Probability Of Pocket Pair

Brian Alspach

Lotus asia casino ndb codes. Poker Digest Vol. 5, No. 1, December28-January 10, 2002

Every now and then in hold'em, a table is treated to two players who'vebeen dealt the same pocket pair. Determining the probability of thishappening is an easy application of inclusion-exclusion. Let's gothrough the steps outlining the method but excluding the details.

Since we are interested only in the outcome of two players being dealtthe same pocket pair and not which players are dealt the pairs, we arediscussing player ``semideals'. The number of player semideals in a10-handed game is easy to determine. There are ways ofchoosing the 20 cards for the players to receive. Once the 20 cardsare chosen, there are 19!! ways of breaking the 20 cards into 10 handsof two cards each. Recall that .

Then subtract the probability that more than one opponent has a higher pocket pair. N = Number of players remaining in the hand Probability that several opponents have a pocket pair, for which Probability that exactly n players have a pocket pair, for which. Probability of facing several higher pocket pairs. Take the number of higher pairs, multiply by the number of other players, and divide by 2. That is the percentage probability that there will be at least one higher pair. For example, with a pair of jacks in a 10 player game the probability of a higher pocket pair is 3.9/2 = 13.5%. Using that formula you get the following for all situations. Two Pairs 123,552 4.75% One Pair 1,098,240 42.3% Nothing 1,302,540 50.1% Wait, how did I compute the probability of getting “Nothing”? How would you answer the question: “What is the probability of getting Three of a Kind or better?” 3.

The product of and 19!! is then the number of player semideals.This product is 82,492,346,176,096,206,475,125 and is the last huge numberI'll throw at you in this article.

In order to get the exact probability, and the reason we useinclusion-exclusion, we have to take into account the fact there couldbe more than two people having the same pocket pair as someone else. Infact, it is possible (though highly unlikely) that five sets of identicalpairs have been dealt. Let's count how many ways this could happen.

There are choices for the ranks. Each set of four cardsmay be split intotwo hands in three ways. Thus, we multiply by and obtain312,741 player-semideals with five sets of identical pairs. Use to denote 312,741. The probability of such a player-semideal has 17 zerosfollowing the decimal point. I said it was highly unlikely.

We do one more example to illustrate the procedure.

Let's determine howmany player semideals have exactly four sets of identical pairs. Thereare choices for the ranks of the pairs; there are three ways tosplit the four cards of each rank into two hands; there are choices for the remaining four cards; and there are three choices forsplitting the four remaining cards into two hands. Multiplying all thesenumbers gives us the number of player semideals with at least four setsof identical pairs.

However, any player semideal counted in iscounted five times in the preceding product. Thus, we subtract and obtain player semideals with precisely foursets of identical pairs.

We determine and in the same way. We then take thesum to obtain the number ofplayer semideals with at least two people being dealt pairs of the samerank. We divide by the total number of player semideals given above andobtain a probability of .00215899 that, upon randomly dealing two cards toeach of ten players, there is at least one set of pairs of the same rankdealt to two players. This is about once in every 463 deals.

The preceding numbers have no restriction on the rank of the pairs. Whatcauses the most drama at the table is when two players both are dealtpocket aces or pocker kings. So let's work out the probability that twoplayers are dealt pocket aces.

Here we can simply use proportionality arguments. The hardest work isobtaining the numbers , but once they have been obtained we can usethem as follows. Now is the number of player semideals with precisely one rank for which two players are dealt pairs of that rank.Since each rank occurs equally often, we divide by 13 to obtainthe number of player semideals with two players being dealt pocket acesand no other rank with two players being dealt pairs of rank .

Continuing, is the number of player semideals with exactly tworanks for which four players have pairs of these two ranks. There are78 combinations of two ranks chosen from 13, and there are 12 combinationsof ranks including ace. That is, 2/13 of the combinations involve ace.So we multiply by 2/13 to get those which involve two players withpocket aces.

Probability Of Flopping A Set With Pocket Pair

Similarly, we multiply by 3/13, by 4/13, and by 5/13.Doing this, then summing, and then dividing by the total number ofplayer semideals gives us a probability of .000166 that two players havebeen dealt pocket aces (the same is true for any fixed rank). This meansabout once every 6,025 deals two players should be dealt pocket aces.

