 Ask the Wizard: |
Texas Holdem Poker probabilities |
Hello, Wizard. I read your Texas Hold 'Em questions, and I noticed you calculated a 59.85% chance of seeing an ace or king on the board, while holding pocket Queens. How did you come up with that figure? — Jacob from Atwater, CA
There are combin(50,5)=2,118,760 combinations of five cards out of the remaining 50 in the deck. 42 of those cards are 2-Q. The number of combinations of 5 cards out of 42 is combin(42,5)=850,668. So, the probability of not getting a king or ace is 850,668/2,118,760 = 40.15%. Thus, the probability of getting at least one ace or king is 1-40.15% = 59.85%.
An alternative calculation is 1 - pr(first card in flop is not ace or king) ×
pr(second card in flop is not ace or king) ×
pr(third card in flop is not ace or king) ×
pr(fourth card in flop is not ace or king) ×
pr(fifth card in flop is not ace or king) = 1 -
(42/50) × (41/49) × (40/48) × (39/47) × (38/46) = 59.85%. January 27, 2009
In the 2008 World Series of Poker Motoyuki Mabuchi's quad aces were
beaten by Justin Phillip's Royal flush. Here's the YouTube video. I have a simple question about the odds of this occurring. ESPN and others quoted it as 1 in approximately 2.7 billion. It appears to me that they simply took the published odds of quads occurring, and multiplied them by the odds of a royal flush occurring. Is this the correct method of calculation?
— Wade
I disagree with the 1 in 2.7 billion figure too. As you said, they seemed to calculate the probabilities independently for each player, for just the case where both players use both hole cards, and multiplied. Using this method I get a probability of 0.000000000341101, or about in 1 in 2.9 billion. Maybe the one in 2.7 billion also involves compounding a rounding error on both player probabilities. They also evidently forgot to multiply the probability by 2, for reasons I explain later.
There are three ways four aces could lose to a royal flush, as follows.
Case 1: One player has two to a royal flush, the other has two aces, and the board contains the other two aces, the other two cards to the royal, and any other card.
Example:
Player 1: K J
Player 2: A A
Board: A 9 Q T A
In most poker rooms, to qualify for a bad-beat jackpot, both winning and losing player must make use of both hole cards. This was also the type of bad beat in the video; in fact, these were the exact cards.
Case 2: One player has two to a royal flush (T-K), the other has one ace and a “blank” card, and the board contains the other three aces and the other two cards to the royal.
Example:
Player 1: K Q
Player 2: A 5
Board: A A AJ T
Case 3: One player has one to a royal flush (T-K) and a blank card, the other has two aces, and the board contains the other two aces and the other three cards to the royal flush.
Example:
Player 1: K 5
Player 2: A A
Board: A A QJ T
The following table shows the number of combinations for each case for both players and the board. The lower right cell shows the total number of combinations is 16,896.
| Bad Beat Combinations |
| Case |
Player 1 |
Player 2 |
Board |
Product |
| 1 | 24 | 3 | 44 | 3,168 |
| 2 | 24 | 132 | 1 | 3,168 |
| 3 | 704 | 3 | 1 | 2,112 |
| Total | | | | 8,448 |
However, we could reverse the cards of the two players, and still have a bad beat. So, we should multiply the number of combinations by 2. Adjusting for that, the total qualifying combinations is 2 × 8,448 = 16,896.
The total number of all combinations in two-player Texas Hold 'Em is combin(52,2) × combin(50,2) × combin(48,5) = 2,781,381,002,400. So, the probability of a four aces losing to a royal flush is 8,448/2,781,381,002,400 = 0.0000000060747, or about 1 in 165 million. The probability of just a case 1 bad beat is 1 in 439 million. The simple reason the odds are not as long as reported in that video is that the two hands overlap, with the shared ace. In other words, the two events are positively correlated.
September 29, 2008
In your Ultimate Texas Hold 'Em return table why is a large raise recommended for two-card hands listed in the table that have a negative expected return? For example, suited K/2.
