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I feel I was cheated in a poker game. According to my math,
AA vs. KK will happen once every 45,000 hands heads up, but it happened to me 3 times in 400 hands. Is this unlikely enough to suspect something?
— Rafael from Los Angeles The probability of being on the losing end of KK vs. AA is (combin(4,2)/combin(52,2))*(combin(4,2)/combin(50,2)) = 0.000022162, for each opponent at the table. That is once every 45,121 hands, so your math was right. The expected number of times that would happen in 400 hands is 400 × 0.000022162 = 0.008865084, per opponent. The following table shows the probability of 3 or more instances of having KK against AA in 400 hands, by the number of opponents.
So, yes, I would say this looks fishy. The fewer the players, the more fishy it looks. I would be interested to know where this game was. March 24, 2008
I have looked all over the net about the probability of at least getting a pair by the river card in hold’em if you have two different cards dealt to you. I have tried to work it out using a probability tree but my answer seems too high. Also on the net I have read different answers some suggesting it is around 1/3 or 2/5 or 1/2. What is the probability of at least pairing and is it possible to work this out using a probability tree? Your help will be much appreciated thanks. – Nathan S. from New Plymouth For those not familiar with the hold’em terminology you are asking for the probability of at least a pair in six cards, given that the first two are (the hole cards) of different ranks. I hope you’ll forgive me if I just do the proability of getting exactly a pair, including hands that also form a straight or flush. The number of ways to pair one of your hole cards is six (2 hole cards * 3 suits remaining). The other three cards must all be of different ranks from the 11 left. There are combin(11,3)=165 ways to choose 3 ranks out of 11. For each of these there are four suits to choose from. So the number of ways to pair one of your hole cards is 6*165*43=63,360. Now let’s look at the number of ways to get a pair outside of the two hole cards. There are 11 ranks to choose from for the pair. Once the pair is chosen there are combin(4,2)=6 ways to choose 2 suits out of 4. For the other two cards there are combin(10,2)=45 ways to choose 2 ranks out of the 10 fully intact ranks left. For both of those ranks there are 4 possible suits. So the total combinations for a pair, not including the hole cards, is 11*6*45*42=47,520. The total number of ways to choose 4 cards out of the 50 left in the deck is combin(50,4)=230,300. So the probability of getting exactly a pair in six cards is (63,360+47,520)/230,300 = 48.15%. October 17, 2006
Hello, thank you for a very interesting and informative site. I have a question of my own that I hope you can answer for me.
As a Texas hold'em player, I pay special attention to pocket pairs, and have particular interest towards 10 10 or J J or similar, as on the surface they appear strong but can be beaten easily. My question however, is how do you work out the probability of there being at least one person on your table holding a higher pocket pair to yours? - Andrew from St. Albans
The math of this gets very messy due to the possibility of more than one playing having a higher pair, including the same type of pair. For example if you have pocket kings two players could have pocket aces. However it is easy to show the expected number of players who will beat you. This would be n*r*(6/1225), where n is the number of opponents, and r is the number of higher ranks. The following table shows the average number of players who will have a higher pocket pair according to your pocket pair (left column) by the number of opponents (top row).
To get the probability that at least one player will beat you I will make the not entirely correct assumption that the number of players with a higher pocket pair is a Poisson random variable with a mean in the above table. Given that assumption the probability that at least one player will beat you is 1-e-µ, where µ is the mean. For examle if you have pocket queens and there are 9 other players the expected number of players with a higher pocket pair is 0.0882, so the probability of at least one player having a higher pocket pair is 1-e-0.0882 = 8.44%. The table below shows those probabilities.
So my approximation of the probability of at least one higher pocket pair is 1-e-n*r*(6/1225). P.S. After this column appeared one of my fans, Larry B., wrote a brute force combinatorial program to solve this problems. Here are his results.
Later Stephen Z. suggested a simple approximation. Take the number of higher pairs, multiply by the number of other players, and divide by 2. That is the percentage probability that there will be at least one higher pair. For example, with a pair of jacks in a 10 player game the probability of a higher pocket pair is 3*9/2 = 13.5%. Using that formula you get the following for all situations.
What are the odds of having pocket Aces and pocket Kings both being dealt in the same hand? -- Jake from Loveland, CO
The odds of a specific player having aces is combin(4,2)/combin(52,2) = 6/1326. The odds of the next player having a pair of kings is combin(4,2)/combin(50,2) = 6/1225. However in a ten-player game there are 10 possible players who could get the aces, and 9 possible players for the kings. So a strong approximation would be 10*9*(6/1326)*(6/1225) = 0.001995, or 1 in 501. This answer is slightly too high, because it double counts the situation where two players have aces, or two have kings, or both.
