In
an earlier puzzle, you were handed two envelopes, one of which contained twice as much money as the other. After opening one, you were given the chance to swap. At first glance, it appeared that the your chance of getting more money could only increase each time the envelopes were swapped, but clearly this was nonsense: since there is no probability distribution which allows all real numbers to have the same probability, some values would have to have been more likely than others.
Suppose instead envelopes contain the nonnegative integer sums 2^{n} and 2^{n+1} with probability q(1 − q)^{n} for some fixed q < 1/2
Now of course if the envelope you open contains a 1, you know the other must contain 2, so you ought to swap.
But you can do even better than this. Suppose you open an envelope and find an amount of money 2^{k}
What would the expected value of the second envelope be?
Does this lead to the same paradox?
I just started to look at it from a different point of view. Knowing that the first envelope is some value, 2x, we can see that in order to choose the second envelope, we would end up with x if we lost, or 4x if we won. So, if we keep the original envelope every time, we get:
1.0(2x) = 2x
If we always choose the second envelope, and the odds of the envelope being larger or smaller are equal, we would, after infinite tries, get:
.5(x) + .5(4x) = 2.5x
Under this theory, we would be far better off choosing the second envelope.
Another way of looking at this is that you are essentially wagering half of the first envelope, x, to either lose your bet, the smaller outcome, or to earn 3x, totalling 4x, the larger outcome. So, while the odds of winning the bet is 11, the payoff is 21. Las Vegas would certainly not want to go for this kind of game ;).
We are only working with the understanding that the first envelope is an even amount, 2x. If it were odd, we would obviously want to choose the second envelope. Also, we have literally nothing to lose, so by choosing another envelope, we still make out in the end. The only way to make this game fair would be to make it a "doubleornothing" choice. Then the expectation would be:
1.0(2x) = 2x
.5(0) + .5(4x) = 2x
Then again, we would want to keep the first envelope in order to guarantee a positive outcome.

Posted by Ken
on 20050926 21:56:42 