Author: Amir Ban
Date: 03:26:17 06/09/00
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On June 08, 2000 at 16:10:16, blass uri wrote: >On June 08, 2000 at 15:25:22, Amir Ban wrote: > >>Ok, here's my two cents: >> >>The most likely estimate for the game outcome is the overall match result >>(33.5/41). The probability for any *particular* sequence of 10 games (say the >>last 10) to be 10-0 is (33.5/41)^10 = 13.3%. >> >>If you want probability for any sequence anywhere, the probability is larger, as >>you point out. >> >>If you suspect the 10-0 result, you can argue that the most likely estimate >>should discount this result, so this gives you a probability of (23.5/31)^10 = >>6.3%. > >If you use the most likely estimate >The probability for 10:0 is smaller than (23.5/31)^10 because there are draws. > >The result was +20, =7.-4 so if you use the most likely estimate you get >probability (20/31) for Junior to win >7/31 for a draw and 4/31 for a loss > >(20/31)^10=0.012493... so the probability is less than 2% > >I ignored the fact that the white probability is different from the black >probability but not ignoring it does not change much the result because you get: > >(10/15)^5*(10/16)^5 that is 0.012558... > >If you do not ignore the last 10 games and ignore the colours you get >(30/41)^10=0.043992... > >Uri All right, you are right about the draws. My presentation of the problem was more complicated than necessary. Here is the question in its simplest form: A match of 41 games concluded with 30 wins, 7 draws and 4 losses. It is unknown in which order these wins/draws/losses occurred. Out of all possible permutations with the same result, what percent of the permutations have a sequence of at least 10 wins ? This is clear cut question for the combinatorics gurus in this newsgroup. There seems to be gross difference of opinion about the result of this, so until we get an answer, we can have an opinion poll about the result. My guess: 40%. Amir
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