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发帖数: 474 | 1 Assuming P(n,k) is the probability that there are 2k+1 white balls
at the n'th step, then the recursive relation is:
P(n,k) = (2n-2k-1)/(2n)*P(n-1,k) + (2k-1)/(2n)*P(n-1,k-1)
Let P(n,k) = f(n,k)/D(n,k), where D(n,k)=(2n)!!/((2k-1)!!*(2n-2k-1)!!).
Here !! is double-factorial, i.e., m!! = m*(m-2)*(m-4)*(m-6)..., thus
f(n,k) = f(n-1,k) + f(n-1,k-1). We can easily prove (by induction) that
f(n,k) = q * C(n,k), where q is a constant, and C(n,k) is
the combinatorial number = n!/(k!*(n-k)!).
(recall C( |
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