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winds
05-31-2010, 10:32 PM
I read this as a solution to a problem, but I'm not sure how to get it:

Say we have k balls, of which 1 is red and k-1 are black, and k is odd. If we split them into two "half" piles of m and n balls, with m = n + 1, then the probability that pile "m" will have the red ball is m / (m+n).

I understand the other part, when k is even, then probability is 0.5, because it's equally likely to end up in either pile. And I understand, in the odd case, that the probability will be greater for the bigger "m" pile. But why will it be exactly the ratio given above?

Any insight much appreciated. Thanks.

BGM
06-01-2010, 03:23 AM
You have a total of m + n balls
You split them into two piles, one pile with m balls
and another pile with n balls
So I think the answer \frac {m} {m+n} is quite intuitive.
The probability is proportional to the number of balls in the pile

To generalize this idea, you can read about the hypergeometric distribution
\frac {{m \choose 1} {n \choose 0}} {{m+n \choose 1}}
= \frac {(m)(1)} {m+n} = \frac {m} {m+n}

winds
06-01-2010, 06:56 PM
It seemed intuitive, but I wasn't sure how to derive it. Reading up on the hypergeometric distribution helped me work it out. Thanks very much for the pointer!