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Probability Question

Posted By: Timothy Chow
Date: Wednesday, 30 November 2011, at 6:26 p.m.

In Response To: Probability Question (Art Grater)

Art Grater wrote:

Ray lists a question in each of his first two lines. The first one is the one I addressed

And Ray's first question was:

Is there a simple answer to David's question: given a draw of 64, what is the probability that four specific players are each in a different quarter of the draw ?

But the answer you supplied was p = 4/64 x 3/63 x 2/62 x 1/61, which is the answer to David Startin's original question ("In a 64 player tournament involving players of equal ability, what are the chances of 4 named (before the draw is made) players all reaching the semi-finals?") and not the answer to Ray's first question (quoted above).

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