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?

I missed the fact of the "elimination" in a tennis tournament.

Letīs try again:

In the first round (N players), I have a probability to play with X of 1/(N-1).

Assuming that we both won our games, in the second round there will be N/2 players, and I have a probability to play with X of 1/(N/2-1) = 2/(N-2).

Assuming the same, in the third round there will be N/4 players yet, and the probability is 1/(N/4-1) = 4/(N-4).

To a fair tournament, N must be a power of 2, so N = k^2, and there will be k rounds, k = sqrt(N).

So the whole probability in k rounds will be:

P = 1/(k^2-1) + 2/(k^2-2) + 4/(k^2-4) +....+ 2^(k-1)/((k^2)-2^(k-1))

Or P = SUM(from 1 to (k-1)) of {2^(k-1)/(( k^2) - 2^(k-1))}

Iīll try to simplify this.

 

Posted by pcbouhid on 2005-08-03 16:04:48