Re: Probability Question
- From: ol3@xxxxxxxxx (Oscar Lanzi III)
- Date: Sun, 1 May 2005 13:07:48 -0500
Define P(m, n) as the probability that Adam wins n times in m plays. In
a winning match for Adam he wins the last game and eight of teh others,
and the number of games before the last one is in [8, 16]. Thus we st
compute:
P(1,1)*sum(m = 8 to 16) P(m, 8)
P(1,1) = 0.55. P(8, 8) = 0.55^8. Now for each increment of m in P(m,
8), we have:
P(m+1,n) = (m+1)/(m-n+1) * P(m, n) * P(1,0)
The last two factors correspond to n Adam wins in m tries combined with
one additional loss, and the first factor corresponds to how much the
permutations of losses is increased by mixing the additional loss into
the pot. P(1,0) = 0.45, n = 8 here, and we already have the initial
condition P (8, 8) = 0.55^8. Therefore all the terms in the summation
may be calculated in a recursive and straightforward fashion.
--OL
.
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