Author: Michael Neish
Date: 19:04:28 01/19/00
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>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.
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