Ask the Wizard: Poker

Questions about Poker

Note that I reversed the order of the full house and flush. November 23, 2006

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. Nov. 22, 2005

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." Nov. 2, 2005

The number of ways to make a 4-card straight flush is 4*(9*46 + 2*47) = 2,032. There are 3,744 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. Nov. 2, 2005

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. The probability of this happening twice in a row is 1 in 270,725² = 1 in 73,292,025,625. Aug. 28, 2005

What are the chances in a heads up game of Texas Hold'em that each player gets KK. Then on the very next hand both players get KK. We can't even get a close estimate. If you can figure it out please respond, thanks.

The probability for any given hand is (combin(4,2)/combin(52,2))*(1/combin(50,2)) = 1/270,725. So the probability of this happening twice in a row is 1 in 270,725² = 1 in 73,292,025,625. Aug. 28, 2005

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

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 = 1,584. There are 2,598,960 combinations in total, so the probability is 0.0609%. April 3, 2005

I'd like to how much does a casino poker chips weigh and better yet do you know where is the best place you can purchase poker chips that feel and sound (when you drop them) are as close to the real deal as possible?

The standard is 11.5 grams. Casino-quality chips are made of a clay composite. Most poker chips sets are the same weight but the material is not as high quality and feels more like plastic. If you really want the best you could go to a casino and purchase a large quantity of $1 craps/poker chips from the cage at face value. If the casino changes the style, or goes the way of the dodo bird completely, the chips should go up in value. However, for most recreational purposes there are always lots of sets available on eBay for about $50 for a 500-chip set. If you do get generic chips I would recommend true Paulson chips (there are many imitations), which are the same quality as casino chips. However, Paulson no longer makes generic chips so the price will be significantly higher. If the price pushes $1 a chip, which it often does, I would just get actual casino chips instead. Jan. 16, 2005

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.

From probability 101 we know Pr(A or B) = Pr(A) + Pr(B) - Pr(A and B). Therefore, 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. Jan. 16, 2005

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? Thanks, 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. Jan. 2, 2005

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

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. I hesitate because I'll get asked about every bad beat jackpot in every poker room in the whole world. Jan. 2, 2005

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.

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 4⁷ = 16,384 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 1,020.6952. Dec. 27, 2004

Dear Wizard, You advised a previous poster how one can learn to shuffle poker chips (Aug 25, 2003). Becuase of your instruction, my shuffling abilities give me a strong edge (my poker skill, on the other hand, is another story...). I just want to give a few extra tips...

1. Try starting with a stack of 6 chips instead of 10. Even those lacking any dexterity (like me) will have an easier time getting the feel of it.
2. It is much harder to practice on a hard table. One cannot as easily get one's fingers underneath the stack as on a felt table. If you don't have access to a casino table (other than in the casino itself), practice on something soft, such as a mousepad or even a folded newspaper.
3. Warm-up your hands before shuffling (especially if you are new at it). Shuffling puts your fingers in an odd position. It takes a while for them to get use to it.
4. Learn to use both hands. It becomes much tougher when using your weaker hand, but it makes you look twice as intimidating when you show your fellow players that you are good with either hand.

Thanks again for the great site! -John

Thanks, I never thought to try with my left hand. Dec. 20, 2004

A flush is 100% possible on the 17th card dealt no matter what. When is a straight 100% possible on what # card dealt?

A straight is guaranteed only with 45 cards. For example your could deal every A, 2, 3, 4, 6, 7, 8, 9, J, Q, and K for a total of 44 cards and still not have a straight. Dec. 20, 2004

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

1/combin(52,4) = 1 in 270,725. Dec. 13, 2004

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., 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) = 1,980. 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 1,980+45+41 = 2,066. The total number of ways to choose 3 card out of the 47 left is combin(47,3) = 16,215. So, the probability of improving the hand to three of a kind or better is 2,066/16,215 = 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. Nov. 19, 2004

Do you think online poker rooms are"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.

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 an 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. Aug. 23, 2004

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

(12/52)*(11/51)*(10/50)*(9/49)*(8/48) = 0.00030474, or about 1 in 3,282. May 22, 2004

Are the probabilities for the various hands the same in Texas Hold 'em as in Seven-card stud or are they different somehow due to the community cards? Could you please explain why or why not?

Yes, the probabilities are the same. Seven random cards out of 52 have the same odds regardless of how they are taken out of the deck or whom you share them with. May 22, 2004

How does an online poker site protect its customers from collaboration?

