The Binomial Coefficient in Poker

Even the inattentive observer can guess that the effort of counting the possible combinations of some number of cards when taken from a universe of some greater number of cards can become complicated. You might think, for example, that a computer might think that two hands with identical cards would be different hands if they were dealt out in a different order. In other words, we need to be sure that a draw of two cards (say, ace of hearts and king of spades) is not counted as a different hand from the same cards issued in reverse order (like king of spades and then ace of hearts).

The solution to this is a specific function. It is called the binomial coefficient or "combinatory" function. It automatically weeds out all issues related to the order of the cards and focuses only on the number of possible combinations, irrespective of how they got there.

The binomial coefficient answers the question, how many distinct sets of "k" cards (say, 5) can be made out of a universe of "n" cards (say, 52) without replacement (that is, without putting any back) The calculation is contemporaneous, not sequential (which is why "order" does not matter - there is none.). It does not consider first one card, then another card, then another card, up to "k."

The number of combinations of five cards that are possible from a deck of 52 is an important number for calculating the probability of five-card hands. It is the denominator for the calculation of the frequencies reported in the table of frequencies of five-card hands in an initial deal. That number is 2,598,960.

To calculate reliable probabilities for specific subsets of the universe of all possible combinations, all possible desired combinations must be included in the numerator. If any are missing, the resulting answer can greatly underestimate the real chances for a successful deal or draw. This means accounting for all cards in the deal or draw, including any cards that don't make a contribution to the result desired, but are there to fill out the requisite number of cards. These are called "other" cards from here on out.

The binomial coefficient is itself a function with a specific, conventional notation. "Math-phobes" need not panic. The next few paragraphs are important, but they probably can be picked through without injury to the liberal arts psyche. Just focus on how to use the function and skip over the part that says exactly what it is.

The notation for a binomial coefficient is a large parenthesis with n over k. In print it is sometimes rendered as "(n;k)." The definition is:

For this definition to work, both "n" and "k" must be positive integers or zero, and k must be less than or equal to n. This is always the case in dealing with poker probabilities.

This whole expression is reduced to a single function in spreadsheets, where it is usually called the "combin" function in English-language programs. It has two parameters, "n" and "k". The easiest way to set this up is to put "n" and "k" in two neighboring cells, like A1 and A2, and then put the formula in A3, indicating that it is to be "combin(A1;A2)." If A1 is 6, for example, and A2 is 2, then A3 will be 15, exactly the number of different ways that two cards can be drawn from a universe of 6. To test the "combin(n,k)" function, calculate the formula above for (6;2). The numerator is 720 (six factorial). The denominator is 2 (two factorial) times 24 (four factorial) or 48. 48 goes into 720 exactly 15 times. Magic.

Now going back to how the function is used, the objective is to come up with a fraction or ratio in which the numerator is the count of all successful combinations, and the denominator is the count of all possible combinations. This calculation results in the probability of a successful outcome, whatever it is.

The denominator of the fraction to be calculated is usually the easiest part. It is the number of different combinations possible by drawing "k" cards out of an "n-card" deck. For example, if we are looking at a draw of two from a deck with 47 unseen cards remaining, then the denominator is "combin(47;2)." For calculating two- and three-card draws in a game of five-card draw, the numbers "combin(47;2)" and "combin(47;3)" come up all the time. They are 1081 and 16,215 respectively. The denominator mentioned above for all five-card hands that can come out of a 52-card deck is "combin(52,5)." It may be worth a try in the spreadsheet to put 52 in A1 and 5 in A2 and see if A3 gives the right answer (2,598,960). If you get asterisks, make the column wider.

The numerator for these calculations is usually the source of angst, if there is any angst at all. Sometimes counting the number of instances that will satisfy the requirement or constraints placed on the deal or draw will be tricky.

