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?

>>

 

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