I'm not sure, but they swear that they have tests for this. They wouldn't want to explain exactly what they test for, lest the ,take countermeasures. However, an easy sign would be when 2 or more players always play at the same time. Feb. 27, 2004

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?

The odds are the same. Jan. 31, 2004

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

No! Absolutely not! Jan. 31, 2004

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

There are four possible suits for the royal. There are five possible gaps for the missing card. 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 (52 × 51 × 50 × 49 × 48)/(1 × 2 × 3 × 4 × 5) = 2,598,960 total combinations. So the probability is 940/2,598,960 = 1 in 2,765. Jan. 20, 2004

I have one simple question. I know you are the wizard of odds but I need help. I play Craps and Texas Hold'em in the casinos. I feel I could be an intimidating force (THUS RAISING MY ODDS OF WINNING) if I could just figure out how to shuffle my poker chips. I have practiced but just can't get it. I was hoping you could point me in a direction to learn this. Thank you for your time. - Joseph

You asked the right person! I'm quite good at shuffling poker chips. Unfortunately, I don't often get to show it off because when I do play with chips it is usually counting cards or hole card reading, and in either case I don't want to look like a pro. Anyway, start with a stack of 10 chips. Then cut them in half, putting two 5 stack chips side by side. Think of the two stacks as an 8. Put the 8 at about a 45 degree angle to your plane of symmetry. Put your thumb on the bottom of the 8, your index finger where the two stacks come together, and the other 3 fingers at the top of the 8. All five fingers should be barely touching the table. Then using your index finger gently lift both stacks while the other four fingers gently push the stacks together. After your index finger is about a quarter inch from the bottom of the table, quickly move it away and keep pushing the stacks together with your other four fingers. It takes practice. I would recommend getting 10 nice clay chips and practicing at home or work. During commercials or whenever you have a minute to spare you can work on it. Before you know it you you'll be riffling chips like a pro and casting fear in your fellow poker players. Aug. 25, 2003

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

The probability of a single player getting a 7-card flush is 4*combin(13,7)/combin(52,7) = 1 in 19,491. The probability of at least one player out of 7 getting a 7-card flush is about 1 in 2,785. Aug. 2003

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

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%. June 27, 2003

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 = 1,081 ways to arrange the other two cards. 4*1,081 = 4,324. 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. Feb. 20, 2003

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 Dye 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. Feb. 4, 2003

If I hold just the queen of clubs what are the odds (ten million to one etc.) of drawing to a royal flush? - Bradford R. Schmalfuss from Houston, USA

There are combin(47,4) = 178,365 ways to choose 4 cards out of the 47 remaining. Only one way will result in the three cards you need. So, the probability is 1 in 178,365. Sept. 24, 2002

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) = 5,108/2,598,960 = 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)* 0.0101836²*(1-00.0101836)³ = 0.001006, or about 1 in 994. Sept. 24, 2002

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 Keaton from Mahomet, Illinois

The probability of any two specific players both having a four of a kind is (13*COMBIN(12,3)*4³*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. Mar. 24, 2002

What are the odds of making a royal flush in Texas hold-em on the river? - John Hinrichs from Trinity, Texas

For those readers who don't know, the river is the fifth and last community card in Texas hold-em. The player must make the best poker hand between his own two cards and the five community cards. So you're asking what is the probability that a player will form a royal flush in seven cards, and that the seventh card dealt will be part of the royal. The probability of forming a 5-card royal flush out of 7 cards, before considering card, is 4*combin(47,2)/combin(52,7) = 4,324/133,784,560, or 1 in 30,940. The probability that the seventh card will be part of the royal flush is 5/7. So the final probability is 21,620/936,491,920, or 1 in 43,316. Jan. 15, 2002

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 Schneider 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. Sept. 20, 2001

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 Norgaard from Montgomery, USA

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

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. Aug. 6, 2001

What are the statistical odds of getting a flush in Texas hold 'em. Is it easier to get a flush in 7-card stud or in holdem as a player? - Kevin from Richmond, USA

You can refer to my section on probabilities in poker to see the probability is 3.03%. The odds are the same in both Texas hold 'em and 7-card stud. June 13, 2001

