# Poker - FAQ

What are the odds in drawing three cards to a pair and getting a full house at five-card draw poker?

Nick

There are two ways to get a full house in this situation: (1) draw a three of a kind or (2) draw one more to the pair and another pair. I'm going to assume you discard three singletons.

First, lets work out the number of combinations under (1). There are 3 ranks with only 3 suits left (remember you discarded 3 singletons) and 9 ranks with 4 suits left. The number of combinations is thus 3*combin(3,3)+9*combin(4,3) = 3*1 + 9*4 = 39.

Next, let's work out the number of combinations under (2). There are 2 suits left to add to the existing pair. There are combin(3,2) ways to form a pair from the 3 ranks with 3 cards left and combin(4,2) ways to form a pair to from the ranks with 4 cards left. So the total combinations under 2 is 2*(3*combin(3,2)+9*combin(4,2)) = 2*(3*3 + 9*6) = 126. The total number of ways to arrange a full house is the sum under (1) and (2), or 39+126=165. There are combin(47,3)=16,215 ways to arrange the 3 cards on the second draw. The probability of drawing a full house is the number of ways to draw a full house divided by the total combinations, or 165/16,215 = 0.0101758, or about 1 in 98.

For more information on the combin() function, please see my section on probabilities in poker page.

I started playing poker with my friends once a week (five-card draw, stud, seven-card stud). We have seven players at the table. It seems to me the probability of getting the hands would be reduced dramatically due to the numbers of players being allocated cards from a 52 card deck. Do you have a mathematical formula that could direct me in the proper direction?

Tim from Santa Rosa, California

No, the probability of getting any given hand is the same regardless of how many other players are at the table. An unseen card is an unseen card, it doesn't matter if another player has it or it is still in the deck.

I was recently told a story that I could not believe!! A friend of mine told me that at a friendly poker game at his house, he and his friend both pulled a natural straight flush in the same hand without drawing any cards!! (in five card draw) I find this hard to believe and from your site I computed the odds of one straight flush to be approx 65,000 to one, what would the odds of 2 in one hand be with 6 players in the game (without drawing any cards??)

R.E. from New York

I'm going to give an approximate answer by assuming that each player was dealt a hand from a separate deck. This should not change the odds much. The probability of any one player drawing a straight flush as found in my section on probabilities in poker is 36/2,598,960. Lets call this probability p. The probability of two players drawing a straight flush is combin(6,2)*p^{2}*(1-p)^{4} = .0000000028779. In other words, the odds against this happening are 347,477,740 to 1.

Love your site, I have a math degree and I am a blackjack counter who has made numerous trips to Vegas and I want to start to use my math skills to play poker. I have watched poker from a distance in Vegas and would appreciate any advice/explanations of the rules of poker. Can I gain an advantage similar to counting in blackjack?

Kal from Chicago

First, let me say that I'm not an expert on poker. It is no big secret that Texas Hold 'Em is the most popular form. In this game there are five community cards and only two down cards per player so a person good at calculating probabilities has more to go on. However, even the best math genius may make a bad poker player if he can't read the other players or the other players can easily read him (both of which I think are true in my case).

What are the odds of being dealt a Royal Flush? Of being dealt a SEQUENTIAL Royal Flush (forwards or backwards)?

Ric from Torrance, California

The probability of any royal flush is the number of possible royals, which is four (one for each suit), divided by the number of ways to **choose** 5 cards out of 52, which is combin(52,5)=2,598,960. So, the answer is 4/2,598,960 = 0.00000153908, or 1 in 649,740.

The probability of a sequential royal flush equals (number of suits) * (number of directions) / (total permutations of 5 cards out of 52) = 4 * 2 / permut(52,5) = 8 / 311,875,200 = 8 /
number of possible royals, which is four (one for each suit), times the number of directions it can be in divided by the number of ways to **pick** 5 cards out of 52, which is permut(52,5)=311,875,200. So, the answer is 4/311,875,200 = 0.00000002565, or 1 in 38,984,400 .

Where did you get your odds software for seven-card stud?

Ron from Orlando, U.S.

I wrote a program in C++ to test all combin(52,7)=133,784,560 ways to arrange 7 cards out of 52. For each one I formed all combin(7,5)=21 ways to arrange 5 out of 7. Then I scored each one of these hands. The highest score of the 21 ways was the value of the seven-card hand. So, overall, I had to score over 2.8 billion hands, this took the computer all night, if I remember correctly.

I'm a little confused on what beats what in five- and seven-card poker. For example, flush beats a straight and so on. Can you please help me out and let know the full list of what hands beat what in poker. Thanks!

James from USA

Here are the hands from highest to lowest, for both five- and seven-card poker: straight flush, four of a kind, full house, flush, straight, three of a kind, two pair, pair.

Is it still tough for an outsider to get in a poker game in Las Vegas with out running into "teams"? I hear a lot of casinos are shutting down their poker rooms.

