Re: Random choice of numbers

"Stephen J. Herschkorn" <sjherschko@xxxxxxxxxxxx> writes:

Dante wrote:

I have problems with writing down a formal solution of the following
problem:

===
Two boys - Chris & Michael - are playing the following game: Both of them
are selecting randomly numbers from [0,1] interval (with uniform
distribution). Chris is taking two numbers (each one is independent of
another), and Michael is taking only one number (once again this is
independent event). Next, Chris is multiplying both of his numbers, and
Michael is calculating a square of his number. The boy with a larger
result is a winner. Who has more chances of winning this game - Chris or
Michael?
===

For me this is obvious, that Michael is the one. (It's important that they
are randomly selecting numbers from [0,1]!)

But how we can prove this result in a formal way?



Others have shown how to show the probability that Chris wins is 4/9.
(Ludovicus's response is incorrect.) Another way to do the
computation is via the integral of the uniform density on the unit cube
over the region {(x,y,z) in [0,1]^3: xy >= z^2}, which is integral(z =
0..1, y = z^2..1, x = z^2/y..1, 1).

.... and more generally, if the random numbers are selected from the interval
[a, 1], I get that Chris's probability of winning is

4/9 (1 - 2 a^(3/2) + a^3)/(1-a)^3

which >= 1/2 if a >= 0.0445338838868045196876255090994 approximately.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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