Wizard Recommends
Texas Hold ’em  Probability  General
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*x^{2}*y*(1xy)^{47}. The answer comes out to .0000421845, or 1 in 23,705.
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 7card stud.
John from Trinity, Texas
For those readers who don’t know, the river is the fifth and last community card in Texas holdem. 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 5card royal flush out of 7 cards, before considering card, is 4*combin(47,2)/combin(52,7) = 4324/133784560, or 1 in 30940. The probability that the seventh card will be part of the royal flush is 5/7. So the final probability is 21620/936491920, or 1 in 43316.
Mr. D from Cherry Hill, USA
For the benefit of my readers, 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 1combin(43,2)/combin(47,2) = 0.164662. With two gaps the probability is 4^{2}/combin(47,2)=0.0148.
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) = 36/1176 = 0.0306.
"Anonymous" .
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.
"Anonymous" .
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.
"Anonymous" .
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 0.117857.
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 1086088/2118760 = 51.26%. Thus the probability you will pair up is 151.26% = 48.74%. This is pretty close to 1 in 2.
"Anonymous" .
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 = 3432. 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 3432/3,679,075,400 = 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 5card hand and two 2card hands is 9!/(5!*2!*2!) = 756. Following are the ways those 9 cards can fall.
Four of a Kind Bad Beat Combinations
36
Player 1 
Player 2 
Flop 
Mirror Patterns 
Combinations per Pattern 
Total Combinations 
AA 
BB 
AABBC 
2 
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 (3432/3,679,075,400)*(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.
"Anonymous" .
The number of ways you can choose 3 cards out of the 40 nonface cards is (40*39*38)/(1*2*3) = 9880. The number of ways you can choose 3 cards out of 52 is (52*51*50)/(1*2*3) = 22100. So the probability of not getting a face card is 9880/22100 = 44.71%. Thus the probability of getting a face is 55.29%. His side of the bet had a 10.58% advantage.
Mark from Milford
Before considering your own cards the probability is 4×combin(13,3)/combin(52,3) = 5.1764706%.
Another way to look at it is the probability second card in the flop will match the first in suit is (12/51). The probability third card in the flop will match it is (11/50). (12/51)×(11/50)=5.1764706%.
The odds change a little if you consider your own cards. If you have two cards of the same suit, then the probability of a suited flop is pr(flush in same suit) + pr(flush in a different suit) = combin(11,3)/combin(50,3) + 3×combin(13,3)/combin(50,3) = 5.2193878%.
If you have two cards of a different suit, then the probability of a suited flop is pr(flush in suit in common) + pr(flush in a different suit) = 2×combin(12,3)/combin(50,3) + 2×combin(13,3)/combin(50,3) = 5.1632653%.
Eric from Toronto
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 am 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) = 165/19600 = 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 commulative 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%.
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 = 1081. So the probability is (18+9)/1081 = 2.50%.
Shane from Santa Rosa
Thanks for the kind words. The probability you will get pocket aces in any one hand is 6/1326, or once every 221 hands. According to my 10player Texas Hold ’em section (/games/texasholdem/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 10player 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.
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) = 342/1081 = 31.64%.
