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Basic Probability for Poker Players

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By Tony Guerrera
Poker is a game of skill in which the players who make the best decisions will end up making the most money in the long run. A large part of making good poker decisions is being comfortable with probability, the math of random events. This article, the first of a two-part series here on Poker Helper, will help you with becoming comfortable.

Probability Defined

The following equation concisely defines probability:

In this equation, P(Outcome of Interest) refers to the probability that an outcome of interest will happen. In general, P(A) refers to the probability of A happening. Let’s put this definition to work by considering some examples.

Suppose you have a deck of cards. You draw one card. What’s the probability that it’s an ace? Well, there are 52 possible outcomes, and of those, 4 are outcomes of interest (i.e. there are 4 aces in the deck). Therefore, the probability that you draw an ace is . If you wish, you can simplify this fraction down to . Examples like these, in which you’re drawing a single element from a set, are the easiest to do. Let’s consider something a little bit more difficult now.

Suppose you have two standard six-sided dice. What’s the probability of rolling an 11? Our working definition of probability tells us that . One possible answer is that since there are 11 possible rolls (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12), and only one of those is 11. However, this is wrong because each possible roll is not equally likely.

To see why each number is not equally likely, we must break down each possible roll to be (a,b), where a is the number on the first die and b is the number on the second die. The set of all possible outcomes for rolling two dice is the following: {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. 36 equally likely rolls exist, and two of them result in an 11: (5,6) and (6,5). Therefore, the probability of rolling an 11 is actually .

Notice how we had to enumerate every possible roll in such a way that each possible roll was equally likely. Doing this properly is the key to doing probability problems correctly; probability is really all about fancy counting. When dealing with things such as rolling two dice, listing all the possible outcomes is possible. Unfortunately for us poker players, listing all the possible outcomes having to do with a deck of 52 cards is impractical. We poker players need a more convenient way of counting.

Permutations

A permutation is an ordering of elements in a set. For example, let’s take the set {A♣, K♥, Q♠}. The permutations of this set are the following: {A♣, K♥, Q♠}; {A♣, Q♠, K♥}; {K♥, A♣, Q♠}; {K♥, Q♠, A♣}; {Q♠, A♣, K♥}; {Q♠, K♥, A♣}. Since a flop in hold’em consists of three cards, the permutations of {A♣, K♥, Q♠} can be considered to be all the ways that we can see a flop containing the A♣, the K♥, and the Q♠ (consider the first element in a set to be the first card on the flop, the second element in a set to be the second card on the flop, and the third element in a set to be the third card on the flop).

To calculate the number of permutations of objects in a set, take the number of ways to choose the first element and multiply that by the number of ways to choose the second element. If there is a third element, you then multiply by the number of ways to choose the third element, and you repeat this process until there are no more elements to choose. Following this procedure for the example above, we see that there are 3 ways to choose the first element (A♣, K♥, or Q♠). Once the first element is chosen, there are only two elements left, and once the second element is chosen, the last one is completely determined since there are only three objects. By doing the multiplication out as described, we get that there are permutations of the set {A♣, K♥, Q♠}.

How many permutations of hole cards are there? There are 52 ways to get dealt the 1st card and 51 remaining ways to get dealt the 2nd card. The total number of permutations is . Notice that order matters in this calculation * for example, A * K * and K * A * are two different permutations even though they actually correspond to the same hand.

Let’s now calculate the probability of being dealt AKs (written as P(AKs)). From our working definition of probability, we have that .

We just calculated that there are 2,652 permutations of hole cards. To calculate the number of AKs permutations, note that there are 4 aces and 4 kings that we can receive for the first card, meaning that there are 8 possibilities for the first card. Given that the first card is an ace or king, there is only one possibility for the second card, the card of the same suit as the first card. Suppose the first card is the K * . The only possible second card that gives us AKs is A * . The number of AKs permutations is , and the probability of being dealt AKs is therefore .

Combinations

Sometimes, card order doesn’t matter. As an example, it often makes sense to treat A * K * and K * A * as the same outcome if we’re thinking of how many ways an opponent can have a certain hand. To think in these terms, we must to shift from permutations to combinations,a combination is an unordered sampling of objects. A few ways exist to count combinations.

Using AKs as an example, we know that there are 4 aces in the deck, and for each ace, there is only one king of the same suit. Thus, there are 4 combinations of AKs. Notice how the counting here is different than when we counted permutations of AKs. Since AKs and KAs are now the same, we can simply consider one of the orderings.

Another way to find the number of AKs combinations is to start with the fact that there are 8 permutations of AKs. To find the number of AKs combinations, we need to divide out the permutations leading to doubly counted combinations (for example, we don’t want to double count A * K * and K * A * since they are the same combination). There are two cards, meaning that there are permutations for each combination. Using the 8 AKs permutations and the fact that there are 2 permutations of 2 objects, we get that there are combinations of AKs, just like we did before. In general, we can find the number of combinations by taking the total number of permutations and then dividing by the permutations of the number of elements comprising a permutation.

Combinations can be used to find probabilities much in the same way as permutations. Suppose you hold JJ, and you put your opponent on a distribution of {AA-TT, AK}. What’s the probability that your opponent is holding a better pocket pair? 6 combinations each exist for AA-QQ and TT. (For example, with AA, there are 4 aces available for the first card and three aces for the second card., and then you need to divide by 2, the number of permutations of 2 objects, to eliminate double counts). For JJ, there’s only one combination (only two jacks left in the deck since you hold two jacks in your hand). For AK, there are 16 combinations (four aces and four kings, and . In total, your opponent has 41 combinations, and 18 of those are higher pocket pairs. The probability of being against a higher pocket pair is therefore .

Summary

The basic definition of probability, along with an understanding of how to use permutations and combinations to compute probabilities, will go a long ways towards unraveling the numbers that operate beneath the surface. Having a command of these numbers and knowing a few approximations will help you make profitable decisions when playing poker. The next article in this two piece series of Poker Helper, Advanced Probability for Poker Players, will give you the analytical tools necessary for doing very sophisticated analysis when you’re away from the table.

Tony Guerrera is the author of Killer Poker By The Numbers and co-author of Killer Poker Shorthanded (with John Vorhaus)

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