Newsgroups: rec.puzzles,sci.math
From: jscholes@kalva.demon.co.uk (John Scholes)
Subject: Tilly's Trolling question - proof.
Organization: Kalva
Date: Fri, 15 Jul 1994 14:17:33 +0000

In article <2uslfj$ed8@dartvax.dartmouth.edu>
           Benjamin.J.Tilly@dartmouth.edu "Benjamin J. Tilly" writes:

> You have it right. Suppose that you have two different real numbers.
> The experiment is that you hand me one at random, and I try to guess
> whether I have the larger. The curious fact is that there is a strategy
> which gives me better than even odds of being right--guarenteed.
> 
> The strategy is for me to pick a random number from a normal
> distribution and pretend that it is the number that you still have. 

I set out to write a clear post disproving this, but to my surprise
proved it instead.

Suppose my numbers and X, Y with X<Y.
As I understand it you are choosing a fixed distribution f before you see
my number. f is effectively a non-zero density over the entire line.
You use f to pick Z. You say X is larger iff X>Z.

Suppose f gives prob(Z<X)=p, prob(X<Z<Y)=q, prob(Y<Z)=1-p-q.

If I send X, then you guess correctly with probability 1-p.
If I send Y, then you guess correctly with probability p+q.
So overall your chance of guessing correctly is 0.5+0.5q and q
is guaranteed >0 because f is non-zero over the entire line.

[Of course, I do not know f and do not know X and Y. But a third party
could calculate your advantage 0.5q as soon as he saw f and the two numbers
X and Y in front of me.]

A good strategy for me might be to pick as X and Y: 10^100 + n.10^-100
where I pick two small numbers n at random. This will give you a negligible
advantage for a single trial. For repeated trials I could increase the
exponent each time. But even for 1 trial "negligible" is something of an
understatement, so the result has no practical importance. It is neat
though.

John Scholes