Let's take another viewpoint now and ask the following: Suppose you areplaying a 10-handed hold'em game and upon looking at your two hole cards,you find A-A. What is the probability another player also has beendealt pocket aces?

This is now a question involving conditionalprobability and you must be careful. Even though the probability, as wejust saw, for two people to be dealt pocket aces is small, the result maybe quite different once the condition that someone has aces is given.

The total number of completions to player semideals is given by theproduct of and 17!! because we are choosing 18 cards from 50to be dealt to the other nine players, and there are 17!! ways topartition the 18 cards into nine hands of two cards each. Then set asidethe remaining two aces to go into another hand. For the other eight hands,we are choosing 16 cards from 48 and partitioning the 16 cards in 15!!ways to form completions.

This gives us completions toplayer semideals with another pair of pocket aces. Dividing the secondnumber by the first gives us a probability of 9/1,225 that someone elsealso has been dealt pocket aces. Expressed as a decimal, the probabilityis about .00735, and expressed as an Egyptian fraction, the probabilityis about 1/136.

Therefore, if you are dealt pocket aces, the odds against another playeralso having pocket aces is about 135-to-1.

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last updated 4 December 2002
mathman9
I have some very difficult combinatorial questions. Hopefully someone can provide me with insight into how to attack these questions and answers. What is the probability of exactly one player being dealt a pocket pair, at a table consisting of 10 players (Pre-flop in Texas Hold-Em, each of the 10 players is dealt 2 cards from a 52 card deck)? Then I would like to know the probability of this event happening to exactly 2 players of the 10. Then the probability of this event happening to 3 players of 10. And so on, until the probability of all 10 players being dealt pocket pairs. So basically I am asking 10 questions. Secondly, am I correct in figuring the probability no player is dealt a pocket pair out of 10 players is the complement of the sum of the 10 answers to my questions above?
Probability of being dealt a pocket pair
AceTwo
This seems quite difficult to calculate combinatorially.
Maybe a sim would be easier. Or maybe such calculations are already available on the internet.
COMBINATORIAL
First you need to calculate the total combs.
Total Combs = { Comb(52,2)*Comb(50,2)*Comb(48,2) .... Comb(34,2) } / 10! = 8.25 x 10^22.
Note that the above is for 10 time (10 hands) and then divide by 10! to convert from permutations to combinations (order of hands does not matter).
Easiest to calculate is for all 10 players to get pair (and each pair to be different)
10 pairs (all different pairs) = Comb(4,2)^10 *Comb(13,10) = 1,73 x 10^22
Not exactly what you asked but the only that I can thing that can be easily calculated.
For the rest you will need to do a lot of computations.
mathman9
so that gets you the number of the event? (the numerator in the probability)… what about the denominator (number of sample space)?
kubikulann
Just to be sure: is there something special about 'pocket' pair, or is it just a pair?
kubikulann

Secondly, am I correct in figuring the probability no player is dealt a pocket pair out of 10 players is the complement of the sum of the 10 answers to my questions above?

Yes. That covers all the possible cases.
Reperiet qui quaesiverit
AceTwo

so that gets you the number of the event? (the numerator in the probability)… what about the denominator (number of sample space)?


Probability Of Being Dealt A Pocket Pair

denominator (number of sample space) = Total Combs = { Comb(52,2)*Comb(50,2)*Comb(48,2) .... Comb(34,2) } / 10! = 8.25 x 10^22
Sample space (denomiantor) is easy, see above.
Number of the event is the difficult part.
PocketIt is diifficult , or more correct time-consuming because I think there is no short formula.
Always in these combinatorial calculations you are trying to find the shortest formula.
The formulas would kind of be a big excel table.
For example for 1 Pair the Algorithm would be like:
First Hand = Pair That's easy
Comb(4,2) *13 = 78 MULTIPLY by
Second hand = 50 cards to chose 2 . 48 of these can be thought as Comb(4,1) for 12 ranks and 2 are Comb(2,1) for 1 rank.

