Don reble wrote:

[Guy Haworth wrote:

> 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?

126*124*122*120*...*92

Could it be ---------------------- ~= .20822833 ?

127*126*125*124*...110

I wrote what amounts to

109*108*107*...92*91

--------------------,

127*125*123*...93*91

which is ~= .20822833 also,

so I presume Don's answer is another way to calculate

the same probability. But I don't see why yet. Can

someone please explain his method to me?

--Dan