Ask the Wizard: Probability - Cards
I was playing baccarat online and out of 75 hands the banker won 52 and the player 23. This is a difference of 29, what is the probability of that happening?
|Specific Number Odds|
in Liar's Poker
The next table shows the probability of every possible type of bill, categorized by the number of each n-of-a-kind. For example, the serial number 66847680 would have one three of a kind, one pair, and three singletons, for a probability of 0.1693440.
|General Probabilities in Liar's Poker|
|8 o.a.k.||7 o.a.k.||6 o.a.k.||5 o.a.k.||4 o.a.k.||3 o.a.k.||2 o.a.k.||1 o.a.k.||Probability|
o.a.k. = "of a kind"
For more information, see my page on liars poker.
With a 52-card deck, what are the odds of drawing a pair of Jacks?
— Rick from Gardnerville, USA
What is the probability of getting a three pair in Pai Gow Poker? Are the chances lesser or greater than three of a kind?
— Alex from Long Beach, Mississippi
Not counting a three of a kind and two pairs, the following are the ways to get a three pair and number of combinations.
No wild card: combin(13,3)*10*63*4 =2471040
Wild card used to compete pair of aces: combin(12,2)*10*62*42 = 380,160
Wild card used as singleton ace: combin(12,3)*63 = 47,520
The total number of combinations is 2,898,720. This is less than half of the 747,0676 combinations for a three of a kind.
The odds according to your formula for a royal flush is 4/2,598,960 = 1/649,740. So, if I was playing Caribbean Stud one-on-one with the dealer, then my hand and the dealers would equal 649,740*2=1,299,480. Therefore, according to the math, after 1,299,480 hands there should be two royal flushes. Please tell me if I understand the odds correctly.
— Bill from Niagara Falls, Canada
The probability of zero royals in 1,299,480 hands is 13.53%.
Hi, I am regular player of Pai Gow Poker, and I noticed your site has a lot of great information on the game. The other day when I was playing with a friend of mine he was dealt a 9-high hand, which I believe is the lowest hand possible. In all the time I had spent playing the game I had only seem it happen once before. Then five hands later he got the exact same hand(2-3-4-5-7-8-9). We couldn't believe it and were wondering what the odds of that happening were so we thought we would ask you. Thanks for your time and your great site.
— Doug from Calgary, Canada
My question has to do with the House edge and element of risk calculations for Casino War for the Casino Niagara Rules (i.e., 3-1 pay out on raise and lose the original wager). How did you come up with these numbers I am currently trying to calculate them? I am having trouble. Thanks for you help.
— Mark from Vancouver, Canada
Could you tell me how the total number of combinations in Caribbean, 19,933,230,517,200 are arrived at? I followed your 5-card poker combinations to get the 2,598,960. From there how do I continue? Thank you in advance.
— Claudio from Punta del Este, Uruguay
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
I understand what the odds of being dealt a royal, straight flush are for any individual on a Caribbean Stud Poker or Let it Ride table are and how they are derived. But my question is this: as a 3rd party watching the game what are the odds of seeing any one of these hands being dealt to a player at the table on any given deal. I must believe it is dependant on the number of hands in play...is it merely the individual’s odds times the number of hands in play. ie. seeing a royal dealt on any particular hand with 4 players on a table means 4*odds of getting a royal? I’m slightly perplexed!
— Amyn from Brantford, Canda
What is the probability of being dealt a natural seven card straight flush in pai gow poker? I work in a casino and just saw this for the first time in 15 years. The lucky patron won $40,000.
— Michael from South Haven, Mississippi
What is the probability of getting two identical straight flushes (in both ranks and suit) two hands in a row in Three Card Poker?
— Ralph from Harpster
In an 8 deck baccarat, what is the probability of getting an Ace and an 8 of Diamonds for both the player and the banker in a same deal?
— Emi from Manila, Philippines
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
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
When you open a new deck of cards they Ace to King of each suit. What are the odds of taking a shuffled deck of cards and reshuffeling it to the Ace to King state it was originally in?
If ten people are each dealt two cards from a single deck what is the probability that two players will get a pair of aces?
If seven players each get seven cards, what is the probability at least one person will get a 7-card flush?
I know that mathematically anything is possible, but the other night at the casino I think that I witnessed something that would be a billion to 1 shot, not that those don?t ever happen. Here is what happened: In the course of 40 hands (40 single 3 card deals about 8 rounds with 5 players) at a let it ride table 3 four of a kinds were dealt. With a four of a kind being about 4100 to 1 what would be the odds of three of them being dealt within 40 deals? Please answer as this is killing me. Long time fan
First I wanted to tell you how much I look at and love your web site, and admire your math skills. I use 6 decks to deal blackjack, and added 3 jokers for reasons I won’t waste your time with but, what are the odds of dealing all 3 jokers to a player right in a row. Thank you very much.