Bob from Longmont, Colorado

If you read Dirty Poker by Richard Marcus, you'll likely be paranoid about collusion whenever you play with strangers. However, poker expert Ashley Adams answers this questions as follows:

* I've played in nearly every public card room in Las Vegas and over 100 others around the country. At the lower limits I have never encountered collusion. Once in a 20/40 stud game, I thought that two players might have been colluding. I have heard that it may exist in the higher stakes games (about 20/40). But that typical tourist, playing 1/2 or 2/5 blind no-limit, or 10/20 or lower limit poker, is unlikely to ever encounter this.*

First let me say, I think your web site is really great. I have told a few people about it, and hope they will try it too. I wish you continued success with it. I also liked the link to WinPoker. I liked WinPoker enough to order it. This is a great program. I have a question that I am hoping you can help me with. I have been trying to figure out the number of times each hand in seven-card stud occurs. I have a copy of your seven-card table, but I am interested in the mathematics to arrive at those numbers. I can figure out the five-card numbers, but the seven-card just baffles me. I would like to send an Excel 2000 file with my numbers. I would also like to know how to figure the number of straights in a 53-card deck with a joker. H E L P ! ! !

Stan from Harahan, Louisiana

Thanks for your kind words. I agree that calculating the numbers for seven-card stud is hard. That is why I do it my computer. My program goes through all possible combinations and scores each one. The number of wild straights in pai gow poker is 11*(4^{4}-4)+10*3*(4^{4}-4)=10332. Combined with the 10200 natural straights the total is 20532.

We play three card guts with a pay-the-pot if you hold the highest hand and no one goes in. We play with straights and flushes. What is minimum hand you should go in on? Ace high? Any pair? a high pair? If you remove the straights and flushes, what would the odds be? Can you also explain how you came to your conclusion. Much thanks old wise one!

Tebo from London, UK

Good question. I’ve been toying with doing a section on guts for years. I have a computer program half-way finished. One problem is there are so many ways to play guts that one analysis would only fit a small percentage of games. The dummy hand also makes things much more complicated. On a related note let me suggest a good guts variation. If nobody stays in then you go again, everyone with the exact same cards. Knowing everyone else has a lousy hand will induce players with a marginal hand to stay in. The first time my friends and I adopted this rule everybody went in on the second round.

I noticed that use posted odds for double down stud. Have you found this at any on-line casino’s. I played it in Kansas City, but Biloxi doesn’t have it and my wife loves that game. Thanks for any help you can give me.

Jason from Montgomery, USA

No, I haven’t seen it at any online casinos. The only place I have seen it is Atlantic City. The game seems to be going the way of the dodo bird.

My friend and I have a side bet going on. I said to him that I think that blackjack has the best odds in a casino, he said to me that he thinks poker has the best odds. In a casino, what game do you have the best odds in winning, blackjack or poker?

Jeff from Chicago, IL

Although they are hard to compare I say blackjack is the better bet. It is easy to be a good blackjack player by learning the basic strategy. It is difficult to be a good poker player. Casino poker rooms are often full of very good players just waiting for an inexperienced player to fleece. However some people may be naturally gifted at poker, so take my answer with a grain of salt.

I play 7-stud in a poker room with a bad beat jackpot. The minimum bad beat hand to win the jackpot is four of a kind beat by four of a kind. What would be the probability of this occurring, and how would you go about calculating it?

Bruce from Mahomet, Illinois

The probability of any two specific players both having a four of a kind is (13*COMBIN(12,3)*4^{3}*9*COMBIN(41,3)+13*12*11*4*6*10*COMBIN(41,3)+13*12*4*11*COMBIN(41,3))/(COMBIN(52,7)*COMBIN(45,7)) = 0.000003627723. There are combin(7,2)=21 ways to choose 2 players out of 7. Avoiding the case of 3 or more four of a kinds the probability would be 0.000076182184.

I recently witnessed a strange event. I was watching five card draw poker, where you could only draw a maximum of 2 cards. One player drew 1 card and completed a heart flush. The dealer drew one card, and drew a spade flush. Naturally, the dealer’s flush was higher. There were 3 other players in the game. What are the odds of having two flushes in the same hand?

Ted from Mandeville, USA

Let’s define the probability of a flush of either getting one on the deal, or drawing to a 4-card flush. For the sake of simplicity we’ll assume a player will draw to a pat pair or straight with 4 to a flush. The probability of getting a flush on the deal (not including a straight/royal flush) is 4*(combin(13,5)-10)/combin(52,5) = 5108/2598960 = 0.0019654. The probability of being dealt a 4-card flush is 4*3*combin(13,4)*13/combin(52,5) = 111540/2598960 = 0.0429172. The probability of completing the flush on the draw is 9/47. So the overall probability of getting a 4-card flush and then completing it is 0.0429172*(9/47) = 0.0082182. So the total probability for a flush is 0.0019654 + 0.0082182 = 0.0101836. The probability that exactly 2 out 5 players receiving a flush is combin(5,2)* 0.0101836^{2}*(1-00.0101836)^{3} = 0.001006, or about 1 in 994.

I need to know the odds of someone getting 4 of a kind during a hand of 7 card stud with five players and one deck of cards? I hope you can help me, and Thank You for your time.

Richard from Saint Joseph, USA

There are combin(52,7)=133,784,560 ways to arrange 7 cards out of 52. The number of 7-card sets including a four of a kind is 13*combin(48,3) = 224,848. The 13 is the number of ranks for the 4 of a kind and the combin(48,3) is the number of ways you can choose 3 cards out of the 48 left. So the probability is 224,848/133,784,560 = 0.0017, or 1 in 595.

How do you get the number 4,324 combinations for a Royal Flush playing Seven Card Stud? Also, can you recommend a good book that explains how to perform these calculations?

