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Topic: Two letters containing money, but... (Read 1584 times) |
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Eigenray
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Re: Two letters containing money, but...
« Reply #25 on: Aug 20th, 2006, 6:03pm » |
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This has all been said before, so I'll be brief:
For this game to be well-defined, there must be some probability distribution on the values in the envelopes: say the pair {x, 2x} occurs with probability p(x). The expected value of switching, given your envelope contains x, is
x*[2p(x) + 1/2 p(x/2)]/[p(x) + p(x/2)],
while the expected value of staying is of course x.
In order for the "always switch" strategy to maximize the expected value, we need
p(x) > 1/2 p(x/2) for all x.
This is certainly possible, but note that if this happens, then the expected value of the unopened envelope in your hand is already infinite (as is the other envelope, of course). What we have here are two random variables X,Y, with E(Y|X=x) > x, and E(X|Y=y) > y. But this just means
[sum]y yp(x,y) > [sum]y xp(x,y) and
[sum]x xp(x,y) > [sum]x yp(x,y),
and summing the first, say, gives only
E(Y) = [sum]x([sum]y yp(x,y)) > [sum]x([sum]y xp(x,y)) = E(X),
not E(Y) > E(X), which would indeed be a contradiction (but the only way to for equality to hold is if both sides are infinite). Whatever X is, Y is better, and whatever Y is, X is better, but there's no contradiction*.
On the other extreme, for "always stay" to maximize your expected value, we need
p(x) < 1/2 p(x/2) for all x,
but this is impossible: p(x/2n) > 2np(x), and so [sum] p(x) will either be infinite or 0, never 1.
So for any instance of the game, there is always at least one x for which you should switch given x, and if the expected value of the game is finite, there's at least one x for which you should stay given x. But of course you can't know the optimal strategy without some information about p. For example, suppose that for some fixed a>0, 0<r<1,
p(2na) = rn(1-r), n>0.
If r<1/2, you should switch iff x=a (and the expected value is a(1-r)(2-r)/(1-2r)). Otherwise, always switch (and the expected value is infinite).
Of course, this is just to maximize the expected value. Your utility function is quite nonlinear. Consider the following "game":
You flip a coin until either (1) you get tails or (2) you decide to stop. If you got a tail, you get $0, otherwise if you got n heads, you get $3n. Now, the expected value of continuing after n heads is at least 3n+1/2 > 3n, so you should never stop . . . which is obviously a pretty bad strategy.
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*But it is a bit odd. The following is an argument against the continuum hypothesis: Let f : [0,1] -> omega1 be a bijection (omega1 being the smallest uncountable ordinal), and consider the following game. Players X and Y pick real numbers x,y in [0,1] with uniform probability distribution. X wins iff f(x) < f(y), otherwise Y wins. For each x, the set of y for which Y wins has cardinality |f(x)| < omega1, which is countable, hence has Lebesgue measure 0. So no matter what X picks, he wins with probability 1. But no matter what Y picks, Y wins with probability 1 too. (By Fubini's theorem, this must be because {(x,y) : f(x)<f(y)} is not measurable, even though each cross-section is.)
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| « Last Edit: Aug 20th, 2006, 6:13pm by Eigenray » |
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Grimbal
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Re: Two letters containing money, but...
« Reply #26 on: Aug 21st, 2006, 4:08am » |
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It changes only insofar that you can compare that amount to what your uncle is likely to give. It doesn't change just because you know the amount.
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