— Charlie Masterson from Quincy, MA
According to my two-player Texas Hold 'Em probabilities, the following are the possible outcomes with suited K/2:
Win 51.24%
Lose 44.82%
Draw 3.94%
My table on Ultimate Texas Hold 'Em shows that the player has the advantage on the Play bet, but a disadvantage on the Ante and Blind bets. In this case, the player is stuck with bad odds on the Ante and Blind. However, his odds are favorable on the Play. So, by making the maximum raise he is getting the most value out of his better than 50% chance of winning. The bad odds on the other two bets bring the overall value under 50%. That value would be even less with a smaller raise. July 21, 2008
In 55,088 hands of poker I had a pair going into the flop 2,787 times. Of those 2,787 I hit a set 273 times. How does that square with expectations? — Linus from Alingsås, Sweden
For readers who may not know, a "set" is a three of a kind after the flop, including a pocket pair.
The probability of not making a set is (48+combin(48,3))/combin(50,3) = 17,344/19600 = 88.49%. So the probability of making a set is 11.51%. In 2,787 pairs you should have made a set 320.8 times. So you are 47.8 sets under expectations. The variance is n × p × (1-p), where n = number of hands, and p = probability of making the set. In this case the variance is 2,787 × .1176 × .8824 = 283.86. The standard deviation is the square root of that, or 16.85. So you are 47.8/16.85 = 2.84 standard deviations south of expectations. The probability of luck this bad or worse can be found in any Standard Normal table, or in Excel as norsdist(-2.84) = 0.002256, or 1 in 443.
September 20, 2007
I play Texas Hold 'Em at Caesars Indiana. They have a Bad Beat Jackpot, which is now quads or better being beat. Both players have to play both hole cards, and there must be four players dealt cards. My question is, what is the probability of any hand being a bad beat hand, assuming all players stay until the end? – Daniel from Louisville
What is the probability of seeing a "broken board" in Texas Hold’Em? That is, five cards on the board where no pair exists, no flush is possible and no straight is possible. Thanks, Tim from Arcata
The number of combinations of five different ranks on the board is combin (13,5)*45 = 1287 × 1024 = 1,317,888.
The probability that these five ranks will represent three suits, two of two, and one of one, is combin(4,2)*2*combin(5,2)*combin(3,2)=360. Combin(4,2) is the number of ways to choose two suits out of four for the suits represented twice. 2 for the two ways to choose the suit represented once. Combin(5,2) for the number of ways to choose two ranks out of five for the first suit of two cards. 45 for the number of ways to choose two ranks out of the three left for the other suit of two.
The probability these five ranks will represent four suits, one of two, and three of one, is 4*combin(5,2)*3*2=240. 4 is the number of ways to choose one suit out of four for the suits represented twice. Combin(5,2) is the number of ways to choose two ranks out of five for that suit of two cards. 3 is the number of ways to choose one rank out of the three left for the first suit of one. 2 is the number of ways to choose one rank out of two for the second suit of one.
There are 45=1024 ways to arrange four suits on five different ranks.
So the probability that no more than two of one suit will be present is (360+240)/1024 = 600/1024 = 58.59%.
There are combin(13,5)=1287 ways to arrange 5 ranks out of 13. The number of these combinations in which no three ranks are within a span of 5 is 79. There is no easy formula for this one. I had to cycle through every combination. So the probability the ranks will be sufficiently spaced apart is 79/1287 = 6.14%.
So, the probability of a broken board is (1317888/2596960)*(600/1024)*(79/1287) = 1.825211%.
I have been challenged on my number of broken straights. Here is a list of all 79 possible.