John

There are 4 suits for the royal and 47*46/2 = 1081 ways to arrange the other two cards. 4*1081 = 4324. The other hands get much messier. I had to use a computer to play through all 133,784,560 ways to choose 7 cards out of 52. Sorry, can’t recommend a book either.

What are the odds in Omaha that at least three of the up cards will be of the same suit?

Anonymous

For those not familiar with the rules there are five up cards. So the question is asking what is the probability that in 5 cards dealt from a single deck, without replacement, that at least three will be of the same suit. There are combin(52,5)=2598960 ways to deal 5 cards out of 52. The number of ways to deal 4 of the same suit is 4*combin(13,5)=1144. The number of ways to deal 4 of a suit is 4*combin(13,4)*39=111540. The number of ways to deal 3 of a suit is 4*combin(13,3)*combin(39,2)=847704. So the total combinations is 960388 and the probability is 36.95%.

If seven players each get seven cards, what is the probability at least one person will get a 7-card flush?

Anonymous

The probability of a single player getting a 7-card flush is 4*combin(13,7)/combin(52,7) = 1 in 19491. The probability of at least one player out of 7 getting a 7-card flush is about 1 in 2785.

What is the probability of being dealt four to a royal?

Anonymous

There are four possible suits for the royal. There are five possible missing cards. The fifth card can be one of 47 other cards. So there are 4*5*47=940 ways to get four to a royal. There are combin(52,5) = 2,598,960 total combinations. So the probability is 940/2,598,960 = 1 in 2,765.

Do you think the Jacks or Better strategy on your site would work well in live poker?

Anonymous

No! Absolutely not!

In a 5 card draw game if a player is sitting out and the dealer accidentally deals him in. Do the odds change? Or since the cards are random the odds are still the same?

Anonymous

The odds are the same.

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?

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.

What is the probability of getting all face cards in five card stud?

Anonymous

(12/52)*(11/51)*(10/50)*(9/49)*(8/48) = 0.00030474, or about 1 in 3282.

In four-card poker, which is more likely a straight or a flush?

Anonymous

Not counting straight flushes and royal flushes the probability of a straight is 1.02% and of a flush is 1.04%. So a flush is slightly more likely.

You specifically mention that the A2345 is the second-highest straight. I saw one hand where the dealer had it but lost to another straight that wasn’t AKQJ10. I didn’t want to ask and cost a player his winnings. Is this a hard-and-fast rule that the dealer just overlooked, or do houses sometimes exclude it from their rules?

Anonymous

There are some casinos that treat A2345 (known as "the wheel") as the lowest straight but most still treat it as the second highest straight. I will make a note that this rule is a generality and not always the case.

Do you think online poker room is "fair" in general? Yes? Maybe? Or don’t ever touch it. I figured it is almost impossible to find out if the casino or other players are cheating you.

Anonymous

I doubt the casino would cheat, why would they? The bigger concern is the other players. It would be very easy for players to collude over the phone or instant messenger. Whether they actually do or not I don’t know. There is probably a greater risk for that at the higher limit tables.

Many Oklahoma Indian casinos can only use "class 2" poker machines, where "skill" is not allowed to be a factor -- does this mean the hands are somehow predetermined? And would the next hand be the same no matter who played it?

Anonymous

Let me explain what a class 2 machine for the benefit of others. It is a slot machine in which the outcome is determined by the draw of bingo balls. If done well (and it often isn’t) the game will play just like a regular slot machine. I have been to two casinos in Tulsa and the closest thing I found to video poker were not class 2 slots but rather "pull tabs." With pull tabs the player makes his bet, presses a button, 5 cards appear on the screen, and a voucher drops if you won anything. You may take that to the cashier. Although there is a pay table for the 5-card stud hand I do not think the cards are dealt randomly. Rather it is just a visual aid to show you how much you won.

What is the probability, over the course of 1 million hands, that there is a royal flush drought extending for 200,000 hands? I'm more interested in the solution than the answer itself.

Anonymous

It isn't often I say this but I tried for hours but the math on this one was simply over my head. So I turned to my friend and math professor Gabor Megyesi. Here is his formula for any "drought" problem.

- Let
*p*be the probability of winning any given hand. - Let
*d*be the length of the drought. - Let
*n*be the number of hands played. - Set
*k=dp*and*x=np*. - If
*k*=1 then let*a*=-1, otherwise find*a*such that*k*=-ln(-*a*)/(1+*a*). (*a*is a negative number, if k>1 then -1 <*a*< 0, if*k*< 1 then*a*< -1, and a needs to be calculated to high accuracy.) [Wizard’s note: This kind of solution can be easily found in Excel using the__Goal Seek__feature under the tools menu.] - if
*k*=1 then let A=2, otherwise let A=(1+*a*)/(1+*ak*). - The probability of no drought of length d in n hands is approximately Ae
^{a}^{x}.

In this particular problem p=1/40391, d=200000, n=1000000, k=4.9516, x=24.758, a=-0.0073337, A=1.03007. So the probability of no drought is 1.03007*e^{-0.0073337*24.758} = 0.859042. Thus the probability of at least one drought is 1-0.859042 = 0.140958.

Here is Gabor Megyesi's full 5-page solution (PDF). Thanks Gábor for your help.

I did a random simulation of 32,095 sets of one million hands. The number with at least one drought was 4558, for a probability of 14.20%.