What are the odds of passing out 13 cards each to four players using a 52 card deck and all four player have a straight from Ace to two? The cards don’t have to be in the same suit.
What is the probability of being dealt four to a royal?
Dear wiz, say a deck of 52 cards is shuffled, and we draw 18 of the 52 cards at random, placing the 18 cards into 6 piles of 3 cards each. What is the probability that one of the piles contains exactly 3 (of the 4) aces?
What is the probability of getting all face cards in five card stud?
What is the probability of getting two four of a kinds in a two hour period playing Let it Ride?
In four-card poker, which is more likely a straight or a flush?
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.
- 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 Aeax.
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 Gábor 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%.
After I posted Gábor’s solution Joshua Green supplied a shorter one by another method.
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?
Ace Probabilities - Random Hand
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 is states 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
This shows the probability of at least one more ace is 0.221369.
However suppose you want to solve it the same way as hand A, using Bayes’ Theorem. 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%.
Okay, I believe your numbers but it still doesn’t make sense to me. I should think the probabilities would be equal. What difference does the suit make of the one ace you are given?
To make another example suppose you go to Jiffy Lube and they offer two deals for the same price. Deal A is they will check four parts and replace only the first defective one found. Deal B is they will check only one problem and fix it if found. Wouldn’t you rather take deal A? Your car came in with the same number of expected bad parts but the probability of finding a problem is greater under deal A, and thus you will leave with a small number of expected defective parts under that plan. Likewise a test for any ace will probably turn up the only ace, while a test for the ace of spades does not check for the other three suits, leaving them more likely to be aces.
How did you arrive at 2072 for the number of straight flushes utilizing 4 cards out of 5 in Four Card Poker?
A flush is 100% possible on the 17th card dealt no matter what. When is a straight 100% possible on what # card dealt?
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.
From a single deck if I deal 4 cards what are the odds that at least 1 card is a spade?
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.
What are the odds of being dealt the jack of diamonds 27 hands in a row in a six-card game?
What is the probability of getting the "dead man’s hand", a two pair of aces and eights?
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
What are the odds/probability of each possible five-card hand that could be dealt for from a standard 52-card deck where the suit of a single, unduplicated card in a hand is considered generic? For example, consider the hand A♠ A♣ A A 2. In this hand the suit of the 2 is disregarded and would represent any of the four 2’s in the deck. Another example is A-J-8-6-5. In this hand the suit of all 5 cards is disregarded so that only one such combination can occur. Another example: 3♠ 3 7♣ 7♠ Q. In this hand the suit of the two 3’s and two 4’s is not disregarded because there is more than one in the hand but the suit of the Q is generic. In other words the suit of any card that is not duplicated in a hand is disregarded and that hand is considered to be one of the possible hands even though there would be many possibilities of the hand re-occuring if the suit of each card was not disregarded. Thus a straight or flush made up of five specific cards; say Q J♣ 10 9 8♠ or A♠ J♠ 8♠ 7♠ 3♠ could only occur once since any other combination of those same cards even though in a different suit would be a duplication. Therefore, using this criteria, what is the O/P of any single hand being dealt? In other words how many such five-card hands exist in a std 52 card deck? Thanks for your input.
— Mike from Lavallette, NJ
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|
|Five of a kind||13||0.000002||13|
|Four of a kind||3900||0.000472||13*12*COMBIN(5,4)*5|
|Three of a kind||214500||0.025969||13*COMBIN(12,2)*COMBIN(5,3)*5^2|
Note that I reversed the order of the full house and flush.
How many five-card combinations of a standard playing card deck have cards from exactly two suits?
— Samantha from Belize
What are the odds of being the dealt 2-3-4-5-7 unsuited? Thanks a lot, the site’s great!
— Kevin from Massapequa
I was at Foxwoods the other day, watching the final two tables of the Foxwoods Poker Classic. When Vince Van Patten (one of the World Poker Tour hosts) came in to watch, he started making all kinds of prop bets with some of the poker pros who were hanging around. He was offering anyone 20 to 1 if they could flip through an entire deck of cards cycling through the ranks and saying aloud while peeling each card Ace, 2, 3, 4, and so on up to King and starting again at Ace without ever having the card they’re announcing come up. Nobody made it all the way through and Vince won a few hundred dollars in about 10 minutes before everybody gave up. I know this has to be possible, but I have a suspicion Vince has quite the hustle going offering only 20 to 1 on this. What are the odds of actually getting through the whole deck?
— Matt from New Britain
Accoring to G.M., who is a better mathematician than I am, the actual probabiliy is 1.6232727%. The reason for the difference is the outcome of each pick is positively correlated to previous picks.
Playing last night, one of the players, an old crafty scruffy aggressive player, was challenging the table to make even money side bets on the flop. This old curmudgeon was betting that one of the three cards on the flop would be either an ace, deuce, or jack (sometimes he would change the 3 identifiable cards). What are the odds of this bet? Your sage wisdom would be greatly appreciated.