2378Q
2378K
2379Q
2379K
237TQ
237TK
237JQ
237JK
237QK
2389K
238TK
238JK
238QK
2479Q
2479K
247TQ
247TK
247JQ
247JK
247QK
|
2489K
248TK
248JK
248QK
257TQ
257TK
257JQ
257JK
257QK
258TK
258JK
258QK
267JQ
267JK
267JA
267QK
267QA
267KA
268JK
268JA
|
268QK
268QA
268KA
269JA
269QA
269KA
278QK
278QA
278KA
279QA
279KA
289KA
3489K
348TK
348JK
348QK
358TK
358JK
358QK
368JK
|
368JA
368QK
368QA
368KA
369JA
369QA
369KA
378QK
378QA
378KA
379QA
379KA
389KA
469JA
469QA
469KA
479QA
479KA
489KA
|
March 29, 2007
Last night I played a hand where three players all hit sets on the flop. Luckily for me I had AA against QQ and 22. What is the probability that three players hit sets on the flop? Cheers – Gareth H. from Auckland, NZ
The probability of three different ranks in the flop is combin(13,3)×4 3/combin(52,3) = 0.828235. There are combin(10,3)=120 ways you can choose three players out of ten. Of the three, the probability the first will have a set is 3×combin(3,2)/combin(49,2) = 0.007653061. The probability the second will have a set is 2×combin(3,2)/combin(47,2) = 0.005550416. The probability the third will have a set is combin(3,2)/combin(45,2) = 0.003030303. Take the product of all this and the probability is
0.828235 × 120 × 0.007653061 × 0.005550416 × 0.003030303 = 0.00001279, or 1 in 78,166.
March 18, 2007
Dear Mr. Wizard, I have been recently trying to calculate the probability of getting a flush in Texas Hold 'Em if dealt two suited hole cards? My answer keeps on coming out to be 5.8% but this seems ulimately incorrect, your help would be much appreciated thanks - Nathan S. from New Plymouth
The probability of making a flush, with exactly three cards to the same suit as your hole cards, is combin(11,3)×combin(39,2)/combin(50,5) = 122265/2598960 = 0.057706. The probability of making a flush, with four more cards to the same suit as your hole cards, is combin(11,4)×combin(39,1)/combin(50,5) = 2145/2118760 = 0.001012. The probability of making a flush, with five more cards to the same suit as your hole cards, is combin(11,5)/combin(50,5) = 462/2118760 = 0.000218. The probability of making a flush on the board in another suit is 3×combin(13,5)/combin(50,5) = 3861/2118760 = 0.001822. Add this all up and you get 0.057706 + 0.001012 + 0.000218 + 0.001822 = 0.060759.
December 26, 2006
Consider the 6,5 suited starting hand in hold em. According to your published odds, in a 2-player game, this hand is rated 128th out of 169 possible hands. But in a 10-player game, this hand is rated 61st out of 169 possible hands. I am very curious as to why this is the case. Thanks for your help. – Kevin from Seattle
Much of the value of suited connectors comes from the relatively high probability of forming a straight, flush, or straight flush. These are premium hands, which usually win against any number of players. In a two-player game you only are rewarded with one person’s money, but in a 10-player game you can milk it for a much larger pot.
August 24, 2006
If the flop comes up three of the same suit and I do not have a suit that matches the flop, and there are ten players left at the table, what is the probability of someone having a flush? – David from Seattle
The probability of any one player having a flush is combin(11,2)/combin(49,2) = 55/1176 = 4.68%. Assuming independence between hands, which is not the case, the probability of 9 players not having a flush is (1 – 0.0468%) 9 = 64.98%. So the probability of at least one player having a flush is 1-0.6498 = 35.02%. This is just a quick estimate. If I did a random simulation I think the probability would be just a little bit higher, because of the dependence between hands.
August 24, 2006
Your site is awesome, I've learned so much from it. On the flop in holdem poker and you have 20 cards that give you the winning hand. Why is the percentage of you making your hand by the river 67.5% and not 86% (20/47 + 20/46)? I've looked everywhere and can't seem to figure this out, even though I know it's probably very elementary. I'd appreciate the help, Thanks!! – Kevin from Vernon, NJ
Thanks! Your way is double counting getting two of the 20 cards you need. The probability you get one of your 20 needed cards on the turn is 20/47 = 0.4255. The probability you don’t get it on the turn and then do get it on the river is (27/47)*(20/46) = 0.2498. So the total probability is 0.4255 + 0.2498 = 0.6753.
August 9, 2006
Great site. Asking this for my own personal knowledge. I was playing a 4 person game of texas hold'em. I was dealt pocket aces. I got the royal flush on the river. I was wondering what the odds are of making the royal flush on the river with aces to start? – Rhythmic from Hoquiam, WA
Thanks. Assuming the royal consists of one of your two aces, the number of ways to make a royal by the river is 2*46=92. This would be the two suits in your pocket aces and the 46 possibilities for the extra card. There are combin(50,5) = 2,118,760 ways to deal 5 cards out of 50. So the probability is 92/2,118,760 = 1 in 23,030.