Suppose you have two five-card poker hands dealt from separate decks. You are told hand A contains at least one ace. You are told hand B contains the ace of spades. Which hand is more likely to contain at least one more ace?

Anonymous

The following table shows the probability of 0 to 4 aces in a totally random hand.

### Ace Probabilities — Random Hand

Aces | Formula | Combinations | Probability |
---|---|---|---|

0 | combin(48,5) | 1712304 | 0.658842 |

1 | combin(4,1)×combin(48,4) | 778320 | 0.299474 |

2 | combin(4,2)×combin(48,3) | 103776 | 0.03993 |

3 | combin(4,3)×combin(48,2) | 4512 | 0.001736 |

4 | combin(4,4)×combin(48,1) | 48 | 0.000018 |

Total | 2598960 | 1 |

Take the sum for 1 to 4 aces we see the probability of at least one ace is 0.341158. The probability of two or more aces is 0.041684.

The probability of there being at least one more ace, given there is at least one, can be restated per Bayes' theorem as probability(two more aces given at least one ace) = probability (two or more aces)/probability(at least one ace) = 0.041684/ 0.341158 = 0.122185.

For those rusty on Bayes' Theorem, it states that the probability of A given B equals the probability of A and B divided by the probability of B, or Pr(A given B) = Pr(A and B)/Pr(B).

The next table shows the combinations and probability for each number of other aces given that the ace of spades was removed from the deck.

### Ace Probabilities — Ace Removed Hand

Aces | Formula | Combinations | Probability |
---|---|---|---|

0 | combin(3,0)×combin(48,4) | 194580 | 0.778631 |

1 | combin(3,1)×combin(48,3) | 51888 | 0.207635 |

2 | combin(3,2)×combin(48,2) | 3384 | 0.013541 |

3 | combin(3,3)×combin(48,1) | 48 | 0.000192 |

Total | 249900 | 1 |

This shows the probability of at least one more ace is 0.221369.

For fun, let's solve it the same question using Bayes' Theorem. Assume that random hands are dealt until one is found containing the ace of spades. The probability of at least one additional ace, given the hand contains the ace of spades, could be rewritten as probability(at least two aces given ace of spades is in hand). According to Bayes' theorem, this equals Probability(hand contains ace of spades and at least one more ace) / Pr(hand contains ace of spades). We can break up the numerator as Probability(2 aces including ace of spades) + Probability(3 aces including ace of spades) + Probability(4 aces). Using the first table this equals 0.039930×(2/4) + 0.001736×(3/4) + 0.000018 = 0.021285. The probability of the ace of spades is 5/52 = 0.096154. So the probability of at least two aces given the aces of spades is 0.021285/0.096154 = 0.221369.

So the probability of two or more aces given at least one ace is 12.22% and given the ace of spades is 22.14%.

How did you arrive at 2072 for the number of straight flushes utilizing 4 cards out of 5 in Four Card Poker?

Anonymous

First I separated the straight flushes into two types, those with four consecutive suited cards and those with five. The number of five cards straight flushes is the number of suits * number of spans (ace to 10 as lowest card) = 4*10 = 40. Four the four card straight flushes there are 11 different spans (ace to jack as the low card). In the case of the A234 and JQKA straight flushes the fifth card can be one of 47 (52 less the 4 cards already removed and the fifth card which would form a 5 card straight flush, which were already accounted for). So there are 4*2*47=376 straight flushes of span A234 or JQKA. Of the other nine there are 46 possible cards for the fifth card (52 less the 4 cards already removed and two for cards that would form a five-card straight flush). So the number of straight flushes of span 2345 to TJQK are 4*9*46=1656. So the total number of 4-card straight flushes are 40+376+1656=2072.

First of all, if there’s a better site on gambling on the web, I sure haven’t seen it! It was also cool to put a name to a face when watching the Travel Channel. We always bring this question up at my monthly game, and decided it was time for an answer. In a 5-card draw game of "trips to win", where you must have 3 of a kind or better to win the pot, if I get dealt 2 pair, is it better to keep only one of the pair, and get 3 new cards to try and match the first pair, or should I keep the 2 pair and get one card to try to match either pair? Assume 6 players at the table, no wild cards, that players can draw 3 cards, four with an Ace, and experience shows that any 3 of a kind will probably win the hand, making pulling the full house not that much more advantageous than just the 3 of a kind. Thanks!

Dave A. from Cincinnati, Ohio

Thanks for the kind words. I’m familiar with this game. Let’s assume your intial hand was JJQQK and you keep the two jacks. The number of ways to get one jack and two other cards on the draw is 2*combin(45,2) = 1980. The number of ways to get two jacks on the draw is 45. The number of ways to get a three of a kind on the draw is 10*4+1 = 41. So the number of ways to improve the hand to a three of a kind or better is 1980+45+41 = 2066. The total number of ways to choose 3 card out of the 47 left is combin(47,3) = 16215. So the probability of improving the hand to three of a kind or better is 2066/16215 = 12.74%. If you kept the two pair the probability of improving to a full house is 4/47 = 8.51%. So assuming a three of a kind would probably win I agree that keeping just one pair (the higher one) is the better play.

What is the probability of getting four aces in four-card stud?

Anonymous

1/combin(52,4) = 1 in 270725.