To look at this more closely, recall the question posed of how likely is it to draw a third king in three cards when there's already a pair in the hand. The denominator is fairly easy. There are 47 cards left unseen in the deck and the draw is for 3 cards. So the "combin(47,3)" calculation yields 16,215. For the numerator, ask how many of the possible draws will be "winners." Well, there are two kings left, and only one is required. The other two cards are unconstrained, except that they can not be a king, or else they would not be "other" cards. How many cards will be "other" cards (that is, "non-king cards" left in the deck)? The answer is 47 minus the two kings, or 45. How many "other cards" are needed in the draw? Two.

That is all the information that is required. The requirement is for combin(2;1) "good" cards times combin(45;2) "other" cards. All three cards have been accounted for between the "good" and the "other" categories. The answer, then, is their product. There are 1980 possible three-card draws that have one king in them. For each king in a successful three card draw there are 990 [combin(45,2)] combinations of "other" cards that will still yield the result of one king. Dividing this number by the universe of possibilities (16,215) yields 12.21%.

This calculation did not cover the possibilities of both kings' being in the draw. How can you tell that? The "combin(2;1)" term does not admit the possibility of both. This function returns the number 2, meaning it is either one king or the other, but only one. The "combin(45;2)" term for "other cards" is "45" because both kings were taken out of the competition. Adding in the possibility of getting both kings to the first question of getting only one king will improve the picture ever so slightly. There are combin(2;2) combinations of both kings, and there are combin(45;1) "other" cards in the draw. This number reduces to 45 instances. Intuitively, this should make sense, as there is only one way both kings can appear in the draw, and the third card can be any one of the remaining 45 cards in the deck. Adding 45 to 1980 one-king cases yields 2025 "hits" out of 16,215, or 12.49%, which reflects the chance of "at least one" king.

Beware of two things: (1) This calculation includes more possibilities than kings. In addition to what has been calculated, it is possible that some of the 990 combinations of "other" cards will be pairs, meaning that the draw will result in a full house. This is not a bad thing, presumably, but it is important to realize that getting a king and another pair is also part of this probability, as the combinations creating a full house were not separately identified. It turns out that there are 126 of these occurrences in the set. How that is determined is taken up in a minute. (2) Also note that the many non-qualified combinations of "other" cards will also contain pairs and three-of-a-kind draws of other ranks, which may do nothing for the kings, but can also create two pair and full houses of their own. This calculation does not address any of those circumstances. Presumably it does not need to, as we're playing poker, not doing scholarly work, and, well, more firepower is always better than less firepower

Nevertheless, whether any given answer is "correct" depends a lot on how the question is framed. The 12.49% number reports the chances of "at least one king in a draw of three." The 12.21% answer refers to the chances of winding up with exactly three kings. Deducting out the possible full houses yields a percentage of 11.43%, which says, "only three kings, and nothing more." There is yet another calculation to represent "two kings" and nothing more, or "two kings and another pair" or "a full house of some other trip and kings." Most people are glad to keep the higher hands in the calculation, as their real question is something even vaguer, like, "what is the chance that I will improve my hand?"

But now suppose you really wanted to know about the other way this hand can "improve" on a three-card draw. What are the odds of drawing a second pair, or a trip, but no third king?

It turns out that the odds are better of drawing a second pair in three cards than they are of drawing the third card to the existing pair. There is a non-obvious detail in performing this calculation. Ask yourself, "How many draw combinations will satisfy this requirement?" There is a tendency to think, well, there are 12 ranks left to worry about (other than the kings), so the number of combinations will have something to do with 12 times the number of combinations that 2 cards (the second pair) can be taken out of 4 available cards (the number of each rank in the deck, one for each suit).

The non-obvious detail is that when you made the draw, you discarded three cards, and they now become relevant to the calculation. They are "seen cards" and no longer in the "unseen" deck.

It is true that there are 12 ranks of interest; however, only 9 of them have four cards in the deck, and 3 of them have only three left. This is because you had the fourth one of each of these three in your hand before you sloughed them.

The number of combinations is thus arrived at by adding two separate pieces together. The first piece, that of the three ranks with only three cards still available, is, for each rank involved, combin(3,2) times the number of "other" cards in the deck. This number is multiplied by three again, because three ranks are involved. The second piece, that of the nine ranks with four cards each in the deck, will be combin(4;2) times the number of "other" cards in the deck, and then times nine because nine ranks are involved.