July 31, 2006
In Texas Hold 'em, I'd like to know the probability that another player may have been dealt two suited cards to make a higher flush than my own when there are exactly 3 suited community cards at the river. So, for example, if I've got 89 of spades and the community cards include the 3 7 and K of spades, there are 4 spades larger than my 9 in the deck (TJQA). What's the chance that someone was dealt a hand with two spades including at least one of the higher spades (assuming a 9 handed game)? What if there are only 2 (or 3 or x) spades larger than my largest hole card? Thanks for your help and the great site. – Bob B from Scottsdale
The following table shows the probability for 1 to 8 higher ranks and 2 to 10 players, including yourself. In the case of your example of 4 higher ranks and 9 total players, the probability is 16.89%. The way I calculated these probabilities assumed independence between hands, which is not a correct assumption, but the results should be a close estimate.
| Probability of Higher Flush: Higher Ranks (down) by Total Players (across) |
| Higher Ranks | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 1 | 1.29% | 1.93% | 2.57% | 3.2% | 3.82% | 4.45% | 5.06% | 5.68% | 6.29% |
| 2 | 2.39% | 3.56% | 4.72% | 5.87% | 7.00% | 8.12% | 9.23% | 10.32% | 11.4% |
| 3 | 3.3% | 4.91% | 6.50% | 8.05% | 9.58% | 11.09% | 12.57% | 14.03% | 15.46% |
| 4 | 4.03% | 5.98% | 7.90% | 9.77% | 11.61% | 13.41% | 15.17% | 16.89% | 18.59% |
| 5 | 4.57% | 6.78% | 8.93% | 11.04% | 13.1% | 15.11% | 17.07% | 18.99% | 20.86% |
| 6 | 4.93% | 7.31% | 9.62% | 11.88% | 14.08% | 16.23% | 18.32% | 20.36% | 22.35% |
| 7 | 5.11% | 7.57% | 9.97% | 12.3% | 14.57% | 16.78% | 18.94% | 21.04% | 23.08% |
| 8 | 5.11% | 7.57% | 9.97% | 12.3% | 14.57% | 16.78% | 18.94% | 21.04% | 23.08% |
July 22, 2006
An increasingly common side-bet I'm seeing in Texas Hold 'Em games is for players to bet on the total "value" of the flop, where the value of a card is assigned via Blackjack rules. (ie, A=11, KQJ10=10, others are face) Players have the option to pick one or more total value numbers by putting a side bet in the pot. Play continues until the flop hits a value that is covered by a player (so, a flop of KK4 would send the pot to a player that bet on 24). Mathematically, what is the best number to bet on? Most games I have played in have the stipulation that 30 cannot be bet on, which would make you think it would be the "best" number to bet on, but considering 30 has a very limited number of ways it can be made (three 10-value cards, A-9-10, or A-A-8) I'm not sure that that's true. Also, hands where more players have ten-value and Ace cards are more likely to see a flop. Would you be interested in doing an analysis on this side-bet? – Scott from Philadelphia
I’m going to assume that if nobody wins on the flop that either the bets are refunded or the next flop is used to resolve them, as opposed to the turn and river cards being used. I am going to ignore the rule that if everybody folds then the bet is not resolved. Clearly this rule helps the lower totals but to factor that in the analysis would get complicated and subjective. That said, the table below shows the probability of each total. As you can see, the best bet would be on a total of 23, with a probability of 8.3982%.
