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Solution and related problem

Posted By: Bob Koca
Date: Tuesday, 9 April 2019, at 3:58 p.m.

In Response To: Probability puzzle (Ray Kershaw)

for 8 players:

1/7 time they play each other in first round and then no chance of split.

2/7 time they are in same half of bracket without being first round opponents and then a split if and only if both lose first game so 1/4 chance.

4/7 time they are in opposite half of bracket and then split if and only if both win exactly 0 or both win exactly 1 game. 1/4 + 1/16 = 5/16 chance.

(2/7) (1/4) + (4/7) (5/16) = 28/112 = 1/4.

For 8 players:

7/15 in same half of bracket and then we have the 8 player situation.

8/15 in opposite half and then split if both win exactly 0 games, or both win exactly 1 game or both win exactly 2 games. 1/4 + 1/16 + 1/64 = 21/64 chance.

(7/15)(1/4) + (8/15)(21/64) = 7/24 chance of a split.

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Suppose this problem is done for number of players equalling 2^N. What is the limit of the split probability as N goes to infiinity?

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