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 16.45%. 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 FlushHigher Ranks (down) by Total Players (across)Wizard Estimate
Higher Ranks  2  3  4  5  6  7  8  9  10 
1  0.71%  1.41%  2.11%  2.80%  3.49%  4.17%  4.85%  5.52%  6.19% 
2  1.31%  2.61%  3.89%  5.15%  6.40%  7.62%  8.84%  10.03%  11.22% 
3  1.82%  3.60%  5.36%  7.08%  8.77%  10.43%  12.05%  13.65%  15.22% 
4  2.22%  4.40%  6.52%  8.60%  10.63%  12.61%  14.56%  16.45%  18.31% 
5  2.53%  4.99%  7.39%  9.72%  12.00%  14.23%  16.39%  18.50%  20.56% 
6  2.73%  5.38%  7.96%  10.47%  12.91%  15.29%  17.60%  19.85%  22.03% 
7  2.83%  5.58%  8.25%  10.84%  13.36%  15.81%  18.20%  20.51%  22.76% 
8  2.83%  5.58%  8.25%  10.84%  13.36%  15.81%  18.20%  20.51%  22.76% 
Probability of Higher FlushHigher Ranks (down) by Total Players (across)Miplet Simulation
Higher Ranks  2  3  4  5  6  7  8  9  10 
1  0.69%  1.41%  2.12%  2.83%  3.54%  4.25%  4.96%  5.66%  6.36% 
2  1.28%  2.60%  3.91%  5.21%  6.48%  7.77%  9.05%  10.31%  11.56% 
3  1.79%  3.61%  5.41%  7.19%  8.92%  10.68%  12.41%  14.11%  15.79% 
4  2.19%  4.40%  6.58%  8.75%  10.85%  12.94%  15.01%  17.04%  19.04% 
5  2.50%  5.00%  7.47%  9.92%  12.28%  14.63%  16.94%  19.21%  21.43% 
6  2.71%  5.40%  8.06%  10.69%  13.23%  15.74%  18.23%  20.65%  23.03% 
7  2.81%  5.61%  8.36%  11.08%  13.70%  16.30%  18.86%  21.37%  23.82% 
8  2.81%  5.61%  8.36%  11.08%  13.70%  16.30%  18.86%  21.37%  23.82% 
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.
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 10.6498 = 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.
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.
Tim from Arcata
The number of combinations of five different ranks on the board is combin (13,5)*4^{5} = 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. 4^{5} 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 4^{5}=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 
Danielle from Louisville
My new Bad Beat Jackpot section shows the probability of this kind of bad beat in a 10player game to be 0.0000108, or about 1 in 93,000.
Charlie Masterson from Quincy, MA
According to my twoplayer 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.
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:
Player 2:
Board:
In most poker rooms, to qualify for a badbeat 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 (TK), 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:
Player 2:
Board:
Case 3: One player has one to a royal flush (TK) 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:
Player 2:
Board:
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 twoplayer 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.
You are absolutely right, according to the paper Telling the Truth about New York Video Poker. The player’s outcome is indeed predestined. Regardless of what cards the player keeps, he can not avoid his fate. If the player tries to deliberately avoid his fate, the game will make use of a guardian angel feature to correct the player's mistake. I completely agree with the author that such games should warn the player that they are not playing real video poker, and the pay table is a meaningless measure of the player's actual odds. It also also be noted these kinds of fake video poker machines are not confined to New York.
 How did you come up with the percentages found in the charts?
 If you used a computer program, how did you develop it and how long did it take?
 You stated that you started the Wizard of Odds as a hobby. Did experimenting change as your site became more wellknown? Why or why not?
"Anonymous" .
 The twoplayer table was done by a bruteforce looping program, that cycled through all 1225 possible opponent cards, and 1,712,304 possible community cards. For three to eight players, looping would have taken a prohibitive amount of time, so I did a random simulation.
 I write almost all my programs in C++, including both programs I just mentioned. The rest are in Java or PERL. I mostly copied and pasted code from other pokerbased programs. The new code only look about a day to write.
 Yes, I started my site as a hobby in June 1997. It wasn’t until January 2000 that I accepted advertising, and tried to make a business out of it. It has gone through three different domains over the years. Here is what it looked like in May 1999. The purpose of the site has always remained the same, a resource for mathematicallybased gambling strategy. Through the years, I have just been adding more games and material. One experiment was providing my NFL picks for the 2005 season, which was an abject failure.
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 2Q. 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 140.15% = 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%.
Jack H. from Duncanville, TX
There are combin(11,2)=55 ways to get two more cards of the same suit, and 39 for the unsuited card. There are combin(50,3)=19,600 total possible combinations of cards on the flop. So, the probability of having exactly four to a flush after the flop is 55×39/19,600 = 10.94%.
