The Magic Cafe Forum Index » » Magical equations » » A Full House (0 Likes) Munseys_Magic Special user 508 Posts  Posted: Jul 12, 2006 10:35 pm 0 Fact: There are 3744 ways that a full house can be formed if one is holding a five card hand in poker (where, of course, the position/order that the cards are held in the hand is irrelevant). Keep in mind that (KH, KS, KD, 7D, 7C) is a different Full House than (KH, KS, KD, 7D, 7H) To find the probability that a full house would be delt to a player in a five card hand would be 3744/(52!*51!*50!*49!*48!). In Texas Hold 'Em, one can use 7 cards to form a hand. Can somebody tell me how to figure the probability of getting a Full House in Texas Hold 'Em? Jim Munsey Munsey's Magic http://www.MunseysMagic.com Munseys_Magic Special user 508 Posts  Posted: Jul 12, 2006 11:39 pm 0 Wait! I goofed in my 2nd paragraph. There are (52!)/(5!*47!) = 2,598,960 different 5 card hands that one could be dealt, so the probability of a Full House in a 5 card game would be 3744/2,598,960. Sorry for the mistake. Anyways, my original question still stands: Can somebody tell me how to figure the probability of getting a Full House in Texas Hold 'Em? Thanks!! Jim Munsey Munsey's Magic http://www.MunseysMagic.com landmark Inner circle within a triangle 5019 Posts  Posted: Jul 15, 2006 09:26 pm 0 This is my quick analysis. Maybe some of the real math whizzes and poker players can chime in here. As I understand it, Texas Hold 'Em is played with two face down cards and five community cards, and you have to make the best five card poker hand from those seven cards. So first, how many seven card sets of cards contain a full house? If there are 3744 possible full houses then the other two cards of a seven card set tells us to multiply by 47*46. But we don't care what order those two extra cards are in, so divide by 2. Thus there are 3744*47*46/2 =4047264 combinations of seven cards that contain a full house. The total number of seven card hands is 52C7 which equals 52!/(7!)(45!) which equals 133784560 So on a simple level, the probability of your being able to make a full house in Texas Hold 'Em is 4047264/133784560 = .0302 . . . which is about 1 out of 33. BUT of course, be aware that since Texas Hold 'em has community cards, if you can make a full house, then the odds are greater than 1 out of 33 that your opponent can make a full house as well. For example, if all five community cards show a full house, then the probability is 100% that your opponents also have a full house. I'm not really a poker player, but it seems that in an actual game of T.H.'Em, there are a whole lot of much more complicated probability issues. Some quick research turned up this wiki article that may be of use if you're actually trying to play the game: http://en.wikipedia.org/wiki/Poker_proba......ld_'em) Jack Shalom Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years All proceeds to Open Heart Magic charity. Munseys_Magic Special user 508 Posts  Posted: Jul 16, 2006 09:41 am 0 Jack, Thanks for your thoughts. One comment: We *DO* care what those other two cards are. Suppose my FH consists of 3 Kings and 2 Queens. If one of the other two cards is another King, then I no longer have a FH when I show the cards that comprise my hands. I have 4 Kings and a Queen. Again, I appreciate your thoughts on the matter. Jim Jim Munsey Munsey's Magic http://www.MunseysMagic.com Munseys_Magic Special user 508 Posts  Posted: Jul 16, 2006 09:45 am 0 Here's a reply that I just got from a math site from Drexel University. It actually gives the probabilities for all hands! ********************************************************************************* There are C(52,7) total ways of getting 7 cards out of a standard deck. That's 133,784,560 possible 7 card hands. Now, let's go through the number of ways to get each particular hand. Royal Flush: