Poker Probabilities

Last Update: Jul 22, 2010

Five to Nine Card Stud

The following tables show the number of combinations and probability for each poker hand using the best five cards from out of 5 to 10 cards.

Low Ball

Rules vary in low ball whether aces are high or low, and whether straights and flushes work against the player. The following tables look at two different sets of rules.

Omaha

In Omaha the player many use any 2 of his own 4 cards, and any 3 of the 5 community cards, to form the best highest and lowest poker hand. Straights and flushes are not enforced in the low hand. For the low hand aces always count as low. Here are the probabilities for each hand.

Median Hand

I have been asked several times what the median hand is in both 5-card and 7-card stud. First let me review what median means. It is the mid point in a set of values. For example if I gave a test and the scores were 20%, 30%, 40%, 50%, 100%; the median would be 40%. This is not to be confused with the average, which in this example would be 48%.

The median 5-card stud poker hand is ace,king,queen,jack,6 or 7. There are 1298460 hands less than AKQJ6, 1298460 higher than AKQJ7, and 1020 equal to both of either hand.

John Lichtenberg wrote to me to add that if the ace, king, queen, 10, 2 is the hand in which a random hand from the remaining 47 cards has cloest to a 50/50 chance of being higher or lower. He adds that 766919 remaining hands in the deck are lower, 766777 are higher, and 243 are the same.

I have not determined the median 7-card stud hand but a reader (Rocke V.) sent me an e-mail claiming it is jack/jack/ace/10/8. Another reader (David Mitchell) wrote saying he agrees with the 5 and 7-card median hands and adds that the median 6-card hand is 5/5/king/10/7.

Multi-Deck Probabilities

I've been asked several times about the probabilities of each poker hand in multiple-deck games. Although I strongly feel poker based games should be played with only one deck, I will submit to the will of my readers and present the following tables. The first table shows the number of raw combinations, and the second the probability.

Wild Deck Probabilities

The next two tables show the probabilities in 5-card stud with one wild card. The first table is for a partially wild card that can only be used to complete a straight, flush, straight flush, or royal flush, otherwise it must be used as an ace (same usage as in pai gow poker). The second table is for a fully wild card.

The next table shows the combinations and probability with two fully wild jokers. Note that the full house and four of a kind are equal in probability.

5-Card Stud with Two Fully Wild Jokers
Hand	Natural Combinations	One Joker Combinations	Two Jokers Combinations	Total Combinations	Probability
Five of a Kind	0	26	52	78	0.000025
Royal flush	4	40	40	84	0.000027
Straight Flush	36	288	216	540	0.000171
Four of a Kind	624	4992	3744	9360	0.002960
Full House	3744	5616	0	9360	0.002960
Flush	5108	5392	888	11388	0.003601
Straight	10200	20664	3840	34704	0.010974
Three of a kind	54912	164736	13320	232968	0.073666
Two pair	123552	0	0	123552	0.039068
Pair	1098240	339696	0	1437936	0.454682
Nonpaying hand	1302540	0	0	1302540	0.411869
Total	2598960	541450	22100	3162510	1

The next table shows the number of combinations for each hand when a particular rank is wild.

Four Card Stud

The next table is for four-card stud with no jokers.

The next table is for four-card stud one fully wild joker.

The next table is for four-card stud with two fully wild jokers.

Player Must Use Every Card

The following table shows the number of combinations for 2 to 10 cards from a single 52-card deck, with no wild cards. For the purpose of this table, a royal flush, straight flush, flush, and straight must use all cards. A royal flush is defined as an ace-high straight flush. For example, with three cards, a royal flush would be suited QKA.

I would like to thank Miplet for confirming the table above.

Derivations for Five Card Stud

I have been askedso many times how I derived the probabilities ofdrawing each poker hand that I have created thissection to explain the calculation. This assumessome level mathematical proficiency; anyonecomfortable with high school math should be able towork through this explanation. The skills used herecan be applied to a wide range of probabilityproblems.

The Factorial Function

If you already knowabout the factorial function you can skip ahead. Ifyou think 5! means to yell the number five thenkeep reading.

The instructionsfor your living room couch will probably recommendthat you rearrange the cushions on a regular basis.Let's assume your couch has four cushions. How manycombinations can you arrange them in? The answer is4!, or 24. There are obviously 4 positions to putthe first cushion, then there will be 3 positionsleft to put the second, 2 positions for the third,and only 1 for the last one, or 4*3*2*1 = 24. Ifyou had n cushions there would be n*(n-1)*(n-2)*... * 1 = n! ways to arrange them. Any scientificcalculator should have a factorial button, usuallydenoted as x!, and the fact(x) function in Excelwill give the factorial of x. The total number ofways to arrange 52 cards would be 52! = 8.065818 *10⁶⁷.

