Russell Standish has the right idea.  If you knew the range, say the first
number is higher/lower than the third depending on whether the first numbers
is greater than or less than the middle of the range. Since you don't know
the range, the second random number is used instead.  Say higher/lower
depending on whether the first number is higher/lower than the second.

Also, think about it this way. If the middle number is either greater than
or less than both the first and the third, you have a 50% chance of being
right. If it's between the first and the third, the strategy described will
always be right. Presumably there is a non-zero probability that the second
number will be between the first and the third. Therefore one has a greater
than 50% chance of being right.

*-- Russ *


On Wed, Jun 8, 2011 at 10:06 AM, Owen Densmore <o...@backspaces.net> wrote:

> Do you have a pointer to an explanation?
>
>        -- Owen
>
> On Jun 8, 2011, at 12:11 AM, Russ Abbott wrote:
>
> > Although this isn't new, I just came across it (perhaps again) and was so
> enchanted that I wanted to share it.
> >
> > Generate but don't look at three random numbers. (Have someone ensure
> that they are distinct. There is no constraint on the range.) Look at the
> first two. You are now able to guess with a better than 50% chance of being
> right whether the first number is larger than the unseen third.
> >
> > I like this almost as much as the Monte Hall problem.
> >
> > -- Russ
> >
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>
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