Re: [math-fun] Permuation Probabilities ... with apologies ...
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
126*124*122*120*...*92 Could it be ---------------------- ~= .20822833 ? 127*126*125*124*...110
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?
.. but I'm sorry to say I don't perceive your thinking behind your original fraction?
Imagine a table of 128 paired slots. The first (of the 19) can go anywhere. The second can go into 126 of the 127 remaining slots. The third can go into 124 of the remaining 126. ... The 19th can go into 92 of 110. -- Don Reble djr@nk.ca
participants (2)
-
Daniel Asimov -
Don Reble