| Blackjack Points in Flop |
| Total |
Combinations |
Probability |
| 33 | 4 | 0.000181 |
| 32 | 96 | 0.004344 |
| 31 | 504 | 0.022805 |
| 30 | 840 | 0.038009 |
| 29 | 784 | 0.035475 |
| 28 | 920 | 0.041629 |
| 27 | 1108 | 0.050136 |
| 26 | 1264 | 0.057195 |
| 25 | 1472 | 0.066606 |
| 24 | 1652 | 0.074751 |
| 23 | 1856 | 0.083982 |
| 22 | 1800 | 0.081448 |
| 21 | 1508 | 0.068235 |
| 20 | 1408 | 0.06371 |
| 19 | 1336 | 0.060452 |
| 18 | 1196 | 0.054118 |
| 17 | 1080 | 0.048869 |
| 16 | 896 | 0.040543 |
| 15 | 740 | 0.033484 |
| 14 | 512 | 0.023167 |
| 13 | 352 | 0.015928 |
| 12 | 268 | 0.012127 |
| 11 | 200 | 0.00905 |
| 10 | 136 | 0.006154 |
| 9 | 92 | 0.004163 |
| 8 | 48 | 0.002172 |
| 7 | 24 | 0.001086 |
| 6 | 4 | 0.000181 |
July 11, 2006
First of all I'd like to thank you for an awsome site. Now here's my question: We're playing Texas hold'em and flop a flushdraw with two small cards. We all know the % of hitting the flush. but what we really want is the % of winning the hand. And let's say that we are sure that somebody has a higher card of that suit than us. So my question is what are the % of only one card of that suit shows and not two? Regards - Henrik from Sweden
You're welcome. So you have four to a flush with two on the board after the flop. The probability of getting exactly one of the needed suit is 9*38/ combin(47,2) = 342/1,081 = 31.64%.
Feb. 21, 2006
Party Poker has a high hand jackpot awarded to tables showing down a Royal Flush. What are the odds in a 10 handed Texas Holdem game of a table hitting a Royal Flush? - Dan from San Diego
If we assume all hands are played to the end, the probability of any given player having a royal flush is 4* combin(47,2)/combin(52,7) = 1 in 30,940. To make things simple lets assume each hand is independent. The probability of at least one player in 10 having a royal flush would be 1-(1-(1/30940)) 10 = 0.00032, or 1 in 3,094.
Dec. 6, 2005Dear Wiz, I wish I had your brain. In any event, my struggle with stats continues. I am trying to figure out a formula to figure out the probability in Texas Holdem of getting a flush on the flop, by the turn and by the river, crossed by whether my hole cards are suited or not. I tried C(50,2) / C(47,5) but that didn't work out for a suited pair by the river...I should have paid more attention in school! Thanks! Your biggest fan - Eric from Toronto
Ah, shucks. Thanks for the kind words but I'm not that smart. A couple years ago I took the Mensa entrance exam and didn't make the requisite top 2%. I'm still upset that they refused to tell me how well I did do. On January 13, Jeopardy tryouts are coming to Vegas, for which I have an appointment (and I'm sure I'll blow that too). Anyway, to answer your question here you go:
With suited hole cards:
Flush after flop: combin(11,3)/combin(50,3) = 165/19,600 = 0.842%.
Flush after turn: (combin(11,2)*39/combin(50,3))*(9/47) = 2.096%.
Flush after river: (combin(11,2)*combin(39,2)/combin(50,4))*(9/46) = 3.462%.
With unsuited hole cards:
Flush after flop: 0%
Flush after turn: 2*combin(12,4)/combin(50,4) = 0.430%.
Flush after river: (2*combin(12,3)*39/combin(50,4))*(9/46) = 1.458%.
Here are the comulative probabilities.
With suited hole cards:
Flush by flop: 0.842%.
Flush by turn: 2.937%.
Flush by river: 6.400%.
With unsuited hole cards:
Flush by flop: 0.000%
Flush by turn: 0.430%.
Flush by river: 1.888%.
Nov. 22, 2005
Playing Texas Hold'em with 10 players using a standard 52 card deck, after the first two cards are dealt to each player, what are the odds that the "flop" - the next three cards - will all be the same suit? Does it make a difference if my hand has both cards of the same suit and/or each one a different suit? - Mark from Milford
Before considering your own cards the probability is 2*(2/17) = 23.529%, as I just explained in an earlier question. If you have one of each color the probability that the flop is all the same color is 2*combin(25,3)/combin(50,3) = 2*(2,300/19,600) = 23.469%. If you have two of the same color the probability is (combin(26,3)+combin(24,3))/combin(50,3) = (2,600+2,024)/19,600 = 23.592%. Oct. 18, 2005
Party poker has added a side bet: the odds are 7 to 1 on the flop being all red or all black. (You must choose the correct color.) Is this a sucker bet, or should I be asking how bad of a sucker bet is this? Thanks for the great site. - Kerry T. from Austin, TX
Thanks for the compliment. The probability the flop being a particular color is combin(26,3)/combin(52,3) = 2,600/22,100 = 2/17 = 11.765%. The expected return on this bet is (2/17)*7 - (15/17) = -1/17 = -5.882%.
