Probability Of Pocket Pair
Brian Alspach
Although aces form a winning hand 80% of the time, they are prone to losing against better hands. Two pairs will take the pot 31% of the time, while a simple pair will win the game 27% of the time. Three of a kind will win 12% of games, and a straight will win on average 9% of the time. On one specific deal, the chance of getting a pocket pair on the first deal is 1/17 and if you do, the chance of getting the same pocket pair on the next deal is 1/221; so the combined probability is 1/(17.221) = 1/3,757.
Poker Digest Vol. 5, No. 2, January 11-24,2002
You are playing in a 10-handed hold'em game and find that you have beendealt pocket 8s. When thinking about possible hands for otherplayers, it is natural to wonder about the possibility of other playershaving been dealt pocket pairs.
In my previous article, I determined the probability of another playeralso holding pocket 8s. What we learned is that the probabilityof this occurring is about 1/136. Thus, it is an unlikely event.
Given that you have been dealt a pocket pair, contemplating the possibilityof another player holding a pair of the same rank is mostly a matter ofcuriosity. What is far more interesting, and more important, are thechances that one or more opponents have been dealt pocket pairs of larger rank. This is the topic of this article.
The probabilities I am about to present should be of interest to allplayers. As far as I know, these calculations have not been carried outbefore. I cannot be certain no one has done them before, but giventhe general level of probabilistic results commonly available, it isno surprise I haven't seen this information. The methods employed inthe popular poker books are incapable of handling this problem.
If someone has worked on this problem, I would not be surprised to learnsimulations were used. I want to emphasize these probabilities are theoutcome of exact computations. Thus, the results are exact to withinthe decimal approximations of rational numbers.
pair | one rank | two ranks | three ranks |
K-K | .0439 | - | - |
Q-Q | .08412 | .001863 | - |
J-J | .12161 | .005435 | .0000768 |
10-10 | .15519 | .01044 | .0002844 |
9-9 | .1857 | .01662 | .0007109 |
8-8 | .2132 | .02397 | .001367 |
7-7 | .2380 | .03218 | .0023 |
6-6 | .2603 | .04114 | .003538 |
5-5 | .2801 | .05071 | .0051 |
4-4 | .2977 | .06076 | .007 |
3-3 | .3133 | .07118 | .009251 |
2-2 | .3269 | .08186 | .01186 |
The table above contains the information. I'll now describe the entriesof the table. The column headed ``Pair' tells us the rank of the pairheld by the player. The column headed ``One Rank' tells us theprobability there is exactly one larger rank for which either one or twopairs of that rank have been dealt to other players. The column headed``Two Ranks' tells us the probability there are exactly two larger ranksfor which either one or two pairs of each rank have been dealt to otherplayers. There is an analogous meaning to the column headed ``Three Ranks'.
Now consider some examples. Look at the row corresponding to a playerholding pocket kings. We find a probability of .0439 under the columnheaded ``One Rank.' That is the probability there is either one or twoplayers holding pocket aces. This translates into odds of about 21.8-to-1against someone having pocket aces. Note that there are no entries underthe other columns for pocket kings because there is only one rank largerthan king.
Now look at the row corresponding to a player holding 8-8. We see thereis a probability of .2132 that there is exactly one larger rank with oneor two players holding pocket pairs of the larger rank. This is slightlybigger than one-fifth, so that approximately one in every five times youfind 8-8 in your hand, you are going to be going against a pocket pair(or two) of one larger rank.
It's now appropriate to mention one fact about the probabilities givenin the table. I have not bothered to filter out situations when youare facing two people holding pocket pairs of the same larger rank. Thatis, if you hold 8-8, then a deal with one person holding 10-10 against you,and a deal with two people holding 10-10 against you have not been separated.The reason for this is that two people holding pocket pairs of the samerank is rare as we saw in part I. Thus, most of the contribution to theprobabilities given in the table comes from each larger rank having onlyone pair of that rank.
Of course, as a player holding 8-8, for example, having one player with10-10 against you is worse than having two people holding 10-10 againstyou. Two people with 10-10 prevent the larger pair obtaining trips whichis to your advantage. On the other hand, a 10 cannot come on board, sothat your chance for a straight has diminished.
Now look at the row corresponding to 2-2. We see the probability of apair of larger rank having been dealt is .3269. In other words, aboutone-third of the time, a pair of larger rank will be dealt. Not only doespocket deuces have to worry about overcards possibly pairing another player,about one-third of the time he already is up against a larger pocket pair.
Looking in the third column, we find a probability of .08196 that there arepairs of two larger ranks against pocket deuces. Thus, the odds against twolarger pocket pairs is only about 11.2-to-1. This is significant andindicates just how weak pocket deuces are.
Let me conclude with a few words on the derivation of these probabilities.A fairly detailed version is available at my web site(http://www.math.sfu.ca/~alspach) under a file entitled ``Multiple PocketPairs' in the Poker Computations directory. I always am happy to havesomeone check the details since elimination of errors is desirable.
The first step is to calculate the total number of ways of completing agiven fixed player hand. We are choosing 18 cards from 50, and this canbe done in ways. Once 18 cards are chosen, they can be brokeninto nine hands in ways. Theproduct of those two numbers is the number of completions.
The next step is to choose the rank of the pair held by the player. Callthis rank . Now count the number of ranks larger than .
We then calculate how many completions have 1, 2,..., larger rankswith pocket pairs of those ranks. After doing this, we useinclusion-exclusion to get the exact number of completions with 1, 2and 3 larger ranks. Dividing by the total number of completions givesthe entries in the table.
What makes the computations more intricate than normal is the fact therecan be either one or two pocket pairs of the same ranks.
last updated 4 December 2002
By Alton Hardin
Introduction to Odds and Probabilities Poker Strategy
Probability Of Flopping A Set With Pocket Pair
There are numerous odds and probabilities to learn in the game of Texas Holdem, which are essential poker strategy to understand. Below are some basic and essential preflop and flop odds and probabilities to learn.
Preflop Odds
Hand | Probability |
Two Suited Cards | 3.25:1 (24%) |
Any Pair | 16:1 (6%) |
Pocket Aces or Pocket Kings | 109.5:1 (0.9%) |
Ace King Offsuit | 109.5:1 (0.9%) |
Pocket Aces | 220:1 (0.45%) |
Ace King Suited | 330.5:1 (0.3%) |
Flop Odds
Pocket Pairs | Probability |
Pocket Pair Flopping a Set | 8.28:1 (10.8%) |
Pocket Pair Flopping a Full House | 135:1 (0.73%) |
Pocket Pair Flopping Quads | 407.33:1 (0.25%) |
How Many Combinations Of Pocket Pairs
2 Suited Cards | Probability |
2 Suited Cards Flopping a Flush Draw | 8:1 (11%) |
2 Unsuited Cards Flopping a Flush Draw | 44:1 (2%) |
2 Suited Cards Flopping a Flush | 118:1 (0.85%) |
2 Non-Pair Cards | Probability |
2 Non-Pair Cards Flopping One Pair | 2.5:1 (29%) |
2 Non-Pair Cards Flopping Two Pair | 49:1 (2%) |
2 Non-Pair Cards Flopping a Straight | 76:1 (1.3%) |
2 Non-Pair Cards Flopping Trips | 73:1 (1.3%) |
2 Non-Pair Cards Flopping a Full House | 1088:1 (0.9%) |
2 Non-Pair Cards Flopping Quads | 9800:1 (0.01%) |