Re: The Double or One Half Paradox
From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 07/04/04
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Date: Sun, 04 Jul 2004 05:54:50 -0500
On Sun, 04 Jul 2004 05:28:53 -0500, David C. Ullrich
<ullrich@math.okstate.edu> wrote:
>On 3 Jul 2004 19:24:40 -0700, chvol@aol.com (Charlie-Boo) wrote:
>
>>There are two boxes on a table, one of which contains twice as much
>>money as the other. You are allowed to take one. You do so, but
>>before you open it you are allowed to switch boxes. Should you
>>switch?
>>
>>No: The choice was random. The other box is equally random.
>
>That's correct.
>
>>Yes: If the box you have now contains X then the other box contains
>>either X/2 or X*2, for an expected value of 1 1/4 * X, which is > X.
>>
>>The solutions that I have seen either say that the answer is No and
>>don't refute the logic for Yes, or say that the question is
>>meaningless because an unbounded random number with even distribution
>>has no average value or can't exist.
>
>I don't think the question is meaningless for this reason,
>because there's no reason we need to be talking about
>unbounded random variables; we may as well assume that
>one box contains one dollar and the other contains two,
>without changing the problem.
>
>The calculation that seems to show that you should
>switch is wrong, because it ignores the fact that X is
>a _random variable_. It is true that there are two
>things that can happen if you switch:
>
>(i) You double your money, from X to 2X.
>(ii) You halve your money, from X to X/2.
>
>But it's not true that the payoff in case (i)
>is four times the payoff in case (ii), because
>the value of X is different in the two cases!
>
>(To be technical: The calculation would be correct
>if the value of X was independent of the event (i).
>But the value of X is _not_ independent of the
>event (i).)
>
>In fact there are two equally likely possibilities
>if you switch:
>
>(i) You double your money, from 1 to 2.
>(ii) You halve your money, from 2 to 1.
>
>So your expected value is 3/2 if you switch,
>the same as if you don't switch.
PS. I see someone else has posted a link to an analysis that
seems much more complicated. In case anyone objects to
assuming that the amounts are 1 and 2, I could say that
that situation is equally paradoxical and simpler to
analyze, or I could say this:
Suppose that the boxes contain A and 2A, where A has
whatever distribution you like. Then the two
equally likely possibilities if you switch are these:
(i) You double your money, from A to 2A.
(ii) You halve your money, from 2A to A.
And so if you switch your payoff is 3/2 E[A],
same as if you don't switch.
>> I used to believe the latter
>>until I realized the truth. (I believe that Smullyan thinks that it's
>>a genuine paradox, but of course that is so like him, isn't it?)
>>
>>I will allow other to try first.
>>
>>Charlie Volkstorf
>>Cambridge, MA
>
>
>************************
>
>David C. Ullrich
************************
David C. Ullrich
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