Re: fun probability question from an old Putnam (spoiler!)
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Interesting!
Simply prove by induction that if P(k; n) is the probability
of k free throws in the first n attempts, that
P(k; n) = 1/(n-1) for 1<=k<=(n-1) for any n>=2.
The proof is simple: just note that
P(k; m+1) = P(k-1; m)[(k-1)/m] + P(k; m)[1-(k/m)] even
when k=1 (then the (k-1)/m term then is zero).
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Yep; this was the approach I took.
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