There are N players in a tennis tournament. Assuming the initial pairings are done randomly, what are the odds that a certain pair will play each other?

Letīs go, again...

First round : P(1) = 1/(N-1)

Second round : N/2 players ===> P(2) = 1/(N/2-1) = 2/(N-2)

Third round : N/4 players ====>P(3) = 1/(N/4-1) = 4/(N-4)

etc.....

There will be sqrt(N) rounds. So:

P = P(1) * P(2) * P(3) *...* P(sqrt(N))......sqrt(N) terms.

P = 1/(N-1) * 2/(N-2) * 4/(N-4) *...* 2^[sqrt(N)-1]/(N-2^[sqrt(N)-1]

P = PROD (from 1 to sqrt(N)) {2^[sqrt(N)-1]/(N-2^(sqrt(N)-1))} 

2^(sqrt(N)-1) = 2^sqrt(N)/2

N - 2^(sqrt(N)-1)) = N - (2^sqrt(N)/2) = (2N - 2^sqrt(N))/2

P = PROD (from 1 to sqrt(N)) {(2^sqrt(N))/(2N - 2^sqrt(N))}

The numerator is a geometric progression, so its evaluation is easy. But I have to think a little bit more, to simplify the expression above. Iīm too tired...to be continued

 

 

 

Posted by pcbouhid on 2005-08-03 19:44:05