March 13, 2006
I've been a huge fan for many years (even before you got interested in poker and sports betting) and looked forward to every Ask The Wizard column. It's great to see you're doing them again! My question is this: at my local card room, they offer Aces Cracked, Win A Rack during certain hours. That is, if you have pocket Aces in one of their 3-6 or 4-8 Texas Hold 'Em games and you lose the pot, the casino will give you a rack of chips ($100). I'm trying to figure out how often a)I get pocket Aces b)how often they would lose if I played them aggressively as I'm supposed to and c)whether it's not better to just check all the way down and hope to lose, as $100 is usually better than what the pot would have been anyway. Any stats you may have at the ready would be wonderful and forever appreciated! Thanks again and keep up enlightening the masses! - Shane from Santa Rosa
Thanks for the kind words. The probability you will get pocket aces in any one hand is 6/1,326, or once every 221 hands. According to my 10-player Texas Hold'em section (/holdem/10players.html) the probability of winning with pocket aces is 31.36%, assuming all players stay in until the end. However, that is a big if. If forced to make a guess I'd estimate the probability of winning with aces in a real 10-player game is about 70%. So the probability of getting pocket aces and then losing is 0.3*(1/221) = 0.1357%. So, at $100 per incident that is worth 13.57 cents per hand. Over ten people, that costs the poker room $1.36 per hand on average, which cuts into the rake quite a bit. I tend to agree with your strategy of calling, which will keep more players in the hand, and increase your chance of losing.
Feb. 21, 2006
In a Three-handed game of Holdem, what are the chances of AA vs. KK vs. QQ? - Chris from Hampton
Let's call the players A, B, and C. The probability A has a pair of aces is combin(4,2)/combin(52,2) = 6/1,326. The probability B has a pair of kings is combin(4,2)/combin(50,2) = 6/1,225. The probability C has a pair of queens is combin(4,2)/combin(48,2) = 6/1,128. However, there are 3! = 1*2*3 = 6 ways you can arrange three pairs between three players. So the answer is 6*(6/1,326)*(6/1,225)*(6/1,128) = 0.000000707321.
Feb. 1, 2006
Hi Wizard, First off I'd like to say I love the precise, fluff-free answers you give. Anyway if the pocket cards in Holdem are AA and the flop cards are KQ9, what is the probability of completing to a full house? I've worked on this for ages :( and still dont have an answer I trust. - Michael from Perth
You could complete the full house with an ace and a K, Q, or 9. There are 2 aces left and 3 each of K,Q, an 9. So, there are 2*3*3=18 such combinations. The only other way would be a K, Q, or 9 pair. There are 3*combin(3,2)=9 such combination. The number of all combinations is 47*46/2 = 1,081. So the probability is (18+9)/1,081 = 2.50%.
Feb. 1, 2006
Thanks for the help your site has given. You've probably saved me thousands. I was playing in a NL Texas Hold'em tournament on-line recently, and was
dealt pocket kings (at a 10-man table) only to be dominated by pocket aces. I'd like to know the probability, given the condition that you have a pair, of at least one other player at the 10-man table having a higher pair than yours (in other words, having a "dominated pair").
Thanks again! - Dan from Cairo, EgyptThe following table shows estimated probabilities that a pair will be beaten by at least one higher pair according to the number of players (including yourself). These probabilities are not exact because the hands are not independent. However, to find the exact probabilities would get complicated and I think these are pretty close. My formula is 1-(1-r*combin(4,2)/combin(50,2))(n-1), where r=number of higher ranks than your pair and n = total number of players. The table shows the probability of another player having a pair of aces, when you have a pair of kings, in a 10-player game, to be 4.323%. Nov. 9, 2005
What are the odds of a pair showing on the board on the flop? ie., A A 10 or 5 Q 5, etc. - Stephen from Addison
13*12*combin(4,2)*4/combin(52,3) = 3,744/22,100 = 16.941%.