Dear awesome Mr. Wizard of Odds, I am in complete and utter awe of your statistical acumen. Would you by chance be able to calculate for me the probability of a seven card straight - i.e. A,2,3,4,5,6,7 or 2,3,4,5,6,7,8 or 7,8,9,10,jack,queen,king in a seven card stud. We recognize this is not a real poker hand; however it came up when we were playing and we were wondering if it had a lower probability than a normal full house in seven card stud. Cheers, oh knowledgeable one.

Anonymous

How can I refuse after you buttered me up so nicely? First there are combin(52,7) = 133,784,560 ways to choose 7 cards out of 52, without regard to order. There are 8 possible spans for a 7-card straight (the lowest card could be A to 8). If we had 7 different ranks there are 4^{7} = 16384 ways to arrange the suits. Note that this includes all the same suit, which would form a straight flush. So the probability is 8*16,384/133,784,560 = 1 in 1020.6952.

When are you going to do something on bad beat jackpots?

Anonymous

I get asked about bad beat jackpots about once a month. When I have the time I plan to add a section to my site about it. My hesitation is I’ll get asked about every bad beat jackpot in every poker room in the whole world.

I play a weekly social poker game. We have a guy who insists that dealing 2 or 3 or 5 straight cards to each player at a time is equally as random as dealing one card to each player. I assume that if a deck has been shuffled 6 or 7 times (depending on who you listen to) then he would be correct. But, it you have just finished a hand and shuffle only a couple times, dealing cards in groups or clumps like this would not be random. What do you say?

Mark

I agree with you. If the cards are well shuffled then it doesn’t matter. However if they are poorly shuffled then I think the dealer should deal the cards one at a time so any clumps are broken up among the various players.

You are the greatest! I just stumbled across your site a few days ago. My question is in regard to Boston 5 Stud Poker. I saw this game tonight at Mohegan Sun in Connecticut. Their "Ante Bonus" for a Straight is listed at 8 x Ante Bet instead of the 10 x Ante Bet on your pay table. How will this affect the overall odds for this game? Thanks again, keep up the great work!

Anonymous

Thank you for all the kind words. If you lower the bonus on the straight from 10 to 8 the house edge increases from 3.32% to 3.48%.

In a single-deck game, what is the probability of getting at least one ace and deuce in four cards? This is useful to know for the game of Omaha.

Anonymous

From probability 101 we know Pr(A or B) = Pr(A) + Pr(B) - Pr(A and B). So Pr(A and B) = Pr(A) + Pr(B) - Pr(A or B). Let's let A be getting an ace and B be getting a deuce. Pr(A) = Pr(at least one ace) = 1-Pr(no aces) = 1-combin(48,4)/combin(52,4) = 1-0.7187 = 0.2813. The probability of no deuces would obviously be the same. By the same logic pr(A or B) = Pr(at least one ace or deuce) = 1-Pr(no aces nor deuces) = 1-combin(44,4)/combin(52,4) = 1 - 0.501435 = 0.498565. So the probability of getting at least one ace and deuce is 0.2813 + 0.2813 - 0.498565 = 0.063962.

What is the probability of getting the "dead man’s hand", a two pair of aces and eights?

Anonymous

There are six ways to arrange two suits out of four for each pair. Then there are 44 cards for the singleton. So the number of successful combinations is 6*6*44 = 1584. There are 2,598,960 combinations in total, so the probability is 0.0609%.

Tequila Poker looks like fun. I know you get a lot of questions but I just had to ask: what would you do if you got dealt four aces?

Bill from Columbia, Maryland

The expected value of playing High Tequila is 115.904, while Tequila Poker is only 16. So you definitely play High Tequila.

I have tried to get the exact odds of getting a royal straight flush in a 7-card game. I hit one at Foxwoods the other day.

Randy from North Kingstown

If you mean a 5-card royal and any two other cards the probability is 4*combin(47,2)/combin(52,7) = 4,324/ 133,784,560 = 1 in 30,940.

Have you ever evaluated Spin Poker and does it pay off comparable to regular multi-hand video poker? What is unique about spin poker is that while it is a multi-hand game, is that on the draw, once a card is drawn it is gone and can not come up on another line. While I have done well at this game, I’ve been very uneasy about this aspect of it.

Jeff from San Diego, California

The same can be said about standard video poker, once a card is discarded it can not come back on the draw. Thus the expected return in Spin Poker is the same as conventional video poker with the same pay table.

There’s a variant of 5 card stud pokercalled Soko. It plays just like regular poker, except that there are two additional hand rankings. Above a pair is a 4 card straight, then a four card flush, then two pair. The rankings then proceed normally. Where should a 4 card straight flush rank if added as a hand ranking?

Nathan from Tuscon

The number of ways to make a 4-card straight flush is 4*(9*46 + 2*47) = 2032. There are 3744 ways to make a full house and 624 ways to make a four of a kind. So the four-card straight flush should fall between a full house and four of a kind.

At our (draw) poker game a player held a high card "kicker" to improve his pair on the draw. That is counter-intuitive to me. Does holding a kicker improve your chances of improving a pair (5 card draw poker)?

Jim from Albuquerque, NM

If you hold only the low pair then the probability of improving the hand to two pair or better is 28.714%. If you hold the pair and a kicker the probability of improving to a two pair or better is 25.902%. So the probability of improving to a two pair or better is higher by holding the pair only. However if you assume that you’ll need a *high* two pair or better to win then the probability of achieving that will likely be higher holding the kicker, depending on the specific cards and how you define "high."

