Author: Robert Hyatt
Date: 05:07:28 01/20/00
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On January 19, 2000 at 22:04:28, Michael Neish wrote: > >>Play four games with two equal computer programs. The win expectancy is 0.5, >>hence the odds of getting zero points for program 'a' is about 6% and for >>program 'b' is about 6%. Hence the odds of a complete 4-0 blasting (for one or >>the other) is about 12%. I'll bet if any short tournament were held between two >>approximately equal computers, if a 4-0 result were posted here, we would >>immediately hear cries of "clear superiority" for the winning system. > >That's a very simple and clear way of stating it. I think it's easy to fall for >it if your computer wins 4-0. > >Anyway, here's a quick back-of-the-envelope calculation for 24 games, again, >assuming no draws. > >Both computers are assumed to be of equal strength, and so a 12-12 score would >be the "expected result". > >The total number of possible combinations of results is 2^24 = 16,777,216. The >probability of a certain match result is therefore the number of ways that the >result can occur divided by 2^24. > >The probability of a 12-12 score is given by > > 24! >------- / 16,777,216 = 16.1% >12! 12! > >The probability of a 15-9 score is > > 24! >------- / 16,777,216 * 2 = 15.6% >15! 9! > >We have to multiply the second probability by two since one or the other >computer can win the match. > >Now look again. The probability of a 15-9 score occurring is almost the same as >that of the "expected" 12-12 score. > >If you do the same for other possible scores, there is, for example, a 4.1% >chance of an outrageous 17-7 score. Beyond that, the probabilities quickly >decay. > >This calculation does not take into account draws, or the slightly better >chances for White to win a game. I don't know how draws would affect the result >(apart from making the calculation more complex), but I think the above is >correct to a first approximation. > >It would be nice if anyone could give a more accurate treatment of the above. > >Cheers, > >Mike. Yes, although the reasoning is often twisted a bit. IE yes, there is a probability of almost 1.0 that if you play enough games, that one program will win N in a row (ie 4 in the above). But the probability that this is the _first_ N games played is very low. Somewhat like the "runs test" for random number generators.
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