Suppose that 5 cards are dealt from a 52-card deck and the first one is a king. What is the probability of at least one more king? I saw an Ace problem you were doing like this one but I couldn’t really follow it. Thank you for any help.
— Brian from College Station
Several people have said that the combinatorial function is probably over the heads of the type of people asking these kinds of simple probability questions. I don’t disagree with that, but a major reason for this site is to try to teach my readers something about math. The combinatorial function is extremely useful in probability, and saves a lot of time. However, the question at hand can be easily answered without it.
The probability the second card is not a king is 48/51. That is because there are 48 non-kings left in the deck, and 51 total cards left. If the second card is not a king, the probability the third card is also not a king is 47/50 (47 non-kings divided by 50 cards left). Following this to the end, the probability that none of the other four cards are kings is (48/51)×(47/50)×(46/49)×(45/48) = 77.86%. The probability that this is not the case, in other words at least one king, is 100% - 77.86% = 22.14%.
First, choose 5 cards from a single 52-card deck. Second, add their blackjack values (T,J,Q,K = 10, A = 1). What are the odds the sum is even/odd? I would think that with the over-abundance of even cards, the sum would be much more likely to be even.
— Eliot from Santa Barbara
What are the odds if you draw 3 cards out of a deck that exactly one of them will be a spade?
— Bill from Tempe
Select two random numbers between 0 and 1 (evenly distributed). Now select the smaller of the two. What is the average of the selection? What about the general case of n numbers?
You are playing a game that involves three people: (a) yourself, (b) your opponent, and (c) a referee. Each of you picks a real number between 0 and 1 in secret. Once all the numbers have been selected, they are revealed. The player who guessed closest to the referee’s number, without going over, wins. If you are closer, you win $1. If your opponent is closer, you lose $1. If both players go over, or there is a tie, the game is a tie.
Is there a number you can pick that will maximize your expected return, if the other player picks randomly? What if the other player has a strategy too?
— Andrew from Toronto
If you pick five cards at random from a standard 52-card deck, what is the probability that all four suits will be represented?
— Carl Libis from Richmond
What is the probability that any two chosen ranks, for example queen and king, will appear consecutively in a random deck? Somebody challenged me to an even-money bet that it would happen.
— Rob from Saratoga, CA
Can you recommend a function to map any five cards from a 52-card deck to an integer from 0 to 2,598,959?
— James from Worchester, MA
int GetIndex(int c1, int c2, int c3, int c4, int c5)
return combin(c5,5) + combin(c4,4)+ combin(c3,3) + combin(c2,2) + combin(c1,1);
Where combin returns the traditional value, except if the first value is less than the second value, return 0, as follows:
int combin(int x, int y)
for (i=x-y+1; i<=x; i++)
for (i=2; i<=y; i++)
If you are doing this to access an array element, load the array as follows.
for (c5 = 4; c5 < 52; c5++)
for (c4 = 3; c4 < c5; c4++)
for (c3 = 2; c3 < c4; c3++)
for (c2 = 1; c2 < c3; c2++)
for (c1 = 0; c1 < c2; c1++)
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)×pw×(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)×0.00185013×(1-0.0018501)7 = 0.000000750178, or 1 in 1,333,017.
Two 54-card decks (including two jokers) are shuffled together. A player is given half of them. What is the probability that the player got all four red threes?
If you don’t like dealing with such large numbers, here is an alternative solution. Number the four red threes 1 to 4. The probability that the first red three is in the player’s stack is 54/108. Now remove the first three. The probability that the player has the second red three is 53/107, because the player has 53 cards left, and there are 107 remaining cards. Likewise, the probability that the player has the third red three is 52/106, and the fourth red three is 51/105. (54/108) ? (53/107) ? (52/106) ? (51/105) = 0.059012.
This question was raised and discussed in the forum of my companion site Wizard of Vegas .
In your December 14, 2010 column you wrote that the AAAAAKK hand, which is specifically mentioned in the house way, may have never occurred in the history of the game. According to another dealer, a player got this hand at the Main Street Station in November 2010.
Reason #4 why the Wizard likes Bovada:
Bovada offers the triple crown of gambling: casino, poker, and sports. Many other casinos have tacked on poker as an afterthought, and many poker rooms have tacked on a casino as an afterthought, and the lack of attention shows, sometimes painfully. And very few of these sites let you make sports wagers.
But Bovada doesn’t just offer all three, they do each one well, and everything’s integrated. It’s easy to play all three off one deposit, off just one account.
Another nice thing about Bovada is that you don’t need a separate account to play casino games with fake money. In fact you do not even need an account for that at all, you can just click over there and play. Finally, Bovada usernames are only six or seven characters long making them possible to remember. By contrast some competitors’ usernames are extremely long and cumbersome.