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Combinatorics solution

Posted By: Maik Stiebler
Date: Thursday, 11 April 2019, at 3:41 p.m.

In Response To: Combinatorics solution (Ray Kershaw)

Imagine I am an alien observer of that tournament and don't know which two players are in the side pool. For my convenience, I label the players "1", "2", "3", "4", "5", "6", "7" and "8", where "1", "2", "3" and "4" are the first round losers, "5" and "6" are the second round losers and "7" and "8" the two finalists. After the tournament, you come to me and pose your question: "I know that exactly two of the players entered a side pool, but not who they are. I know that all players are of exactly equal ability. What is the probability that the pool is split?"

I note that there are 56 different possible combinations of two side pool players. Because of the equal ability of all the players, knowing the tournament result does not give me any information about their respective likelyhoods. As far as I am concerned, all of them are equally likely. So to answer your question, I just have to count how many of the 28 combinations will result in a split pool. A split pool happens when both players are either from {"1", "2", "3", "4"} (six combinations) or from {"5", "6"} (1 combination). Thus, 7 of the 28 possible combinations result in a split. Because all combinations are equally likely, my answer is 1/4.

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