With five cards, there are only 4 possible royal flushes (one of each suit). With 7 cards, we have to consider the 6th and 7th cards. After the 5 cards for the royal flush, there are 47 cards left over. There are C(47,2) = 1081 ways of picking the last two cards, or 4324 total Royal flushes possible. Prob(Royal flush) = 4*1081 / C(52,7) = 0.00323% Straight Flush: There are 9 possible straights possible for a straight flush that is not a royal flush, one each for the lowest card in the straight being anything from ace to 9. There are 4 suits. The last two cards can be any card except the card making a higher straight flush. For example, if you had A-2-3-4-5 of spades, you don't want to let one of the other cards be a 6 spades, or you will be counting that possibility twice. That leaves 46 cards for the last two, and C(46,2) = 1035. 9*4*1035 = 37260 straight flushes possible. Prob(Straight flush) = 4*9*C(46,2) / C(52,7) = 0.0279% Four of a kind: There are 13 different unique 4-of a kinds (one for each rank). There are C(48,3) ways of picking the remaining three cards. So, there are 13*C(48,3) = 224,848 possible Four of a Kinds. Prob(Four of a kind) = 13*C(48,3) / C(52,7) = 0.168% Full Houses: This gets a bit complicated. There are 3 separate cases we must consider. Case 1: We could have two sets of triples and an extra card. There are C(13,2) ways of picking which two ranks are the triples. For each triple, there are C(4,3) ways of picking which 3 of the 4 cards of a rank are in the triple. Finally, there are 44 extra cards to be used as the 7th card. That's a total of C(13,2)*C(4,3)*C(4,3)*44 = 54,912 ways of getting this type of full house. Case 2: 1 set of triples, 2 different unique pairs. There are 13 ways to pick the rank of the triple (any of one the ranks). There are C(12,2) ways to pick the rank of the pairs. For the triples, there are C(4,3) ways of picking the suits. For each pair, there are C(4,2) ways of picking the suits of the pair. That's a total of 13*C(12,2)*C(4,3)*C(4,2)*C(4,2) = 123,552 ways of getting this type of full house. Case 3: 1 set of triples, 1 pair, 2 blanks (don't match either the triple, the pair, or each other). There are 13 ways to pick the rank of the triples. There are 12 ways to pick the rank of the pair. And there are C(11,2) ways to pick the ranks of the blanks. There are C(4,3) ways of picking the suits of the triples. There are C(4,2) ways of picking the suits of the pair. And for each blank, there are 4 suits available. So, there's a total of 13*12*C(11,2)*C(4,3)*C(4,2)*4*4 = 3,294,720 ways of getting this type of full house. In all, there are 3,473,184 ways of getting a full house. Prob(Full house) = 3,473,184 / C(52,7) = 2.60% Flush: For a flush, we again have 3 cases. We'll take each one at a time. Case 1: All 7 cards in one suit. There are C(13,7) ways of picking 7 different cards in one suit. But we want to get rid of possibilities for straights. There are 8 ways all 7 cards could form a 7-card straight. There are 47 ways to form 6 card straights, and there are 162 ways to form 5 card straights from the 7 cards. That's a total of 217 ways. So, there are 4*(C(13,7) - 217) = 1499 possible flushes of this type. Case 2: 6 cards of one suit, and a blank. There are C(13,6) ways of picking 6 different cards in one suit. There are 9 ways to get a 6 card straight here and 62 ways of getting a 5 card straight. And 39 ways the blank could come up. That's a total of 4*(C(13,6) - 71)*39 = 256,620 possible flushes of this type. Case 3: 5 cards of one suit, and a blank. There are C(13,5) ways of picking 5 different cards in one suit. We remove the 10 possible straights to get C(13,5) - 10. The last two cards cannot give us a hand better than a flush. So, we compute the probability directly. C(39,2) ways of picking the last two cards. 