# Poker - FAQ

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.

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.

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.

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).

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 .

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.

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.

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.*

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.

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.

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.

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.

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.

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.

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.

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.

"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%.

"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.

"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.

"Anonymous" .

No! Absolutely not!

"Anonymous" .

The odds are the same.

"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" .

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

"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.

"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.

"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.

"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.

"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%.

"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%.

"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.

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.

"Anonymous" .

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

"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.

"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.

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.

"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%.

"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.

"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%.

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.

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.

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.

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.

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."

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.

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.

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.

"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.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.

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.

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.

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.

Kevin from Massapequa

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

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 |

Dave K. from Ohio

Follow this link.

Don from Lihue, HI

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

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.

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.

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.

"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.

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.

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.

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%

"Anonymous" .

It depends on the number of players at the table. The more the better, because you'll have a greater chance of losing with more players. The following table shows the probability of each of the four pairs losing by the total players at the table, including yourself. This assumes that nobody folds. This is obviously an unrealistic assumption, so I would take these probabilities as an upper bound.

### Probability of Losing in Texas Hold 'Em

Players | Jacks | Queens | Kings | Aces |
---|---|---|---|---|

10 | 80.16% | 77.34% | 73.57% | 68.64% |

8 | 74.87% | 71.29% | 66.74% | 60.95% |

6 | 65.95% | 61.70% | 56.68% | 50.49% |

4 | 50.37% | 46.09% | 41.41% | 35.82% |

3 | 38.43% | 34.71% | 30.79% | 21.22% |

2 | 22.85% | 20.37% | 17.88% | 15.07% |

The average win is easily calculated as $100 × (1/6) + $75 × (1/6) + $50 × (1/2) = $62.50. That said, the next table shows the expected value of each of the four pocket pairs whenever they occur, assuming no other players fold.

### Expected Win per Occasion

Players | Jacks | Queens | Kings | Aces |
---|---|---|---|---|

10 | $50.10 | $48.34 | $45.98 | $42.90 |

8 | $46.79 | $44.56 | $41.71 | $38.09 |

6 | $41.22 | $38.56 | $35.43 | $31.56 |

4 | $31.48 | $28.81 | $25.88 | $22.39 |

3 | $24.02 | $21.69 | $19.24 | $13.26 |

2 | $14.28 | $12.73 | $11.18 | $9.42 |

The next table shows the value of this promotion per hand played. It is simply the product of the table above and the probability of getting the required hold cards, which is 6/1326 = 0.90%.

### Expected Win per Hand Played

Players | Jacks | Queens | Kings | Aces |
---|---|---|---|---|

10 | $0.23 | $0.22 | $0.21 | $0.19 |

8 | $0.21 | $0.20 | $0.19 | $0.17 |

6 | $0.19 | $0.17 | $0.16 | $0.14 |

4 | $0.14 | $0.13 | $0.12 | $0.10 |

3 | $0.11 | $0.10 | $0.09 | $0.06 |

2 | $0.06 | $0.06 | $0.05 | $0.04 |

The next table shows the value of this promotion per hour played, assuming a rate of 30 hands per hour. Again, this assume nobody ever folds, so I would take this as an upper bound on the value per hour.

### Expected Win per Hour Played

Players | Jacks | Queens | Kings | Aces |
---|---|---|---|---|

10 | $6.80 | $6.56 | $6.24 | $5.82 |

8 | $6.35 | $6.05 | $5.66 | $5.17 |

6 | $5.60 | $5.23 | $4.81 | $4.28 |

4 | $4.27 | $3.91 | $3.51 | $3.04 |

3 | $3.26 | $2.94 | $2.61 | $1.80 |

2 | $1.94 | $1.73 | $1.52 | $1.28 |

I sat down at a game of Texas Hold 'Em and the high hand during the hand in progress was a straight flush. Two other players at the table remarked it was the third straight flush in a row. What are the odds of that?

heatmap

In a 10-player game of Texas Hold 'Em, assuming nobody ever folds, the probability that the high hand will be a straight or royal flush is 1 in 350.14. The probability of this happening three hands out of three is 1 in 42,926,491.

However, that table may have been going for hours. Perhaps a more realistic question is what is the probability that would happen at least once in an entire day. Assuming a full 24 hours of play and 24 hands per hour, the answer to that question would be 1 in 59,621.

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