Oct. 18, 2005
I would deeply appreciate it if you could answer my question for me. I have e-mailed several poker pro's including the Canadian one (Blount). Not one has answered my question. Most of them never even wrote back, including Blount. My question is - Could you please tell me the formula for figuring out the odds and percentages of getting the first two cards in hold'em and the percentage of that particular hand beating the other hands-assuming you know what they are-like you see on TV. I already know the formula and the easy way of figuring out you making your hand after the flop. I'm aware of the poker calculator but I would like to know the formula for my own knowledge. You hear about the pro's knowing all the odds. I'm beginning to think it's a bunch of bunk because not one person has answered my question. Thank you very much for taking all the time to read this. - Don from Niagara Falls, Ontario
There is no easy formula. Personally my program cycles through all the remaining cards and records how the number of hands that win for each player and takes a percentage based on those totals. I imagine everyone else either does that or uses a random simluation based method. Aug. 21, 2005
[Bluejay adds: As for your doubting that pros really know the poker odds because they didn't write back to you -- didn't it occur to you that another likely explanation is that they didn't care to serve as a free helpdesk to the whole world? Britney Spears must be a fraud because she never wrote back to me, either.]
Last night a player offered me a side wager in Texas Hold 'em. He said that at least one face card (or any three ranks) would appear on the flop and offered even money? Should I have accepted the bet?
The number of ways you can choose 3 cards out of the 40 non-face cards is (40*39*38)/(1*2*3) = 9,880. The number of ways you can choose 3 cards out of 52 is
(52*51*50)/(1*2*3) = 22,100. So the probability of not
getting a face card is 9,880/22,100 = 44.71%. Thus, the probability of getting a face is 55.29%. His side of the bet had a 10.58% advantage. April 3, 2005
A Hold'em tournament starts by high-carding for the button. Highest card wins, and spades beats hearts beats diamonds beats clubs. What is the average card that will win in a 10 person table? I've tried simulating it by assigning a number value to each card, but I can't figure it out for the life of me! Thanks and keep it up! - Stephen K., Atlanta GA
To simplify the question let's say the cards were numbered 1 to 52. The following table shows the probability that the 10th to 52nd card is the highest card. There are combin(x-1,9) ways to choose 9 numbers under x and combin(52,10) ways to choose any number numbers out of 52. So, the probability that x is the highest number can be expressed as combin(x-1,9)/ combin(52,10). The expected column is the product of the probability and the number of balls. The sum of the expected column shows us that on average the highest ball will be 48.18. Rounding to the nearest card, the highest expected card is the king of spades.
