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 agree with Charlie's reasoning. I worked out a special case (2^N) this way:

Assume the original places are numbered 1, 2, 3, 4, ... T, with T=2^N, and our player is at place #1.

If the other player is at place #2 (odds: 1/T) he will meet him at the first round.

If the other player is at places #3 or #4 (odds: 2/T) he will meet him at the second round, only if both win their first round matches (odds: 1/2x1/2) for a total chance of 1/(2T).

If the other player is at places #5 through #8 (odds: 4/T) he will meet him at the third round, only if both win both their first and second round matches (odds: 1/4x1/4) for a total chance of 1/(4T).

...and so on, up to the case when the other player is at places T/2+1 through T, and the chance of their meeting at the N-th round will be 1/(2^N)x1/T.

Summing all chances we get 1/Tx[1+1/2+1/4+...+1/2^(N-1)]= 1/Tx(2-1/2^(N-1)] = 1/2^(N-1)= 2/T, as Charlie found.

Posted by e.g. on 2005-08-02 21:57:38