AlexH wrote:
I'll label the number in envelope A to be X. X is either Y or 2Y but we haven't determined which. For any particular value V:
p(A=V,B=V/2) = .5 * p(Y=V/2)
p(A=V,B=2V) = .5 * p(Y=V)
However our choice of distribution tells us that
p(Y=V) = p(Y=V/2)
so:
p(A=V,B=V/2) = (A=V,B=2V). |
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Agreed, I believe. You are saying that if the gamer is picking values randomly, values of $0.60 and $0.30 are just as likely as $0.60 and $1.20, and $0.40/$0.20 is as likely as $0.40/$0.80, etc., in fact any $X/$2X is equally likely.
But crucially, one set of values is chosen and fixed when the problem starts. So none of that matters. Who cares what A=X and B=2X are, or how they were chosen - they are fixed values now.
p(A=X,B=X/2) = (A=X,B=2X) only when A and B are random variables of the sort you describe. They are not - A and B are fixed and must be treated this way.
To illustrate I am going to say $1 for X, since X is as fixed as $1. What we really have is stuff like:
p(A=$2,B=$1) = 0.5
p(A=$1,B=$2) = 0.5
p(A=$2|B=$1) = 1
p(A=$1|B=$2) = 1
p(A=$0.50|B=$1) = 0
It is not true that p(A=X,B=X/2) = (A=X,B=2X) when A and B are fixed.
on Jul 30th, 2002, 2:27pm, AlexH wrote:
The ONLY thing that I'm aware of that makes this reasoning fail is the fact that our distribution is not well behaved enough to apply probability densities. If there existed a well behaved distribution which still had the property that every pair of envelopes x/2,x was as likely as the pair x,2x then this reasoning would be perfectly correct in that domain and the payoff really would be X/4. Of course there is no such distribution and this is why the problem's logic must fail.
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Yeah, I can agree that that's a way to say it... if given envelopes with $1 and $2, envelopes containing $1/$2 was somehow as probable as envelopes containing $1/$0.50, then yeah, I'm sure this and a lot of other false conclusions follow!
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #13 on: Jul 30th, 2002, 5:17pm » |
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on Jul 30th, 2002, 3:16pm, srowen wrote:Agreed, once more!
But crucially, one set of values is chosen and fixed when the problem starts. So none of that matters. Who cares what A=X and B=2X are, or how they were chosen - they are fixed values now.
p(A=X,B=X/2) = (A=X,B=2X) only when A and B are random variables of the sort you describe. They are not - A and B are fixed and must be treated this way.
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We're getting closer. 
A and B are not "fixed" when we're trying to compute the expected payoff from the switch strategy in the generalized game. There is nothing wrong with treating them as random variables and in fact your approach is also treating them as random variables, but you're conditioning on a certain Y,2Y pair of numbers while I'm conditioning on a certain value of A=X. Either of these approaches is valid, or would be on a well behaved distribution.
To put it another way, your "A and B are fixed" approach is breaking down the set of all possible occurances of the game into the subsets which have the property that A and B are drawn from a particular pair Y,2Y. The alternative of conditioning on A=X gives us the subsets where A happens to be X. As long as we both account for every case it won't matter how we choose to pair them up. Once again the caveat here: if the distribution is sufficiently ugly then there is essentially no answer to the questions we're asking because we can get lots of different answers.
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #14 on: Jul 30th, 2002, 5:51pm » |
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Whats going on here is actually very close to the following.
Consider the numbers
1, -1, 1/2, -1/2, 1/3, -1/3, 1/4, -1/4, .....
What is the sum of this series?
If you choose to group it the obvious way it looks like the sum goes to 0. But I could group it as
1, -1, 1/2, 1/3, -1/2, 1/4, 1/5, 1/6, -1/3, ...
I'd still be hitting each number exactly once and this sum would head to infinity.
