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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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Hey, FK, another answer... | Comment 18 of 32 |

Help me, FK, here I go again.

A fair tennis tournament has a power of 2 players, so N = 2^k, k being the number of rounds. 

First round : P(1) = 1/(N-1) = 1/(2^k-1).

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

Third round : N/4 players ===>P(3) = 1/(N/4-1) = 4/(N-4) = 4/(2^k-4).

etc.....

P = P(1) * P(2) * P(3) *...* P(k)...... k terms.

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

k = 1.......P = 1/1

k = 2.......P = 1/3 * 2/2 = 1/3 * 1/1

k = 3.......P = 1/7 * 2/6 * 4/4 = 1/7 * 1/3 * 1/1

k = 4.......P = 1/15 * 2/14 * 4/12 * 8/8 = 1/15 * 1/7 * 1/3 * 1/1

So, P = 1/1 * 1/3 * 1/7 * 1/15 *.... * 1/(2^k-1)     or

P = 1/1 * 1/3 * 1/7 * 1/15 *....* 1/(N-1).

And I donīt know if this canīt be resumed.

 


  Posted by pcbouhid on 2005-08-04 17:27:21
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