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lotus776 10-10-2005 07:27 PM

for the math majors
How are the percentages (that particular hands are favorites over others) determined? How are the percentages reached? How does one determine what the percentage chances of winning with J6o comparted to K8s or ANY other combination. Sklansky reccommends that "serious" players should purchase a computer program (most likely b/c the computations are elaborate) that allows input of any amount of hands and then pops out a number given the input. However, someone must have programmed this software with an algorithm that determines these numbers. How can I do this?
I understand the work involved must be lengthy to involve a software program but humor me and explain how this is done.
thank you

AaronBrown 10-10-2005 07:48 PM

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.

lotus776 10-10-2005 08:12 PM

Re: for the math majors
thank you, that information is very useful.

what are the names of some of the programs that would perform the first task mentioned (dealing all possible hands) and how can I get a hold of them?

thanks again

yellowjack 10-10-2005 11:25 PM

Re: for the math majors
Try PokerStove (here). It has a few bugs that prevent it from making completely accurate equity computations but it's a good place to start.

On the web, everyone's favorite is I don't think any bugs/errors have been reported for it because it's just that accurate.

droidboy 10-11-2005 01:36 AM

Re: for the math majors
Try PokerStove (here). It has a few bugs that prevent it from making completely accurate equity computations but it's a good place to start.

[/ QUOTE ]

Which bugs are those? If you know of any specific bugs, please let me know and I'll get them fixed.

- Andrew

yellowjack 10-11-2005 01:24 PM

Actually I haven\'t found a bug with PokerStove.
It's documented anyways. The thing I had an issue with was more than one random hand, where a random hand had up to 6% more equity than another random hand. PokerStove lets you set up a hand vs. a range of hands, as well as completely random hands that can exist (i.e. if 2 hands are AA, then a random hand cannot have an ace).

However in PokerStove, the output tells you that the data is unreliable, and to use Monte Carlo instead (ennumerating random boards?). I didn't see this before, my mistake.

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