David T. from Montego Bay
Normally I'm sick of bad beat questions, but this one was too painful to ignore. Before the first card is dealt, the probability of four kings being beaten by four aces, in a twoplayer game, with both players having pocket pairs, is 2*combin(4,2)*combin(4,2)*44/(combin(52,2)*combin(50,2)*combin(48,5)) = 2*6*6*44/(1326*1225*1712304) = 1 in 877,961,175. This was a sixplayer game, so there are combin(6,2) = 15 different player pairs. In a sixplayer game, the probability is 15 times higher, or 1 in 58,530,745. After the indicated hole cards are dealt, and before the flop, the probability is 1 in 38,916 that the hand will finish as it did.
Annie
Let’s say your first flush is in spades. At 35 hands per hour, in five hours 175 hands could be played. You then have 175 hands to make a flush in hearts, diamonds, and clubs. I’m going to assume the player never folds a hand that has a possibility of attaining a flush in one of the suits he needs.
The probability of a flush of a specific suit, let’s say hearts, using both hole cards is combin(13,2)×[combin(11,3)×combin(39,2) + combin(11,4)×39 + combin(11,5)]/(combin(52,2)×combin(50,5)) = 10576566/2809475760=0.003764605. In the next 175 hands the probability of missing a heart flush would be (10.003764605)^{175}=0.51682599.
It would be incorrect to say the probability of failing to make the other three suits would be pr(no heart flush)+pr(no dimaond suit) + pr(no club flush), because you would double counting the probability of faling to make two of them. So you should add back in pr(no heart or diamond flush) + pr(no heart or club flush) +pr(no club or diamond flush). However, that would incorrectly oversubtract the probability of not making all three flushes. So you should add back in pr(no club, diamond, or heart flush).
The probability of going 175 hands and never get either of two specific suits is (12×0.003764605)^{175}=0.266442448.
The probability of going 175 hands and never getting any of the three suits left is (13×0.003764605)^{175}=0.137015266.
So the answer is 13×0.51682599 + 3×0.266442448  0.137015266 = 0.111834108.
I would like to thank dwheatley for his help with this problem. It is discussed on my bulletin board at Wizard of Vegas.
Jonathan F.
Given two cards of different ranks, the probability of making a full house are 1 in 121.6. The odds of making it on the river are 1 in 207.
The odds of making such a hand on the river two out of two times is 1 in 43,006.
The odds of this happening with the same two starting cards, in rank only, are 1 in 3,564,161.
The odds of this happening with exactly 102 both times is 1 in 295,379,826.
AZDuffman
The answer and solution appear in the following spoiler tag.
There are four possible states you can be in at any given time:
 State 1: The first hand or any hand where the last hand was not a pocket pair.
 State 2: Last hand was a pocket pair.
 State 3: Last two hands were the same pocket pair.
 State 4: Three of the same pocket pairs in a row has already been achieved.
If you're in state 1, you can advance to state 2 with a probability of 3/51. Otherwise, you stay in state 1.
If you're in state 2, you can advance to state 3 with a probability of (4/52)×(3/51). Otherwise, you go back to state 1.
If you're in state 3, you can advance to state 4 with a probability of (4/52)×(3/51). Otherwise, you go back to state 1.
If you're in state 4, you stay there.
That said, you can create your transition matrix, T, as follows:
0.941176  0.058824  0.000000  0.000000 
0.941176  0.054299  0.004525  0.000000 
0.941176  0.054299  0.000000  0.004525 
0.000000  0.000000  0.000000  1.000000 
There are 120 total hands played, so find T^120.
0.941044  0.058549  0.000265  0.000141 
0.941025  0.058548  0.000265  0.000162 
0.936786  0.058284  0.000264  0.004666 
0.000000  0.000000  0.000000  1.000000 
The upper right cell shows us the probability that starting at state 1 will lead us to state 4 after 120 starting hands in a threehand sequence, which is 0.000141471.
Take the inverse of that number, the probability is 1 in 7068.605131.
This question is raised and discussed in my forum at Wizard of Vegas.