While the =0 way of grouping them is certainly asthetically appealing, but its not actually any more correct than the diverge to infnity way. The sum of the series is just not well defined (it doesn't converge absolutely).
Any distribution satisfying p(Y) = p(2Y) has this kind of messiness in spades and its just not sensible to talk about the payoff or expectations from such a game.
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #15 on: Jul 30th, 2002, 7:33pm » |
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I think I get the gist of your argument, that if you start the other way and assume the analysis in the problem is correct, you find that the scenario it must be supposing cannot be reconciled with the riddle. Eh?
Mostly I think you are imagining harder problems here more worthy of your skills. If I'm misunderstanding then feel free to hit me one more time... I'll lay off and wait for a third unsuspecting person to weigh in.
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #16 on: Jul 30th, 2002, 7:47pm » |
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Thats about correct. The problem's logic makes certain assumptions about the distribution which if true force the distribution to be so wacky that things like payoff aren't defined.
You're completely right that this only applies to the generalization of the problem. In a single concrete case of 2 envelopes and 2 numbers then the "distribution" is simply a 100 percent chance of some particular pair Y,2Y and that certainly doesn't have the p(V) = p(2V) property which I needed to make the logic sound.
Enough time spent here ... on to the other puzzles!  Well, maybe to sleep first but there is always tomorrow.
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JimP
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Re: HARD: ENVELOPE GAMBLE
« Reply #17 on: Aug 2nd, 2002, 1:03pm » |
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I'm intrigued as to why you guys think a probability distribution is applicable only to a repeated scenario. Are you saying that multiple single gambles should yield a different result to one multiple gamble?
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AlexH
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Re: HARD: ENVELOPE GAMBLE
« Reply #18 on: Aug 2nd, 2002, 4:02pm » |
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If you consider the numbers as coming from a possibly ugly distribution then you can make things like expectation value undefined. The key thing isn't that you're doing multiple runs but that you're considering the distribution of the numbers. Whenever its well defined the expectation is 0 but you could pick a distribution for which its not well defined. The payoff of a single run if you consider a bad distribution can be undefined. Really I only brought the issue up because I misread the problem as being this more general case rather than the concrete case of 2 fixed numbers in the envelopes and then srowen and I got talking about it.
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James Fingas
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Re: HARD: ENVELOPE GAMBLE
« Reply #19 on: Dec 20th, 2002, 10:02am » |
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S.Owen,
Many months after reading your analysis, I have finally come to the conclusion that I don't agree.
If you've got X in your envelope, and it's 50/50 that the other envelope has X/2 or 2X, then you'd better switch! Your expected gain is very easy to compute, and the problem statement computed it correctly.
Let's suppose I were to try this trick a number of times, just to convince you. Each time, I would give you an envelope that contained X dollars. Each time, there'd be a 50/50 chance of the other envelope containing 2X or X/2 dollars. You can switch as many times as you want. Believe me, you will do better switching!
The exact problem with your argument is where you say:
0.5(2X-X) + 0.5(X-2X) = 0
This equation is correct, but doesn't apply to the situation at hand. In the first term, 0.5(2X-X), X is the amount in your envelope, and 2X is the amount in the other. In the second term, 0.5(X-2X), you re-definite X so that 2X is the amount in your envelope. You're using two different definitions of X in the same equation.
Now let me illustrate why Alex is bang on with the distribution argument. You know that when the host of the show picks the two amounts of cash, he is not picking uniformly on zero to infinity. You may say "that doesn't matter", but in fact it does. Think about it--the larger the amount you're given, the less likely it is that the other envelope has twice that much. That is exactly why the expectation is zero.
If you get $50, then the other envelope probably has $100. If you get $50 000, then the other envelope could have $100 000. But if you're offered $1 000 000, then it's very unlikely that the other envelope has $2 000 000, right? If you got $1 000 000 000 000, then the other envelope probably doesn't have $2 000 000 000 000. And these numbers are very small on the scale of zero to infinity! This is where the distribution the numbers are chosen from is really important in practical terms.