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 I am hoping you can help me with. I have been trying to figure out the number of times each hand in 7 card stud occurs. I have a copy of your 7 card table, but I am interested in the mathematics to arrive at those numbers. I can figure out the 5 card numbers, but the 7 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 Abadie 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 on my computer. My program goes through all possible combinations and scores each one. The number of wild straights in pai gow poker is 11*(4⁴-4)+10*3*(4⁴-4)=10,332. Combined with the 10,200 natural straights the total is 20,532. May 1, 2001

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 Cinea from Longmont, Colorado

I haven't played much casino poker myself but I understand that experts tend to congregate at certain tables and wait for "fish" to come in and then eat them up. I don't know if this is causing poker rooms to shut down. Feb. 18, 2001

With a 52-card deck, what are the odds of drawing a pair of Jacks? - Rick from Gardnerville, USA

My section on probabilities in poker indicates the probability of drawing a pair in five card stud to be 42.256903%. To get the probability of drawing a pair of jacks divide this by 13 to get 3.250533%. Sept. 10, 2000

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 Mularz from the USA

Here are the hands from highest to lowest: straight flush, four of a kind, full house, flush, straight, three of a kind, two pair, pair. Sept. 10, 2000

How do you get your odds software for seven card stud? - Ron Thixton 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 7 card hand. So, overall I had to score over 2.8 billion hands, this took the computer all night if I remember correctly. Aug. 27, 2000

What are the odds of being dealt a Royal Flush? Of being dealt a SEQUENTIAL Royal Flush (forwards or backwards)? - Ric Parks of Torrance, California

I think was asked in an earlier column but since this is an easy one I'll answer again. The odds of being dealt any royal flush are 4 in combin(52,5)=2,598,960. There are 5!=120 ways to order 5 cards and two of them result in a sequential royal (counting both forwards and backwards). So the odds of being dealt a sequential royal are the product of 4/2,598,960 and 2/120 = 1 in 38,984,400. July 9, 2000

Great Site! Clearly the Wizard rules! I am an avid fan of playing some of the free online casino games. My current favorite is "quick draw" poker with a growing jackpot on gamesville.com. First, could you recommend the best strategy for this game? If you assess the rules, players are allowed 17 hands, and may bet 0, 1, 2, 5, and 10 points a hand. The object is to increase your original point value of 150 to 500 in the 17 hands. Second, could you recommend other free sites, and the best strategies? - Phil Brown of Goleta, California

I see from your e-mail address you are with U.C. Santa Barbara. That is where I went to college. BA in mathematics/economics, class of '88. For a fellow Gaucho I spent about an hour playing this game to hopefully help you out. I got up to 426 points one round. For the benefit of other readers, the game in question has 17 rounds of jacks-or-better video poker, with a much different pay table than found in the casinos. Most notably a high pair pays 2:1 and two pair pays 4:1.

With the generous paytable of this game, the player should bet the maximum unless the cards are total garbage. With garbage, the expected return is about 0.7X money bet. With a single high card the return is 0.94X money bet. With a low pair the return is 1.53X money bet. Betting the maximum on a single high card may sound like a bad bet but in this game you need to be aggressive. Towards the end, the player should bet the maximum even on garbage because a long shot is the only hope to get 500. I would not sacrifice pairs for 4 to a straight or flush. It will likely take getting a 4 of a kind at some point to get the 500. Good luck! July 2, 2000

Love your site, I have a math degree and I am a bj 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 fo the rules of poker. Can I gain an advantage similar to counting in BJ? - Kal Badwan from Chicago

First let me say that I'm not an expert on poker. However, I do know that you can gain an edge over the game. However, the edge in poker is more difficult to estimate and will vary depending on the skill of the other players. I would suggest studying Texas Hold 'Em. 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. If you give it a try please keep me updated on your progress, I'm thinking of taking up Texas Hold 'Em myself.

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)*p²*(1-p)⁴ = .0000000028779. In other words the odds against this happeneing are 347,477,740 to 1. April 2, 2000

I started playing poker with my friends once a week (5 card draw, stud, 7 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 McAllister 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. Mar. 25, 2000

In high-low poker, can a pair of aces be a lower hand than a pair of threes, or other pair? - Isaac Shalom of New York, New York

Yes, I believe so. In low ball aces are the lowest rank, thus a pair of aces would beat a pair of threes. Mar. 18, 2000

What are the odds in drawing 3 cards to a pair and getting a full house at 5 card draw poker? - Nick from ?

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 lets 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 see my section on probabilities in poker. March 4, 2000

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