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Pair Playing Probability (Posted on 2005-08-02) Difficulty: 3 of 5
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?

See The Solution Submitted by Old Original Oskar!    
Rating: 3.0000 (2 votes)

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Some Thoughts The Solution??? | Comment 29 of 32 |

I can follow your logic, OOO, but should you not take into account that each player in the selected pairing has only a 50% chance of surviving each round. For example, in a tournament with 8 players:

A. Chance of a particular pair (X and Y) playing in Round 1 = 1/(N-1) = 1/7

B. Chance of Player X advancing to Round 2 = 1/2

C. Chance of Player Y advancing to Round 2 = 1/2

D. Chance that both Player X and Player Y advance to Round 2 = B * C = 1/4

E. Chance that Player X and Player Y meet in Round 2 = D * (1/(N-1)) = 1/4 * 1/3 = 1/12

F. Chance of Player X advancing to Round 3 = 1/2

G. Chance of Player Y advancing to Round 3 = 1/2

H. Chance that both Player X and Player Y advance to Round 3 = F * G = 1/4

I. Chance that Player X and Player Y meet in Round 3 = D * H * (1/(N-1)) = 1/4 * 1/4 * 1 = 1/16 

So, chance that X and Y play each other at some stage in the tournament = Chance of meeting in Round 1 + Chance of meeting in Round 2 + Chance of meeting in Round 3 = A + E + I = 1/7 + 1/12 + 1/16 = 51/112.

Your solution only works if the question was "what are the odds that a certain pair did play each other" where you are looking at the results after the tournament. You asked "what are the odds that a certain pair will play each other", so the pair must be selected before the tournament starts.

I am new to this sort of stuff, so please enlighten me if you think I am wrong.


  Posted by Paul on 2005-08-29 02:01:47
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