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #20 on: Dec 20th, 2002, 2:20pm » |
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on Dec 20th, 2002, 10:02am, James Fingas wrote:
Let's suppose I were to try this trick a number of times, just to convince you. Each time, I would give you an envelope that contained X dollars. Each time, there'd be a 50/50 chance of the other envelope containing 2X or X/2 dollars. You can switch as many times as you want. Believe me, you will do better switching!
The exact problem with your argument is where you say:
0.5(2X-X) + 0.5(X-2X) = 0
This equation is correct, but doesn't apply to the situation at hand. In the first term, 0.5(2X-X), X is the amount in your envelope, and 2X is the amount in the other. In the second term, 0.5(X-2X), you re-definite X so that 2X is the amount in your envelope. You're using two different definitions of X in the same equation.
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I believe you are saying that the riddle is equivalent to this:
I give you an envelope with X dollars. I then show you another envelope, which is equally likely to have 2X or X/2 dollars. Do you switch?
Obviously you would switch in this case, and the point of the riddle is that it's trying to convince you that this is the same as:
I let you choose one of two envelopes - one has X dollars, the other 2X dollars. Then I ask if you'd like to switch to the other because of blah blah blah. Do you switch?
It's not though, of course - in the first scenario, switching clearly has a positive expected payoff, while in the second, it clearly has a 0 payoff.
I do see what you mean about distribution - generally, smaller values are more likely than larger values. So, on the whole, switching is probably slightly more likely to lose you money, right? I agree with this, given the reasonable assumption about the distribution of amounts in the envelopes. But the riddle is trying to convince you that switching is always profitable! I don't think you need any such assumption to show that *that* is not the case.
In any event I don't think that is the distribution argument that AlexH was making, though I confess that his argument is too subtle for my brain.
Rather than make the obvious case about why the riddle's argument is wrong, he advances a subtler argument, which probably makes fewer assumptions:
If the problem's reasoning is correct, and switching, astonishingly, always has a positive payoff, then there must be something impossible going on with the distribution of values in the envelopes... there is some contradiction waiting there.
This probably works too but it's heady stuff!
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fenomas
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Re: HARD: ENVELOPE GAMBLE
« Reply #21 on: Dec 21st, 2002, 12:44am » |
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on Dec 20th, 2002, 10:02am, James Fingas wrote:S.Owen,
If you've got X in your envelope, and it's 50/50 that the other envelope has X/2 or 2X, then you'd better switch! Your expected gain is very easy to compute, and the problem statement computed it correctly.
Let's suppose I were to try this trick a number of times, just to convince you. Each time, I would give you an envelope that contained X dollars. Each time, there'd be a 50/50 chance of the other envelope containing 2X or X/2 dollars. You can switch as many times as you want. Believe me, you will do better switching!
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James, with how similar this is to our dilemma in the littlewood thread, I think I can help show how you are off here, without involving distributions. In your version, you're given an envelope with X dollars, and you can switch to another than contains either 2X or X/2 dollars. However, this is very different from the actual situation, because you've been changing the definition of "X" in the middle of the problem.
The proper way to think of the problem is like this:
You are offered two envelopes, one containing X dollars, and the other containing 2X dollars. You choose one of them. Set Y = the value of the envelope you chose. Now, you have the option of switching to the other envelope, which contains either Y/2 or 2Y dollars (not X/2 or 2X!!). But if you switch, you don't get Y/2 or 2Y with equal odds- the question of which you get is entirely determined by which envelope you chose before. If you chose the lesser envelope (Y=X), then switching gets you 2Y, or 2X dollars. If you chose the higher (Y=2X), switching now gets you Y/2, or X dollars. So before you chose, you had an expectation of 1/2(X+2X) dollars, and switching gets you an expectation of 1/2(X+2X) dollars.