I understand you have already answered the probability of getting the "dead man’s hand", a two pair of aces and eights, is 0.0609% on April 3, 2005, but I believe the dead man’s hand is "two black Aces, two black eights and the Queen of clubs" what is the probability of drawing that exact hand from a single standard deck?

Sett from Gold Coast

There is only one way to get that exact hand. So the probability would be 1 in combin(52,5) or 1 in 2,598,960.

I’ve noticed more and more casinos are swapping decks of cards after large payout hands (full house, 4 of a kind). Yesterday one swapped after a straight less than a 1/2 hour after the previous swapping. In Laughlin, they even swapped decks after I hit two 3 of a kinds in a row. Is this typical or are they responding to my betting? The probability theoretically doesn’t change, so are they essentially chasing me away?

Paul from Kent, Washington

I can think of three reasons that a supervisor would swap decks after a big win. The first is that the decks were worn and due to be swapped anyway. The second is they are concerned the deck is flawed somehow. The third is they are "sweating the money" and incorrectly think swapping decks will change your luck. I would bet that the third explanation is the most likely.

What is the probability of getting a four of a kind in Omaha?

Kevin from Calgary, Alberta

For my readers who may not know, a hand of Omaha has nine cards. If the player is allowed to use any nine cards the probability would be (13*combin(48,5)-combin(13,2)*44)/combin(52,9) = 0.00605. However, if the player is forced to use exactly two of his four hole cards the probability is

(13*combin(4,2)*combin(48,2)*combin(2,2)*combin(46,3)-combin(13,2)*combin(4,2)*combin(4,2)*combin(2,2)*combin(2,2)*44)/(combin(52,4)*combin(48,5)) = 0.00288Note that these formulas adjust for the possibility of getting two four of a kinds.

Putting ethical considerations aside for the moment, what is the best way to go about colluding in poker (cash games and tournaments)?

Anonymous

As I have said many times, poker is one of my weakest games when it comes to gambling. For this one I turned to Tony Guerrera, author of Killer Poker by the Numbers, to be published January 2007.

Tony’s response was two pages long. To summarize one technique is to build up a pot with the two colluders reraising each other, in the interests of pulling more money into from other players or driving other players out. In tournament play another technique is to dump chips to just one player. For more details please see Tony’s reply in its entirety.In my regular home game, players often call many different wild games. Usually, there will be 2 wilds (baseball, follow the queen where both the queen and the next card are wild, football) and occasionally only 1 (our version of 3-5-7, follow the queen with only the next card wild.) In these games with 4-8 wild cards potentially out there, which is statistically less likely? 5 of a kind or a straight flush? There is constant argument over this and I would love for a reputable and universally respected source such as yourself to settle the matter. Thank you in advance.

Myles from Valencia

The five of a kind is less likely. I just added a table to my section on poker probabilities detailing the probability of each hand according to each individual rank as wild.

Last night I was playing a friendly hand of poker with relatives and I was more or less trying to get a response out of my husband when I asked, " Do you have a nine?" When all of a sudden the hostess jumped my ass and said I was soliciting hand advice to which I responded no I wasn’t I was just table talking everyone agreed with her but I think they were just pissed because it was my husband. Is there a rule about soliciting hand advice or directly asking a person what they have? I thought you could say whatever you wanted to at a table unless there were rules about cussing. Please let me know.

Rachelle from Lafayette

I believe it isn’t breaking any rules to ask, but to answer the question certainly would. I’m not making any accusations in your case but in general when a couple plays poker together in a home game, rules against collusion are often broken, causing sore feelings among everybody. The usual infraction is guy advising the gal after he already folded. When I lived in California it got so bad with one couple that when I hosted the game I made a rule that they couldn’t both be in the game room at the same time. So maybe the hostess has had trouble with couples playing poker before and overreacted.

I have been watching High Stakes Poker on the Game Show Network and there are two terms that have not been explained by the commentators. One is "Straddle" and the other is "Props". Could you please explain what these terms mean in the context of the Poker Game that is being played? Thanks very Much. By the way, Gambling 101 is a great book. Very nice Job!

Edward from Baltimore, MD

A straddle, often called a “live straddle”, is when the player after the big blind makes a raise before looking at his cards. For example in a $3/$6 game the large blind would be $3 so the straddle would be $6. I asked my friend Jason about the reason for this. He said, “The reason some people do this is to stimulate action in a "tight" game. The person who straddles also has the option to raise after the big blind acts. Card rooms like this and allow it because is almost assures a larger pot and therefore more rake.”

There are two uses of the term "props" in poker. First, a Prop Player is one who is paid by the poker room an hourly wage to play. The reason for this is to keep a certain minimum number of players at each table. For more information this questions is answered in much more depth on poker-babes.com. Second, a Prop Bet is a side bet made among the players, often on the flop. There is an article about it www.fullcontactpoker.com

What are the probabilities in five card stud using a deck with 5 suits instead of 4?

