I've never actually seen the full description in one place on the net, so I thought I'd do a public service.
First, if you're playing a game with extra cards, like Hold'em or 7 stud, you first use recursion thusly: Iterate through the set of cards, removing one at a time and recurse. From the recursion results, save the best hand and return it.
Once you get down to 5 cards, you first take a histogram of the card ranks. That is, for each rank in the hand, count how often it appears. Sort the histogram by the count backwards (high values first).
If the histogram counts are 4 and 1, then the hand is quads.
If the histogram counts are 3 and a 2, then the hand is a boat.
If the histogram counts are 3, 1 and 1, then the hand is a set.
If the histogram counts are 2, 2 and 1, then the hand is two pair.
If the histogram has 4 ranks in it, then the hand is one pair.
Next, check to see if the hand is a flush. You do this by iterating through the cards to see if the suit of a card is the same as its neighbor. If not, then the hand is not a flush. Don't return a result yet, just note whether or not it's a flush.
Next, check for straights. Do this by sorting the list of cards. Then subtract the rank of the bottom card from the top card. If you get 4, it's a straight. At this point, you must also check either for the wheel or for broadway, depending on whether your sort puts the ace at the top or bottom. I would expect most folks to put the ace at the top of the sort, since it is usually a high card. So to check for the wheel, check to see if the top card is an ace and the 2nd to top card is a 5. If so, then it is the wheel.
If the hand is a straight and a flush, it's a straight-flush. Otherwise if it's one or the other, you can return that.
If we haven't matched the hand by now, it's High Card.
Once you know what the hand is, it is fairly straightforward to compare two hands of the same type. For straights, you simply compare the top card. For flushes, you iterate down through a reverse sort of the ranks until you find a higher card or run out. For the histogram-based hands, you start at the beginning of the reverse-sorted count list and compare the ranks (two pair is tricky - you must always evaluate the higher of the two pairs for each hand first - the histogram won't help, then the lower pair, then the kicker).
Checking for a low (for high/low or Razz) is a little easier. Sort the hand by rank. Iterate through the list. If any card is bigger than an 8 or is the same rank as it's neighbor, it's not a low hand.
Wednesday, July 18, 2007
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5 comments:
This is exactly what I was looking for.
I'm trying to build a texas hold 'em game in python (just for fun and learning) and I needed some clue on how to calculate a poker hand.
So, thank you very much for sharing.
I have written such a code that is similar to this (including calculating for 7 card hands). The problem I have with it is that it is very very slow; I use it to calculate odds of making hands given x hands. The down side is that to do this, I have to calculate 16000 hands to get a good representation of odds, and that takes 4 secs or so to do. Compare this to software out there called "pokerstove" and you will see I am way slow. His is able to calculate the odds of 5 different players hands preflop in about 12 seconds using 4.4billion games. I will be happy to share the code in java if interested, but I am looking for a streamlined algorythmn.
That much I'd gotten on my own, but the part I'm having trouble with is finding the odds for partial hands. Say I'm playing 7 Card Stud, and at 4th street I have my four cards, my opponent's two upcards, and a list of all the cards in the deck I haven't seen yet. Is there a good way to find the odds of my hand beating his at showdown, assuming his hole cards and all the cards not yet dealt are chosen randomly from the list of remaining cards? The only way I know is to just try a large number of combinations, but that can't be the best way.
When I have needed to do this, I've used the Monte Carlo method. You simply run the hand 1000 times and see how many times it wins. It's brute force, and it's only an approximation, but surprisingly it doesn't take terribly long to do it with modern CPUs.
By the way, everyone, there is another algorithm out there that uses a giant array of pre-computed hand values that is much, much faster than my method. The description of his method and source code is here.
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