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POGenerating all 5 card poker hands
 

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  1. POGenerating all 5 card poker hands
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    <p>This problem sounds simple at first glance, but turns out to be a lot more complicated than it seems. It's got me stumped for the moment.</p> <p>There are 52c5 = 2,598,960 ways to choose 5 cards from a 52 card deck. However, since suits are interchangeable in poker, many of these are equivalent - the hand 2H 2C 3H 3S 4D is equivalent to 2D 2S 3D 3C 4H - simply swap the suits around. According to <a href="http://en.wikipedia.org/wiki/Poker_probability" rel="noreferrer">wikipedia</a>, there are 134,459 distinct 5 card hands once you account for possible suit recolorings.</p> <p>The question is, how do we efficiently generate all these possible hands? I don't want to generate all hands, then eliminate duplicates, as I want to apply the problem to larger numbers of cards, and the number of hands to evaluate fast spirals out of control. My current attempts have centered around either generating depth-first, and keeping track of the currently generated cards to determine what suits and ranks are valid for the next card, or breadth-first, generating all possible next cards, then removing duplicates by converting each hand to a 'canonical' version by recoloring. Here's my attempt at a breadth-first solution, in Python:</p> <pre><code># A card is represented by an integer. The low 2 bits represent the suit, while # the remainder represent the rank. suits = 'CDHS' ranks = '23456789TJQKA' def make_canonical(hand): suit_map = [None] * 4 next_suit = 0 for i in range(len(hand)): suit = hand[i] &amp; 3 if suit_map[suit] is None: suit_map[suit] = next_suit next_suit += 1 hand[i] = hand[i] &amp; ~3 | suit_map[suit] return hand def expand_hand(hand, min_card): used_map = 0 for card in hand: used_map |= 1 &lt;&lt; card hands = set() for card in range(min_card, 52): if (1 &lt;&lt; card) &amp; used_map: continue new_hand = list(hand) new_hand.append(card) make_canonical(new_hand) hands.add(tuple(new_hand)) return hands def expand_hands(hands, num_cards): for i in range(num_cards): new_hands = set() for j, hand in enumerate(hands): min_card = hand[-1] + 1 if i &gt; 0 else 0 new_hands.update(expand_hand(hand, min_card)) hands = new_hands return hands </code></pre> <p>Unfortunately, this generates too many hands:</p> <pre><code>&gt;&gt;&gt; len(expand_hands(set([()]), 5)) 160537 </code></pre> <p>Can anyone suggest a better way to generate just the distinct hands, or point out where I've gone wrong in my attempt?</p>
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    1. USNick Johnson
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    1. COgood question, what do you need it for? if you want to use it to compute chances of one hand against second hand you may use monte-carlo method
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    2. COI'm precalculating all heads-up matchups. Monte-carlo is for people who don't have enough computing power. ;)
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    3. COThis is a very interesting problem. I don't have time to play with it right now, but a couple of thoughts have occurred to me which may or may not be useful. The first is to work from high to low – that is, each card's rank must be less than or equal to that of the previous card (A-9-9-8-2, for example). Secondly, I believe it's possible to only draw a club as the first card, a club or diamond as the second card, and a non-spade as the third card. (I haven't got my head around your bitwise code, so it's possible you're already doing these things.)
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