Hopefully, this makes your fallacy clear. another way to say it is: You can't wind up with X/2 dollars because there was never an envelope containing X/2 dollars to begin with!
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fenomas
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Re: HARD: ENVELOPE GAMBLE
« Reply #22 on: Dec 22nd, 2002, 7:52pm » |
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on Dec 21st, 2002, 12:44am, fenomas wrote:
a bunch of stuff about redefining X
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Hoo-ah, just noticed that my previous post just described in more detail what S.Owen said in the very first post!
Oh well, it's still right. 
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Kozo Morimoto
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Re: HARD: ENVELOPE GAMBLE
« Reply #23 on: Dec 23rd, 2002, 7:59pm » |
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Look at:
http://www.wilmott.com/310/today_detail.cfm?articleID=107
Not sure if you need to register, but here is the transcript:
I. Money, money, money
Two different amounts of money are placed into envelopes. One envelope is selected at random and given to you. The other envelope is given to Paul. Neither you nor Paul know the amounts. Paul offers you a bet, which you may take or leave. The bet is that after opening the envelopes whoever has the larger amount of money gives it to the other, leaving him with nothing.
Call the amounts of money in the envelopes X and Y with X>Y. There is a 50% chance your envelope contains X and a 50% chance it contains Y. If you turn down the bet, your expected winnings are (X + Y)/2 and your standard deviation is (X - Y)/2. If you accept the bet you have the same expectation, (X + Y)/2, but the standard deviation is now (X + Y)/2. [You should check these for yourself!] You turn down the bet because it leaves your expected value unchanged and increases your standard deviation.
Q: Is this the correct decision under normal economic utility theory assumptions?
Yes, but see below.
II. The envelope please. . .
You open your envelope and find $100. Paul asks you again if you want to bet. Now you reason that you have a 50% chance of winning or losing. If you lose, you lose $100. If you win, you win more than $100. So your expected value is positive and you take the bet.
Q: Is it correct that your expected value is positive?
Yes, but see below.
Q: Is Paul’s expected value also positive for the same bet?
Yes, but see below.
Q: If so, where does the extra value come from?
This question hurls us into the contentious area of philosophy of probability. The first three questions are answered by using simple tools in standard ways. Now we have to think about why they give apparently inconsistent answers.
There are people who call themselves Bayesians (because they use Bayes Rule a lot, not because Nonconformist Reverend Thomas Bayes was one) who demand consistency in all things. They say probability refers to subjective belief. Observation updates our prior beliefs, and any complete probability calculation must specify a prior distribution, what you believed before making the observation. To a Bayesian your failure to update your probability of winning after opening your envelope implies that your prior distribution of amount of money in the envelopes had infinite expected value. You take the bet because however much money is in your envelope, it is less than the (infinite) expected value of the other envelope. There is no extra value, any more than there is anything left over when you match up all integers with all even integers. Infinity works that way.
Bayesians are a minority, and the majority of statisticians feel no need to name themselves. Bayesians call them “frequentists” or “objectivists.” These people do not demand consistency. To them, the additional value arises from an inconsistency in your probability model. All standard statistical methods have these inconsistencies and you don’t worry about them unless you want people to call you a “closet Bayesian.”
Q: And why would you take a bet after you open the envelope, but not before?
Before you open the envelope your probability of winning is clearly 50%. Once you open it, the probability becomes model-dependent. Models do not always lead to paradox, but they always create the possibility of paradox.
Q: Does it matter if Paul sees how much you had in your envelope?
Now we have a second-order model. We have to figure out what it means if Paul takes the bet. We want to know who has to accept first, or if there is some way for both of us to do it at once. We get inconsistencies even without that nonsense.
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fenomas
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Re: HARD: ENVELOPE GAMBLE
« Reply #24 on: Dec 24th, 2002, 1:03am » |
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Wow, thanks for the info, Kozo.
Now what can that tell us about Littlewood's number game?? 
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