Guy Haworth asks: << re the upcoming FIDE World Chess Championship). In the 'knockout' first round involving 128 players and 64 matches, what is the probability that none of 19 specific players are matched against each other?

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If we assume the first 19 assignments of first-round opponent are to these 19 players, the probability you request must be ((128-19)/128-1) * (128-19-1)/(128-3))* ... *((128-19-18)/(128-37)) since the first player must be assigned to any of 128-19 out of 128-1; the second must now be assigned to any of 128-(19+1) out of 128-3; until the last of the 19 must be assigned to any of 128-(19+18) out of 128-36, which is all that matters. --Dan