| Highest of 10 Cards |
| Highest Card |
Probability |
Expected |
| 10 |
0.000000000063 |
0.000000000632 |
| 11 |
0.000000000632 |
0.000000006953 |
| 12 |
0.000000003477 |
0.000000041719 |
| 13 |
0.000000013906 |
0.000000180784 |
| 14 |
0.000000045196 |
0.000000632742 |
| 15 |
0.000000126548 |
0.000001898227 |
| 16 |
0.000000316371 |
0.000005061939 |
| 17 |
0.000000723134 |
0.000012293281 |
| 18 |
0.00000153666 |
0.000027659882 |
| 19 |
0.00000307332 |
0.000058393084 |
| 20 |
0.000005839308 |
0.000116786168 |
| 21 |
0.000010616924 |
0.000222955411 |
| 22 |
0.000018579618 |
0.000408751587 |
| 23 |
0.00003144243 |
0.000723175884 |
| 24 |
0.00005165542 |
0.001239730087 |
| 25 |
0.000082648672 |
0.002066216811 |
| 26 |
0.000129138551 |
0.003357602319 |
| 27 |
0.000197506019 |
0.005332662506 |
| 28 |
0.000296259028 |
0.008295252787 |
| 29 |
0.000436592252 |
0.012661175306 |
| 30 |
0.000633058765 |
0.01899176296 |
| 31 |
0.000904369665 |
0.028035459607 |
| 32 |
0.001274339073 |
0.040778850337 |
| 33 |
0.001772993493 |
0.058508785267 |
| 34 |
0.002437866053 |
0.082887445794 |
| 35 |
0.003315497832 |
0.116042424112 |
| 36 |
0.004463170158 |
0.160674125694 |
| 37 |
0.005950893544 |
0.220183061136 |
| 38 |
0.007863680755 |
0.298819868684 |
| 39 |
0.010304133403 |
0.401861202713 |
| 40 |
0.013395373424 |
0.535814936951 |
| 41 |
0.017284352805 |
0.708658464999 |
| 42 |
0.022145577031 |
0.930114235312 |
| 43 |
0.028185279858 |
1.211967033891 |
| 44 |
0.035646089232 |
1.568427926212 |
| 45 |
0.044812226463 |
2.016550190844 |
| 46 |
0.056015283079 |
2.576703021634 |
| 47 |
0.069640622206 |
3.273109243697 |
| 48 |
0.086134453782 |
4.134453781513 |
| 49 |
0.106011635423 |
5.194570135747 |
| 50 |
0.129864253394 |
6.493212669683 |
| 51 |
0.158371040724 |
8.076923076923 |
| 52 |
0.192307692308 |
10 |
| Total |
1 |
48.181818181818 |
Although you didn't ask, the median card is the ace of clubs. The probability of the highest card falling under the ace of clubs is 41.34%, exactly on the ace of clubs is 10.60%, and higher than the ace of clubs is 48.05%. Feb. 21, 2005
What is the probability that two players will have different four-of-a-kinds in Texas Hold'em?
Between two players there are 9 total cards. These must consist of two four of a kinds and one singleton. The number of combinations for this is combin(13,2)*44 = 3,432. The total number of ways to pick 9 cards out of 52 is combin(52,9) = 3,679,075,400. So the probability you have the right cards, but not necessarily in the right order, is 3,432/3,679,075,400 = 1 in 1,071,992.
However, just because the cards are AAAABBBBC doesn't mean both players will have different four of a kinds. The number of ways to arrange them into a 5-card hand and two 2-card hands is 9!/(5!*2!*2!) = 756. Following are the ways those 9 cards can fall.
| Four of a Kind Bad Beat
Combinations |
| Player 1 |
Player 2 |
Flop |
Mirror Patterns |
Combinations per Pattern |
Total Combinations |
| AA |
BB |
AABBC |
2 |
36 |
72 |
| AA |
AB |
ABBBC |
4 |
48 |
192 |
| AA |
AA |
BBBBC |
2 |
6 |
12 |
| AA |
AC |
ABBBB |
4 |
12 |
48 |
| AA |
BC |
AABBB |
4 |
24 |
96 |
| AB |
AB |
AABBC |
1 |
144 |
144 |
| AB |
AC |
AABBB |
4 |
48 |
192 |
Of these only the first and the fifth group result in both players having a different four of a kind. So the probability that an AAAABBBBC set of cards results in two different four of a kinds is 168/756 = 22.22%.
So the answer to your question is
(3,432/3,679,075,400)*(168/756) = 1 in 4,823,963. On a
more practical note Party Poker has a bad beat jackpot for a losing hand of four eights. Given that there are two four of a kinds the probability that both are eights or greater is combin(7,2)/combin(13,2) = 21/78 = 26.92%. So the probability that any one hand of two players will result in this bad beat jackpot is 1 in 17,917,577. Dec. 5, 2004
Mr. Wizard, First, let me say you have a terrific site! I've been reading it for a while now. I checked the poker questions, and didn't see this one. Another site claims this, "In Texas Hold'em, the probability of AK dealt pre-flop and hitting an A or K by the river is 1 in 2
(even)." This seems intuitively way too high. What are your
thoughts? Thanks again! - John
Thanks to you too for the kind words. For those not familiar with hold'em this question is akin to asking if a player were dealt an ace and a king plus five random cards from the remaining 50 cards, what is the probability the player would pair up the king and/or ace. Of the other 50 cards 44 of them are not kings or aces. The number of ways to draw any five cards out of 44 is combin(44,5) = 1,086,088. The number of ways to draw any five cards out of all 50 is combin(50,5) = 2,118,760. So the probability of not pairing up the ace and/or king is 1,086,088/2,118,760 = 51.26%. Thus, the probability you will pair up is 1-51.26% = 48.74%. This is pretty close to 1 in 2. Nov. 19, 2004
How would you figure the probability of getting a 4-card flush or better on the flop in holdem, if your hole cards are suited.
