Re: The Double or One Half Paradox

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 07/08/04 

Date: Thu, 08 Jul 2004 05:22:07 -0500

On Sun, 4 Jul 2004 23:47:59 -0400, Carl Cotner
<cfc-usenet@tau.aauetiu.net> wrote:

>[...]

I didn't read the following carefully at first, because
I was wrongly convinced that it was irrelevant. Reading
it carefully it appears there's a problem, although
it seems like just a typo:

>[*] Theorem: Suppose f: (0, oo) -> [0, 1] is a strictly decreasing
>function. The expectation for switching envelopes with probability
>f(x) when the first envelope one chooses has x amount of money in it
>exceeds the expectations of never switching, always switching, or
>switching randomly independent of x. Moreover, this does not depend on
>the distribution of x, nor that x follows a distribution at all.
>
>Proof (ala Hein Hundal): There are four possible events:
>
> E1. You randomly pick the lesser envelope and keep it,
> E2. You randomly pick the lesser envelope and switch,
> E3. You randomly pick the greater envelope and keep it, or
> E4. You randomly pick the greater envelope and switch.
>
>If x is the amount of money in the smaller envelope,

We should note, to prevent confusion, that this "x" is
not the same as the x in the statement of the theorem...

>the expectation
>for any given strategy is
>
> P(E1)(x) + P(E2)(2x) + P(E3)(2x) + P(E4)(x)
>
>which, if one never switches, simplifies to
>
> (1/2)x + (1/2)(2)x = (3/2)x
>
>The expectation for the always switching and the switching at random
>cases is the same.
>
>If one switches envelopes with probability f(x),

Switching with probability x is impossible, because we
don't know what the amount in the smaller envelope
is. But looking at the next few lines it appears that
what you meant was "switch with probability
f(the amount in the envelope we've opened", or
"switch with probability f(the x in the statement
of the theorem)".

>these four events
>have probabilities
>
> P(E1) = (1/2)(1 - f(x)),
> P(E2) = (1/2)f(x),
> P(E3) = (1/2)(1 - f(2x)), and
> P(E4) = (1/2)f(2x)
>
>respectively. Thus the expectation for any strategy that switches with
>probability f(x) is
>
> (1/2)(1 - f(x))(x) + (1/2)f(x)(2x) + (1/2)(1 - f(2x))(2x) + (1/2)f(2x)(x)
>
>which reduces to
>
> (3/2)x + (1/2)(f(x) - f(2x))(x)
>
>and this expression is strictly greater than (3/2)x since f is
>strictly decreasing. This condition can be relaxed to f(x) > f(2x) for
>all x (in the domain of f).

************************

David C. Ullrich

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