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Good solution

Posted By: Bob Koca
Date: Saturday, 5 August 2017, at 11:55 p.m.

In Response To: Chance of one loss player winning twice to win (Mike Clapsadle)

Good solution.

If the favorite has win percentage of 1/2 + q where 0 <=q <= .5 and the favorite is equally likely to be the undefeated player as the one loss player then the chance of one loss player winning is average of (1/2 + q)^2 and (1/2 - q)^2 which equals 1/2 + q^2. If q is not exactly 0 this is greater than 1/4.

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