Thread: for the math majors View Single Post
#2
10-10-2005, 07:48 PM
 AaronBrown Senior Member Join Date: May 2005 Location: New York Posts: 505
Re: for the math majors

There are two ways to answer this question. The first is to describe a simple but tedious algorithm. If you want to find out how often K8s wins over J6o, you would first realize you have three different situations: when the Jack's suit matches the suited cards, when the 6's suit matches the suited cards and when neither do. Let's just start with one, say where neither match the suit. We can then pick our suits arbitrarily, say K[img]/images/graemlins/spade.gif[/img] 8[img]/images/graemlins/spade.gif[/img] versus J[img]/images/graemlins/diamond.gif[/img] 6[img]/images/graemlins/heart.gif[/img].

That leaves 48 cards in the deck. We could deal these one into all 1,712,304 possible five-card flops and see the result with the two sets of pocket cards. This is not a difficult computer program to write.

A simple-mined implementation of this algorithm would be slow, especially if we consider K[img]/images/graemlins/spade.gif[/img] 8[img]/images/graemlins/spade.gif[/img] against any possible other pocket cards, or nine random hands or something like that. So even a computer needs tricks to get the same result faster. Obviously solving it by hand requires even more tricks.

To solve it by hand, I'd begin by figuring the chance for each hand to make a straight or a flush. Then I'd figure the chances of pairing one or more of the pocket cards. The tricky part is combining those calculations, individually they are easy.

For example, getting exactly one pair between the two hands means one of the board card must be selected from the 12 outstanding matching cards (K, J, 8 or 6) and the other four board cards must be selected from the other 36 cards. That can happen in 12*36*35*34*33/(4*3*2*1) = 706,860 ways, out of the 1,712,304 boards. If that happens, and there is no straight or flush, each hand is equally likely to win (whoever pairs wins). Then you can figure the number of two-pair outcomes (in which case K[img]/images/graemlins/spade.gif[/img] 8[img]/images/graemlins/spade.gif[/img] wins 2/3 of the time, when the pairs are K 8, K J, K 6 or 8 6; and loses the other 1/3 when the pairs are J 6 or 8 6).

It might take half an hour with pencil and paper to write down all the possibilities, but it's solveable.