Jason from Egg Harbor Township

### Combinations in Five Suit Poker

Hand | Combinations | Probability | Formula |

Five of a kind | 13 | 0.000002 | 13 |

Straight flush | 50 | 0.000006 | 5*10 |

Four of a kind | 3900 | 0.000472 | 13*12*COMBIN(5,4)*5 |

Flush | 6385 | 0.000773 | 5*(COMBIN(13,5)-10) |

Full house | 15600 | 0.001889 | 13*12*COMBIN(5,3)*COMBIN(5,2) |

Straight | 31200 | 0.003777 | 10*(5^5-5) |

Three of a kind | 214500 | 0.025969 | 13*COMBIN(12,2)*COMBIN(5,3)*5^2 |

Two pair | 429000 | 0.051938 | COMBIN(13,2)*11*COMBIN(5,2)^2*5 |

Pair | 3575000 | 0.432815 | 13*COMBIN(12,3)*COMBIN(5,3)*5^3 |

Nothing | 3984240 | 0.48236 | (COMBIN(13,5)-10)*(5^5-5) |

Total | 8259888 | 1 |

Note that I reversed the order of the full house and flush.

What are the odds of being the dealt 2-3-4-5-7 unsuited? Thanks a lot, the site’s great!

Kevin from Massapequa

Thanks. (4^{5}-4)/combin(52,5) = 1020/2598960 = 1 in 2,548.

What is the probability of getting a straight flush (including the 4 royal) in Omaha poker? Thanks in advance.

Mickey F. from Gambrills, MD

I hope you’re happy, my computer spent five days cycling through all 464 billion possible hands in Omaha. Here are tables for both the high and low hand. For the benefit of other readers, in Omaha the player gets four cards to himself and five community cards. He must use exactly two of his own cards and three community cards to make the best high and low hands. For the low hand, straights and flushes do not count against the player, and aces are always low.

### Omaha High Hand

Hand | Combinations | Probability |

Royal Flush | 42807600 | 0.000092 |

Straight Flush | 368486160 | 0.000795 |

Four of a kind | 2225270496 | 0.0048 |

Full House | 29424798576 | 0.063475 |

Flush | 31216782384 | 0.067341 |

Straight | 52289648688 | 0.112799 |

Three of a kind | 40712657408 | 0.087825 |

Two pair | 170775844104 | 0.368398 |

Pair | 122655542152 | 0.264593 |

All other | 13851662832 | 0.029881 |

Total | 463563500400 | 1 |

### Omaha Low Hand

Hand | Combinations | Probability |

5 high | 7439717760 | 0.016049 |

6 high | 25832342400 | 0.055726 |

7 high | 51687563904 | 0.111501 |

8 high | 76415359104 | 0.164843 |

9 high | 90496557312 | 0.195219 |

10 high | 87800751360 | 0.189404 |

J high | 68526662400 | 0.147826 |

Q high | 39834609408 | 0.085931 |

K high | 13835276928 | 0.029845 |

Pair or higher | 1694659824 | 0.003656 |

Total | 463563500400 | 1 |

You have the odds and combinations listed for five-card stud with one joker fully wild. Would you also post the same for two jokers fully wild, as all decks come with two jokers (1-red, 1-black) and many people play with both used as wilds.

Dave K. from Ohio

Follow this link.

I was recently in a home poker game (Omaha deuces wild) with some family members. There were five players, and went down to two. I was one of the two. The other player had been winning all night. I finally had a good hand. I looked over and in a mocking way called "four sevens." She said that she had four aces, and proceeded to rake the chips. I then corrected myself, and told that I had a straight flush. She then told me that I already called four sevens. I showed my hand and she still insisted that I already called four sevens, and my hand was not good any more. So the question is who won the hand? Clearly the straight flush beats a four of a kind. But did I muck my hand by saying what I said? The money is still on the table.

Don from Lihue, HI

Ultimately, the cards speak. You should have won that hand.

I often play a 6-handed game of Omaha hi/lo. It got me to wondering what the chance is that someone else at the table has both an ace and a deuce given that I also have both an ace and a deuce. If you could figure that probability, it would be much appreciated. Thanks for the great site; I have recommended it to my gambling buddies many times.

David I. from Akron, OH

Let me remind my other readers that in Omaha each player gets four hole cards. I’m going to assume that you had one ace, one deuce, and two other cards of other ranks. Here are the ways and number of combinations another player can get at least one ace and deuce:

1 ace & 1 deuce: 3×3×combin(44,2)=8,514

2 aces & 1 deuce: combin(3,2)×3×44=396

1 ace & 2 deuces: 3×combin(3,2)×44=396

2 aces & 2 deuces: combin(3,2)×combin(3,2)=9

3 aces & 1 deuce: 3×3=9

1 deuce & 3 deuces: 3×3=9

total = 9,321

There combin(48,4)=194,580 total ways to choose 4 cards out of the 48 remaining. So the probability of an opponent getting an ace and deuce is 9,321/194,580 = 4.79%. We can estimate the probability that at least one player out of five opponents will have it as 1-(1-.0479)^{5}=17.83%. This is not exactly right, because the probabilities are not independent among the players.

What is the probability of drawing 3 out of 10 straight flushes, holding three to a straight flush with one gap?

Nick from Tennessee

This is a binomial distribution kind of problem. The general formula is that if the probability of an event is p, and each outcome is independent, then the probability of it happening exactly w out of t trials is combin(t,w)×p^{w}×(1-p)^{t-w}.

