8 Jun
2004
8 Jun
'04
2:56 p.m.
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