4*(C(13,5) - 10)*C(39,2) = 3,785,028 Adding them together, there are 4,043,147 ways of getting a flush. Prob(Flush) = 4043147/C(52,7) = 3.02% Straights: We calculate this in multiple parts as well. Case 1: 7 distinct cards (no paired cards). Above, we calculated 217 ways to get a 5 card straight from 7 cards. All we have to do now is discount flushes. There are 4 ways to choose all 7 cards in any of the 4 suits. For 6 cards in the same suit, there are 4 suits, and C(7,6) ways of picking the 6 cards, and 3 choices for the suit of the remaining card, or 4*C(7,6)*3 = 84 ways of a flush in this situation. For 5 cards in the same suit, there are 4 suits, C(7,5) ways of picking the 5 cards, and 3 choices for suit for each of the remaining two cards, for a total of 4*C(7,5)*3*3 = 756. We subtract 756, 4, and 84 from 4^7, the total number of ways the cards can be suited, 4^7 - 756 - 4 - 84 = 15540. So, with 7 distinctly ranked cards with a straight but without a flush, there are 217*15540 = 3,372,180 ways of making it happen. Case 2: 6 distinct cards (last card pairs one of the others). There are 9 ways of have 6 card straights. We can also have a 5 card straight with the 6th distinct rank separate. If the straight begins with A or 10, there are 7 choices for the 6th card. Otherwise, there are 6 choices. That's a total of 2*7 + 6*8 = 62 ways. In all, there are 71 ways of producing a 5 card straight from 6 distinct cards. There are 6 choices for which distinct card is paired. There are C(4,2) = 6 ways to pick which suits make up the pair. The remaining cards can be chosen from any suit, so there are 4^5 ways to pick the suiting. We now get rid of flushes. There are 4 ways to pick all the 5 non-paired cards all in the same suit. In addition, we can't have 4 of them match the suit of one of the paired cards. There are C(5,4) ways of picking which 4, 2 choices for which paired card is picked for suit, and 3 choices for the suit of the 5th card (that is a different suit) for a total of 4 + C(5,4)*2*3 = 34. So, we subtract this from the total ways to suit the 5 cards: 4^5 - 34 = 990. That makes for 71*36*990 = 2,530,440 ways of a straight of this type. Case 3: 5 distinct cards (must include two pairs or a three of a kind). There are 10 ways to make a straight with 5 distinct cards. First, let's consider the three of a kind. There are 5 different possible three of a kinds. There are C(5,4) for which suits comprise the trips. The suits of the remaining 4 cards can be any of the 4 suits, for a total of 4^4 = 256. But we should eliminate the three possibilities that match the suits with any in the trips (to prevent flushes). So, with trips, that's a total of 10*5*4*(256-3) = 50,600. For the case with two pairs, there are C(5,2) ways of picking which two cards are the two pair. For each pair, there are C(4,2) = 6 ways of picking suits. 6*6 = 36. But now, we have to split the 36 up into cases to get rid of flushes. 6 of them have each pair matching two suits. 24 of them have one suit matched, and 6 of them have no matching suits. There are 4^3 = 64 ways of suiting the three remaining cards in the hand. Of these, if two suits are matched, we don't want either possibility of flush to come up. So, in this case, we have 62 ways of suiting the remaining cards. If only one suit matches in the two pairs, there's only 1 way of matching the suit, and 63 safe ways of suiting the remaining cards. If both two pairs have different suits, all 64 ways of suiting are safe. So, for a 5 card straight with trips or with two pair, that's a total of: 50600 + 10*C(5,2)*[6*62 + 24*63 + 6*64] = 277,400 ways of getting a straight. Adding all 3 cases together, we get 6,180,020 ways of getting a straight. Prob(Straight) = 6,180,020/C(52,7) = 4.62% Three of a kind: A hand that is three of a kind must have 5 distinct ranks of cards (fewer and you have a full house or two pair or something else). There are C(13,5) ways