The "other" cards in each case will be a little different. There are 47 to start with. Then there are the two kings in the deck that don't count, as that would put us back into adding kings to the hand, and these combinations have been counted elsewhere. So that makes 45. For the three ranks with only three cards, those three must come out of "other" because their contribution to the number of combinations is already included in the combin(3;2) term. So that makes 42. For the 9 ranks with four cards in the deck, the "other" cards are reduced from 45 to 41. Thus, for the cards with only three mates in the deck, the number of combinations is 3 * comb(3;2) * 42. This is 378. For the cards with four mates in the deck, the number of combinations is 9 * comb(4;2) * 41. This is 2,214. Thus, there are 378 + 2,214 possible combinations out of a universe of 16,215 that will provide a second pair on a three-card draw, but no full house or third king. In percentage terms, this is 15.99%.

These probabilities are additive, so the chances of obtaining either three of a kind (or better) or a second pair is the sum of 15.99% and 12.21% or 28.2%. Admittedly, this is rough going. But the end is in sight. Before moving to a generalized approach that can be used for all such questions, it may be worthwhile to test your understanding of the methodology by following along through a couple more examples:

Suppose you were dealt three aces. What is your chance of drawing the fourth ace in two new cards? Well, there's only one ace left, and it alone will meet the requirement. There are 46 other cards for the second draw card, so altogether there are 46 combinations of two-card draws that will work for you. The universe is "combin(47;2)" - the total number of two-card combinations in the deck. This number is 1,081. Thus, the answer to the question is 4.26%. If this makes sense, then it is almost time to take on the generalized approach.

It might be helpful to illustrate this approach in the context of seven-card stud and Texas Hold'em. In seven-card stud, it might be worth asking what the chances are of matching one of the first three cards on the next card (Fourth Street). Stud offers a complication in that the number of "unseen" cards will vary according to how many players are in the game. Where you are sitting relative to the dealer also affects the number of cards in the deck when yours is dealt, but not at the time the betting decisions must be taken.

So the number of cards in the deck for stud poker is the number of face-up cards per player, times the number of players, plus two (your hole cards). For the example, let's just assume four players. Suppose you have a jack as one of the first three cards. What are the odds that your next card will also be a jack? There are, presumably, three jacks left in the deck, and the deck has 46 cards in it. This is because there are 4 door cards visible, plus two in your hand, for a total of 6 "seen" cards out of a deck of 52. The probability, then, is 3 divided by 46 or 6.52%. If one of the other visible door cards is a jack, then obviously the numerator goes down by one, dropping the probability to 4.35%. The same thinking is true for each of the other two cards in your hand so far, whether face up or face down. Thus, the probability of a match to one of the three on the next card is three times 6.52% or 19.6%. The odds are between 4:1 and 5:1 against.

Notice that "combin(n;k)" functions did not need to be invoked. That is because only one card was dealt from a universe of 46, so only 46 combinations were possible. Three jacks were available, but only one would be dealt, so there are only 3 ways 1 card can be drawn out of a set of 3. Technically, the calculation was "combin(3;1) / combin(46;1)" but that is the same as 3 / 46.

Finally, try out calculating the odds of getting a king in the flop if five people are playing Texas Hold'em. Presumably you have a king in the hole, which is why you are interested in the answer. Only two cards are known, your hole cards. The number of combinations in the denominator is combin(50;3), that is, a deck of 50 unseen cards, with a draw of three. That means there are 19,600 ways that the flop could go. The number of successful combinations would include the number of combinations with one king, two kings and three kings. With one king the calculation would be combin(3;1)*combin(47;2) or 3245 combinations, yielding a probability of 16.5%. Two kings would add an additional 141 combinations, and three kings would add one. The combined probability is 17.28%.

Some of these combinations will also pair up the other hole card, yielding two pair. How many? There are three of each desired card in the deck, yielding 9 ways that the cards can be paired up. The other flop card can be any one of 44 other cards. That yields 396 combinations, for about a 2% chance of two pair on the flop (not counting a pair within the flop itself).