Author: Dan Homan
Date: 00:43:02 08/15/98
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Bruce, For 2 programs with a draw chance D Pa = chance of program A winning any one game Pb = 1-D-Pa = chance of program B winning any one game For a 4-0 result in a 4 game match, either A or B has to win all games, so the probability, P, that there is a 4-0 result is... P = Pa^4 + Pb^4 So if the programs are equal, Pa = 0.4 and you get a 4-0 (or 0-4) result 5% of the time. If there is a 4-0 result, what are the chances that the stronger program won? Assume A is stronger. The chance that A wins any 4-0 (or 0-4) result is simply Pa^4/(Pa^4+Pb^4). So 4-0 results happen about 5% of the time between nearly
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