The Combinatorial Function

Assume you want toform a committee of 4 people out of a pool of 10people in your office. How many differentcombinations of people are there to choose from?The answer is 10!/(4!*(10-4)!) = 210. The generalcase is if you have to form a committee of y peopleout of a pool of x then there are x!/(y!*(x-y)!)combinations to choose from. Why? For the examplegiven there would be 10! = 3,628,800 ways to putthe 10 people in your office in order. You couldconsider the first four as the committee and theother six as the lucky ones. However you don't haveto establish an order of the people in thecommittee or those who aren't in the committee.There are 4! = 24 ways to arrange the people in thecommittee and 6! = 720 ways to arrange the others.By dividing 10! by the product of 4! and 6! youwill divide out the order of people in an out ofthe committee and be left with only the number ofcombinations, specifically(1*2*3*4*5*6*7*8*9*10)/((1*2*3*4)*(1*2*3*4*5*6)) =210. The combin(x,y) function in Excel will tellyou the number of ways you can arrange a group of yout of x.

Now we candetermine the number of possible five card handsout of a 52 card deck. The answer is combin(52,5),or 52!/(5!*47!) = 2,598,960. If you're doing thisby hand because your calculator doesn't have afactorial button and you don't have a copy ofExcel, then realize that all the factors of 47!cancel out those in 52! leaving(52*51*50*49*48)/(1*2*3*4*5). The probability offorming any given hand is the number of ways it canbe arranged divided by the total number ofcombinations of 2,598.960. Below are the number ofcombinations for each hand. Just divide by2,598,960 to get the probability.

RoyalFlush

There are fourdifferent ways to draw a royal flush (one for eachsuit).

StraightFlush

The highest card ina straight flush can be 5,6,7,8,9,10,Jack,Queen, orKing. Thus there are 9 possible high cards, and 4possible suits, creating 9 * 4 = 36 differentpossible straight flushes.

Four of aKind

There are 13different possible ranks of the 4 of a kind. Thefifth card could be anything of the remaining 48.Thus there are 13 * 48 = 624 different four of akinds.

FullHouse

There are 13different possible ranks for the three of a kind,and 12 left for the two of a kind. There are 4 waysto arrange three cards of one rank (4 differentcards to leave out), and combin(4,2) = 6 ways toarrange two cards of one rank. Thus there are 13 *12 * 4 * 6 = 3,744 ways to create a fullhouse.

Flush

There are 4 suitsto choose from and combin(13,5) = 1,287 ways toarrange five cards in the same suit. From 1,287subtract 10 for the ten high cards that can lead astraight, resulting in a straight flush, leaving1,277. Then multiply for 4 for the four suits,resulting in 5,108 ways to form a flush.

Straight

The highest card ina straight can be 5,6,7,8,9,10,Jack,Queen,King, or Ace. Thus thereare 10 possible high cards. Each card may be offour different suits. The number of ways to arrangefive cards of four different suits is 4⁵= 1024. Next subtract 4 from 1024 for the four waysto form a flush, resulting in a straight flush,leaving 1020. The total number of ways to form astraight is 10*1020=10,200.

Three of aKind

There are 13 ranksto choose from for the three of a kind and 4 waysto arrange 3 cards among the four to choose from.There are combin(12,2) = 66 ways to arrange theother two ranks to choose from for the other twocards. In each of the two ranks there are fourcards to choose from. Thus the number of ways toarrange a three of a kind is 13 * 4 * 66 *4² = 54,912.

TwoPair

There are (13:2) =78 ways to arrange the two ranks represented. Inboth ranks there are (4:2) = 6 ways to arrange twocards. There are 44 cards left for the fifth card.Thus there are 78 * 6² * 44 = 123,552ways to arrange a two pair.

OnePair

There are 13 ranksto choose from for the pair and combin(4,2) = 6ways to arrange the two cards in the pair. Thereare combin(12,3) = 220 ways to arrange the otherthree ranks of the singletons, and four cards tochoose from in each rank. Thus there are 13 * 6 *220 * 4³ = 1,098,240 ways to arrange apair.

Nothing

First find thenumber of ways to choose five different ranks outof 13, which is combin(13,5) = 1287. Then subtract10 for the 10 different high cards that can lead astraight, leaving you with 1277. Each card can beof 1 of 4 suits so there are 4⁵=1024different ways to arrange the suits in each of the1277 combinations. However we must subtract 4 fromthe 1024 for the four ways to form a flush, leaving1020. So the final number of ways to arrange a highcard hand is 1277*1020=1,302,540.

For example, let's find the probability of drawing a jack-high. There must be four different cards in the hand all less than a jack, of which there are 9 to choose from. The number of ways to arrange 4 ranks out of 9 is combin(9,4) = 126. We must then subtract 1 for the 10-9-8-7 combination which would form a straight, leaving 125. From above we know there are 1020 ways to arrange the suits. Multiplying 125 by 1020 yields 127,500 which the number of ways to form a jack-high hand. For ace-high remember to subtract 2 rather than 1 from the total number of ways to arrange the ranks since A-K-Q-J-10 and 5-4-3-2-A are both valid straights.

Here is a good site that also explains how to calculate poker probabilities.

For the benefit of those interested inCaribbeanStud PokerI will calculate the probability of drawing acehigh with a second highest card of a king. Theother three cards must all be different and rangein rank from queen to two. The number of ways toarrange 3 out of 11 ranks is (11:3) = 165. Subtractone for Q-J-10, which would form a straight, andyou are left with 164 combinations. As above there1020 ways to arrange the suits and avoid a flush.The final number of ways to arrange ace/king is164*1020=167,280.

 

Suggested Link: dicedealer.com has a good article on basic casino math, which goes through more slowly and patiently than I do above.