The probability of getting 2 more of the same suit is 39*combin(11,2)/combin(50,3) = 0.109439. The probability of getting 3 more of the same suit is combin(11,3)/combin(50,3) = 0.008418. So the probability of getting at least 2 more of the same suit is the sum of these, 0.117857. Oct. 25, 2004
In the initial two cards can you tell me what the odds are of receiving 7 hands of Ace King or better at hold'em in 35 hands? June 25, 2004
The probability of receiving ace/king is
(8/52)*(4/51) = 0.012066. The probability of receiving
any pair is (3/51) = 0.058824. So the probability of a pair or better is 0.07089. The probability of receiving exactly seven hands of ace/king or better is combin(35,7)*(.07089)^7*(1-.07089)^28 = 0.00772. To work out the probability of 7 or more we would have to go through a total of 7 to 35 one at a time. This adds up to 0.010366551.
Are the probabilities for the various hands the same in Texas Hold 'em as in Seven-card stud or are they different somehow due to the community cards? Could you please explain why or why not?
Yes, the probabilities are the same. Seven random cards out of 52 have the same odds regardless of how they are taken out of the deck or whom you share them with. May 22, 2004
In Texas Hold 'em if the flop cards are all the same suit what is the probability any given player will hold two more cards of the same suit? - Dr. Don
There are 10 cards left of the same suit, and 49 left in the deck. So the probability is combin(9,2)/combin(49,2) = 36/1,176 = 0.0306. Mar. 24, 2002
In Texas Hold em, what are the odds of making a one gap or two gap inside straight by fifth street starting from the flop? - Mr. D from Cherry Hill, USA
For the benefit of my readers unfamiliar with this game or the terms, this question asks what is the probability of filling in a one or two gap inside straight with two more cards, with 47 left in the deck. With one gap the probability is 1-combin(43,2)/combin(47,2) = 0.164662. With two gaps the probability is 42/combin(47,2)=0.0148. March 17, 2002
What are the odds of making a royal flush in Texas hold-em on the river? - John Hinrichs from Trinity, Texas
For those readers who don't know, the river is the fifth and last community card in Texas hold-em. The player must make the best poker hand between his own two cards and the five community cards. So you're asking what is the probability that a player will form a royal flush in seven cards, and that the seventh card dealt will be part of the royal. The probability of forming a 5-card royal flush out of 7 cards, before considering card, is 4*combin(47,2)/combin(52,7) = 4,324/133,784,560, or 1 in 30,940. The probability that the seventh card will be part of the royal flush is 5/7. So the final probability is 21,620/936,491,920, or 1 in 43,316. Jan. 15, 2002
What are the statistical odds of getting a flush in Texas hold 'em. Is it easier to get a flush in 7-card stud or in holdem as a player. - Kevin from Richmond, USA
You can refer to my section on probabilities in poker to see the probability is 3.03%. The odds are the same in both Texas Hold'em and 7-card stud. June 13, 2001
What are the odds of having two four of a kind's and a straight flush dealt to the same player in a hold 'em game with 10 players in 50 hand? ñ Paul from Toronto, Canada
The probability of a four of a kind in seven cards is 0.00168067, the probability of a straight flush is 0.00027851. If x is the probability of a four of a kind and y is the probability of a straight flush then the probability you ask for is combin(50,2)*48*x2*y*(1-x-y)47. The answer comes out to .0000421845, or 1 in 23,705. Dec. 2, 2000
This page covers probabilities only.
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