In this case, there are 2 ways to make the straight flush. You need the 8 of diamonds and another card of either the 6 or J of diamonds. There are combin(47,2)=1,081 ways to draw 2 cards out of the 47 left in the deck. So, the probability of getting a straight flush in any one hand is 2/1,081 = 0.0018501. The probability of making 3 out of 10 is combin(10,3)×0.0018501^{3}×(1-0.0018501)^{7} = 0.000000750178, or 1 in 1,333,017.

Ever since Black Friday (when three major poker sites were shut down by the U.S. government) the forums are loaded with people claiming to having regularly made six-figure incomes for many years. All of a sudden these people have got the whole country asking themselves, "Why not me?" There have to be SOME losers.

Virgi

The newspapers have plenty of stories of professional online poker players lamenting a lack of a way to make a living too. Indeed, you would think everybody is making money from online poker, both players and operators. However, there have to be losers to pay for it all, but I have yet to hear anyone confess to losing.

So, let me be the first. I've played plenty of online poker, usually $1-$2 to $4-$8 structured games, and I don't need to keep track to know I'm down. I don't even know if I'm strong enough to beat the rake. In my opinion, many online poker sites have become infested with bots and pros, aided with player tracking software, making it tough for the recreational players, like me, to have a chance.

If the U.S. government ever does legalize online poker, and I strongly feel it should, I hope a solid regulatory agency will oversee it and make sure it is only human beings playing on a level playing field.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

According to CardPlayer.com, Amir Lehavot, who is one of the nine players to make the 2013 final table in the World Series of Poker, is selling any winnings above the minimum $733,224 for ninth place at a price of $29,248 for each 1% share. Is that a fair price?

Anonymous

First, let's review the chip stacks.

### 2013 WSOP Final Table Chip Stacks

Player | Chips |
---|---|

JC Tran | 38,000,000 |

Amir Lehavot | 29,700,000 |

Marc McLaughlin | 26,525,000 |

Jay Farber | 25,975,000 |

Ryan Riess | 25,875,000 |

Sylvain Loosli | 19,600,000 |

Michiel Brummelhuis | 11,275,000 |

Mark Newhouse | 7,350,000 |

David Benefield | 6,375,000 |

The next table shows the win for each final outcome in the tournament.

### 2013 WSOP Final Table Prize Money

Place | Win |
---|---|

1st | $8,359,531 |

2nd | $5,173,170 |

3rd | $3,727,023 |

4th | $2,791,983 |

5th | $2,106,526 |

6th | $1,600,792 |

7th | $1,225,224 |

8th | $944,593 |

9th | $733,224 |

Assuming each player is of equal skill, the probability of winning could be estimated as the share of the total chip stack. However, it gets more complicated for every position after that. To help answer the question, I developed my poker tournament calculator.

After putting in the information above, you'll see that Amir has an expected win of $ 3,658,046. Then subtract out the minimum prize of $733,224 for 9th place and you get $2,924,822 in expected non-guaranteed winnings. Each 1% share has a value of $29,248.22. This is conveniently the price quoted in the cardplayer.com article.

By the way, Lehavot finished third, for $3,727,023 in prize money. Subtracting the $733,224 guaranteed money for 9th place and dividing by 100, each 1% share returned $29,938. The original cost per share was $29,248, so each share would have seen a 2.36% profit.

This question is discussed in my forum at Wizard of Vegas.

In the 2013 World Series of Poker final table, J.C. Tran was dealt 161 hands and said that not once did he receive a pocket pair and got ace-king only once. What is the probability of getting only one of these premium hands in 161 hands?

Ibeatyouraces

Probability of a pocket pair = 13*combin(4,2)/combin(52,2) = 5.88%.

Probability of AK = 4^{2}/combin(52,2)= 1.21%.

Probability of either = 5.88% + 1.21% = 7.09%.

Probability of NOT getting either = 100% -7.09% = 92.91%.

Probability of getting either once in 161 hands = 161*0.9291^{160}*0.0709^{1} = 1 in 11,268.

This question is discussed in my forum at Wizard of Vegas.

What is the probability that in a 10-player game of Texas Hold 'Em, four players start with ace-king off-suit?

pokerbum

Let's first ask what is the probability that in a four-player game all four players have any ace-king?

The answer to that question would be (4*4/combin(52,2)) * (3*3/combin(50,2)) * (2*2/combin(48,2)) * (1/combin(46,2)) = 1 in 3,292,354,406.

However, it is possible that some of these ace/king hands will be suited. To be exact, the probability that none of them are suited is 9/24. So lower the probability to 1 in 8,779,611,750.

However, it is a ten-player game, and any of the combin(10,4)=210 sets of four players could be the four with non-suited ace-king. So, multiply that probability by 210 and the answer is 1 in 41,807,675.

This question is raised and discussed on my forum at Wizard of Vegas.

In a two-player game of Texas Hold 'Em, which hand has the best odds against pocket aces of unknown suit?

Mike B.

Assuming both hands go to the end, I show the best competing hand is 5-6 suited. If the suit is not represented in the pair of aces the possible outcomes are:

- Win: 22.87%
- Tie: 0.37%
- Lose: 76.76%

If the suit is represented in the pair of aces (lowering the chance of a flush), the possible outcomes are:

- Win: 21.71%
- Tie: 0.46%
- Lose: 77.83%

Overall, the possible outcomes are:

- Win: 22.290%
- Tie: 0.415%
- Lose: 77.295%