of picking the ranks, of which we remove the 10 possible straights. That leaves 1277 possibilities. We can pick any of the 5 ranks for the trips, and there are C(5,4) ways of suiting the trips. Of the remaining 4 cards, we want to get rid of the possibility of a flush. There are 4^4 = 256 ways of suiting the 4 cards and 3 ways of completing the flush. So, there are 253 safe ways of completing the flush. That's a total of 1277*5*C(5,4)*253 = 6,461,620 ways of getting a three of a kind. Prob(Three of a kind) = 4.83% Two pair: Case 1: As above, a hand that is two pair can have 5 distinct ranks, leaving 1277 qualifying hands 5 rank hands. As in the case above for straights, there are 10 ways of picking which two pair and [6*62 + 24*63 + 6*64] ways of not also making a flush. That's a total of: 1277*10*[6*62 + 24*63 + 6*64] = 28,962,360 ways of getting two pair of this type. Case 2: Three pairs plus a blank card. There are C(13,3) ways of picking the rank of the cards. For each pair, there are C(4,2) ways of picking the suits, and 40 cards left over for the blank. That's a total of C(13,3)*C(4,2)^3 * 40 = 2,471,040 ways of getting two pair of this type. In all, there are 31,433,400 ways of getting two pair. Prob(Two Pair) = 23.50% Pair: In a pair, there must be 6 distinct ranks in the hand. Above, we saw there were C(13,6) - 71 = 1645 ways of doing this without straights showing up. There are 6 choices for which rank is paired, and C(4,2) ways of picking the suits of the pair. The suits of the remaining cards can be any of the 4, leaving 4^5 choices. We want to eliminate flushes now. We can't have all 5 cards in the same suit. This can occur in 4 ways. We also cannot have 4 of them match suits with either card in the pair. There are C(5,4) ways of choosing which 4 cards, 3 ways of picking the suit of the extra card, and 2 suits to choose from in the pair, or C(5,4)*2*3 = 30. So, of the 4^5 ways of suiting the 5 cards, 4^5 - 30 - 4 = 990 are safe. In total, there are 1645*6*6*990 = 58,627,800 ways of getting a pair. Prob(Pair) = 43.82% High card: For a high card situation, there must be 7 distinct ranks. There can be no pairs. Getting rid of straights, as above, we are left with 1499 sets of 7 cards without a straight. Now, we have to get rid of flushes. There are 4 ways for all 7 cards to have the same suit. For 6-flushes, there are C(7,6) ways of picking the 6 cards, 4 possible suits, and 3 possible suits for the last card. For 6-flushes, C(7,6)*4*3 = 84. For 5-flushes, there are C(7,5) ways of picking the 5 cards, 4 possible suits, and 3 choices for each of the remaining two cards for C(7,5)*4*3*3 = 756 ways to get a 5-flush. Of the 4^7 possible suitings of the 7 cards, 756 + 84+4 = 844 are bad. So, there are a total of 1499*(4^7 - 844) = 23,294,460 possible ways to get a high card. Prob(High card) = 17.41% If we add up all the possibilities we calculated for each type of hand, we get 133,784,560, which is the appropriate number, giving us an indication our calculations are correct. Here's a chart summarizing the information: Hand Possibilities Probability Royal Flush 4,234 0.00324% Straight Flush 37,260 0.0279% 4 of a kind 224,848 0.168% Full House 3,473,184 2.60% Flush 4,043,147 3.02% Straight 6,180,020 4.62% Three of a Kind 6,461,620 4.83% Two Pair 31,433,400 23.50% Pair 58,627,800 43.82% High Card 23,294,460 17.41% So, you see that in 7 card poker games, high card occurs less than 20% of the time and both two pair and a single pair are much more likely outcomes (a single pair is more than twice as likely as high card in 7 cards). Jim Munsey Munsey's Magic http://www.MunseysMagic.com The Magic Cafe Forum Index » » Magical equations » » A Full House (0 Likes)
 The views and comments expressed on The Magic Café are not necessarily those of The Magic Café, Steve Brooks, or Steve Brooks Magic. > Privacy Statement <   