View Single Post
#3
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636 Re: A Question on rec.gambling.poker

I didn't see Sklansky's solution on RGP; is it there? I did see where he blasted Alspach for insisting on using advanced math where an approximate solution would do. The problem in question was similar to one solved here, where the exact solution requires inclusion-exclusion, but where the assumption of independence is very accurate.

Here's a way to to this problem using the assumption of independence. I assume this is probably how Sklanksy did it too. We assume the probability of any hand occuring is independent of the probability of any other hand occuring.

In order to have 50% probability of every hand occuring, we must have each hand occuring with probability p where p^1326 = .5. So p = .5^(1/1326) = .9994774. The probability of any hand not occuring is
1 - .9994774 = .0005226. But the probability of a hand not occuring is also
(1325/1326)^n = .0005226. If you don't know about logarithms, you can solve this equation by trial and error. If you know about logarithms (and even if you don't) the solution is:
n = log(.0005226)/log(1325/1326) = 10,016 hands.