Ask the Wizard: Texas Hold ’em - Probability - General
What are the odds of having two four of a kinds and a straight flush dealt to the same player in a Texas hold 'em game with ten players in 50 hands?
Four of a Kind Bad Beat Combinations
Combinations per Pattern
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.
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?
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 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%.
Dear 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 of getting a flush on the flop, by the turn and by the river (texas holdem) 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
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%.
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
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
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
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
|Probability of Higher Flush|
Higher Ranks (down) by Total Players (across)
|Probability of Higher Flush|
Higher Ranks (down) by Total Players (across)
| 2378Q |
| 2489K |
| 268QK |
| 368JA |
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?
— Danielle from Louisville
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
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.
In the 2008 World Series of Poker Motoyuki Mabuchi's quad aces were beaten by Justin Phillip's Royal flush. 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?
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.
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.
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.
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|
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.
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.
Hello, I am a seventh grader from Hawaii. I am doing a science fair project on poker and shuffling. I am using your charts about the win percentage of pocket Texas Hold ’Em hands, according to the number of players in the game. I was hoping you could answer a few questions that would help me with my project:
- 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 well-known? Why or why not?
- The two-player table was done by a brute-force 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 poker-based 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 mathematically-based 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.
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, CAThere 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%.
Holding two suited cards in Texas Hold ’em, what are my odds of getting exactly two more cards of the same suit on the flop?
— Jack H. from Duncanville, TXThere 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%.
I was involved in a hand of online poker and would like to know the odds of this happening, please:
— David T. from Montego BayNormally 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 two-player 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 six-player game, so there are combin(6,2) = 15 different player pairs. In a six-player 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.
There is a promotion being advertised by a Las Vegas card room: Make a flush in all four suits and you get $400. You have to use both of your hole cards, and there is a five-hour time limit. Assuming 35 hands per hour, and that the clock starts with the first flush, what is the probability of achieving the other three flushes within five hours? Thanks.
— AnnieLet’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 (1-0.003764605)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 over-subtract 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 (1-2×0.003764605)175=0.266442448.
The probability of going 175 hands and never getting any of the three suits left is (1-3×0.003764605)175=0.137015266.
So the answer is 1-3×0.51682599 + 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 .
Doyle Brunson famously won the Main Event in the World Series of Poker in both 1976 and 1977. Each time he held 10-2 as hole cards, and both times he made a full house on the river. What are the odds of that?
— 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 10-2 both times is 1 in 295,379,826.
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