| Re: <<< Note 1591.0 by HANNAH::OSMAN "see HANNAH::IGLOO$:[OSMAN]ERIC.VT240" >>>
One envelope has X dollars and the other 2*X. Select one with
probability .5. Expectation (given X) is 1.5*X, but we don't
know what X is. Let u(x) be the probility that X=x. Then the
(unconditional) expectation is 1.5*E[X] = 1.5*Sum[x*u(x)] IFF X
has finite expectation. But, if X has finite expectation, then
you can calculate (or approximate) the probability that the
other envelope has $50 or $200. Application of Bayesian theory
will resolve the paradox. The paradox only exists if the
expectation of X is indeterminate. And that only happens in
theoretical cases. You can always find a reason to limit the
expectation. (For example, if I know your weekly salary, I can
be quite sure that you will not offer me 150% of your weekly
salary, just to bother me with this kind of puzzle).
By the way, the only logical conclusion is that both envelopes
have $0 in them, since you are offering either of them to me for
free. Definitely not a win-win proposition, but if you want to
do it, I'm game. :-)
Dick
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| The problem here arises from an implicit assumption that the
distribution of the population from which the envelopes are sampled is
an infinitely wide uniform distribution -- which is not a "proper"
distribution. If you make the population distribution a proper
distribution (which is essentially the same thing as placing a
reasonable prior joint distribution on the contents of the envelopes in
a Bayesian approach) you will find, I think, that the paradox
disappears.
The Bayesian approach doesn't really fix the problem -- it simply makes
what was implicit explicit so that the root of the problem becomes
obvious.
Topher
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| The envelopes contain $x & $2x, with probability p(x). Let's assume
that only whole dollar +ve amounts are possible. When I open an envelope, and
find $y inside, Bayes tells me that the other envelope contains:
$�y with probability p(�y)/[p(�y)+p(y)]
$2y with probability p(y)/[p(�y)+p(y)]
If I keep the envelope I've peeked, my expected win is $y.
If I shift envelopes, the expected win is:
$y * [p(�y)+4p(y)]/[2p(�y)+2p(y)]
I expect to gain by shifting envelopes iff
2p(y) > p(�y).
Example (1): if y is odd, then I will certainly switch envelopes :-)
Example (2): with y = z*2^a where z is odd, suppose that:
p(y) = p(z)/�^a, where � > 1, sum(z odd) p(z) = 1-1/�, and
sum (z odd) z*p(z) < infy]
Here, if � > 2, then the winning algorithm is clear:
"Keep the envelope iff it's even".
[If � =< 2, then the winning algorithm appears to be:
"Always go for the second envelope."
This seems to be a bit peculiar, but expected value of x is:
sum (y) y*p(y)
= sum (z odd) (sum (a=0...) z*2^a p(z) /�^a)
= sum (z odd) (z*p(z)) * sum (a=0...) (2/�)^a
which is not finite if � =< 2. So this is not a reasonable case.]
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