Nov. 2, 2005
I have been playing online poker for a while and I have noticed that premium hands and especially premium pocket pairs (As,Ks,Qs,Js) come up very often. I am aware that you should be dealt pocket aces on average once every 220 hands, but I think it occurs more often than this. I have friends who have (allegedly) tracked their pocket ace occurrence and have an average somewhere in the neighborhood of once every 125-150 hands, over 10,000 hands. I have a feeling this may below standard deviation and may in fact be very unlikely in an honest game. I was wondering if you have ever conducted any research yourself on this subject? Also what are the odds of having an average of pocket aces every 150 hands over the course of 10,000 hands (67 pocket aces in 10,000 hands)? -- Jonah from Davis, California The probability of pocket aces is combin(4,2)/combin(52,2) = 6/1,326 = 1 in 221. Over 10,000 hands you could expect to get pocket aces 10,000/221 = 45.25 times. The standard deviation over 10,000 hands would be (10,000*(1/221)*(1-(1/221)))1/2 = 6.71149. 67 pocket aces would be (67-45.25)/6.71149=3.24 standard deviations above expectations. The probability of results being this skewed or more is 0.06%, or 1 in 1,673. Oct. 3, 2005 In Texas hold em, if two players are dealt a pocket pair pre-flop, what are the odds of each of these players flopping a set(three of a kind)? - Bob from Cincinnati Let's assume you have a pair of aces. Before considering that the other player has another pair the probability of flopping a three of a kind is the [nc(one ace)*nc(two ranks out of 12)*nc(one suit out of 4)2 + nc(any other three of a kind)]/nc(any three cards), where nc(x) = number of combinations of x. This equals [2*combin(12,2)*42+12*combin(4,3)]/combin(50,3) = (2112+48)/19600 = 11.020%. Now lets assume the other player has any other pair, but not the same as yours. Then probability becomes [2*(combin(11,2)*42 + 11*2*4 + 11*combin(4,3)]/combin(48,3) = 11.4477%. Aug. 21, 2005 Hi - Thank you for your web site. I'm wondering if you can tell me what are the odds if you are dealt QQ that any of the remaining 8 people at the table would be dealt AA, AK, KK, or AQ? Thank you! For any given person the probability of having AA is combin(4,2)/combin(50,2) = 6/1,225 = 0.0049 because there are 6 ways to pick 2 aces out of 4, and 1,225 ways to pick any 2 cards out of the 50 left in the deck. The probability is the same for a pair of kings. For AK the probability is 4*4/1,225=0.0131, because there are 4 ways to get an ace and 4 ways to get a king. For AQ the probability is 4*2/1,225=0.0065, because there are 4 aces but only 2 queens left in the deck. So, the probability any given player will have one of these hands is (6+6+16+8)/1,225 = 0.0294. Now the next step is clearly not perfect because if one player doesn't have one of these hands the odds the next player does is a little bit higher. Forgetting this for the sake of simplicity, the probability no player has one of these hands is (1-0.0294)8 = 78.77%. So the probability at least one player has one of these hands is 21.23%. March 10, 2005 In a 10-handed game of Texas Hold 'em, and the flop is three different ranks, what is the probability that three players have a set? For those unfamiliar with the terminology, each player gets two cards to himself and the three flop cards are shared among all players. So this is the same as asking if you dealt three community cards, all of different ranks, and ten two-card hands, what is the probability that three of the two-card hands would be pairs that match one of the three community cards. If ten people are each dealt two cards from a single deck what is the probability that two players will get a pair of aces? First, there are 10*9/2=45 ways you can choose 2 players out of 10. The probability of two specific players getting four aces is 1/combin(52,4)=1/270,725. So the probability of any two players getting a pair of aces is 45/270,725=0.0001662. Aug. 7, 2003 In hold em poker, what are the odds of being dealt pocket aces? And what are the odds of being dealt pocket aces twice in a row? - Adam from Redding, USA There are 52*51/2 = 1,326 ways to arrange 2 cards out of 52. There are 4*3/2=6 ways to arrange 2 aces out of 4. So the answer is 6/1326 = 1/221. The probability of this happening twice in a row is (1/221)2 = 1 in 48,841. Nov. 3, 2002 Thank goodness I just discovered your great site. I have been trying to solve the following, and keep getting different answers. If I am dealt a pocket pair (in Hold'em) what are my chances of getting either three of a kind, or four of a kind on the flop (next three cards). - Elliot Sperber from Harwich, Massachusetts For probability questions I like to take the number of combinations the event you're interested can happen divided by the total number of combinations. First review the combin function in my probabilities in poker section. The number of ways to get a four of a kind is simply the number of singletons in the deck, or 48. The number of ways to get a three of a kind (not including a full house) is the product of the number of ways to get the third card, 2, and the number of ways to get two other singletons, 2*combin(12,2)*42 = 2,112. The total number of ways the cards can come up in the flop are combin(50,3)=19,600. So, the probability of a four of a kind is 48/19,600=0.0024, and the probability of a three of a kind is 2,112/19,600=0.1078. June 6, 2001 With a 52-card deck, what are the odds of drawing a pair of Jacks? -- Rick from Gardnerville, USA My section on probabilities in poker indicates the probability of drawing a pair in five card stud to be 42.256903%. To get the probability of drawing a pair of jacks divide this by 13 to get 3.250533%. Sept. 10, 2000 This page covers probabilities only.I have a separate page for general Texas Holdem questions.
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