Probability Of Pocket Pair


You get some combs to be Comb(4,1)^2 and some Comb(4,1)*Comb(2,1) and we need to also get weights.
For the Comb(4,1)^2 the weight should be Comb(12,2), ie from the 12 ranks chose the 2 ranks
For the Com(4,1)*Comb(2,1) the weight is 12, ie for the 2nd rank
=Comb(4,1)^2 * Comb(12,2) + Comb(4,1)*Comb(2,1) *12 MULTIPLY BY
Third Hand
Here it gets even trickier
It is the same as Second Hand but Cards are reduced by 2 from and we get 2 cases:
a) 1 rank has 2 cards, 2 ranks have 3 cards, 10 ranks have 4 cards (when both cards in Hand 2 are different from Pair)
b) 1 rank has 1 card, 1 rank have 3 cards, 11 ranks have 4 cards (when one of the cards in Hand 2 is same as pair)
So you do teh same exercise as in Second Hand but for each case with the appropiate weight
Fourth Hand Etc
The cases become more and more as you go down the hands and teh calculation increase a lot until you get to the 10th hand.
It can be done in excel table going down the expanding as each hand progresses.
There could be an easier Algoiritm than the above.
beachbumbabs
Administrator

Just to be sure: is there something special about 'pocket' pair, or is it just a pair?


A pocket pair is any pair in hand, your two cards. A pair on the board is a pair in the 5 community cards that can be used by you or anyone else, including a house hand. Pairing the board is one in your hand of two, one on the board of five.
If the House lost every hand, they wouldn't deal the game.
kubikulann
The problem is extremely involved, combinatorially.
To follow the advice of G. Polya (How to solve it) , let us tackle a simpler similar problem first. Let us do it for a 3-person game.
I propose to work with arrangements and not combinations. Indeed, you can always decide beforehand a specific order of the players. So, there will be a 1st player, a 2d player and a 3d player, with those holding a pair ordered first. Also, I consider the cards in order of deal.
Posibilities : 3 different allocations of 'exactly one pair', 3 of 'exactly two pairs, 1 of 'exactly three pairs' (binomial coefficients).
First player has a pair. Probability is 3/51. (I refrain from writing 1/17, as it is easier to follow when somehow 'seeing' the number of cards).
Second player's first card may be of the same rank (p=2/50) or a different (p=48/50). In the former case, the prob of getting a pair is 1/49; in the latter 3/49.
Case A: 2 has a pair.
If it is of the same rank as 1, 3 must receive another rank and his prob of forming a pair is 3/47.
If it is a different pair, 3 may receive a rank already present in the preceding hands (4/48) with a prob of forming a pair equal to 1/47, or a brand new rank (44/48) with a prob of pair equal to 3/47.
Case B: 2 has no pair.
B1. One of his cards is of the rank of 1's pair (p=2x2x48/50/49)
-- 3 may receive a card of 1's rank (1/48): no chance of forming a pair
-- or of the other card of 2 (3/48): 2/47 chance of a pair
-- or a new rank (44/48): 3/47 chance of a pair
B2. 2's cards are both different from 1's (p=48x44/50/49)
-- 3 receives a card of 1's rank (2/48): 1/47 of a pair
-- or of one of é's ranks (6/48): 2/47 of a pair
-- or a new one (40/48): 3/47 of a pair
Phew! You see how complex it would become with 4, 5, etc. players. The number of cases increases exponentially.
Now we have the answers. (Divide all by 51x50x49x48x47)
P(exactly one pair) = 3 x 3 x [ 2x2x48x(1x47+3x45+44x44) + 44x48x(2x46+6x45+40x44)] = 76,380 x 48
P(exactly two pairs) = 3 x 3 x [ 2x1x(0+44x48) + 48x3x(4x46+44x44)] = 19,344 x 48
P(exactly three p.) = 1 x 3 x [ 2x1x(0+3x48) + 48x3x(4x1+44x3)] = 414 x 48
Total prob = 51! / 46! = 5,872,650 x 48
Complement = P(no pair) = 5,776,512 x 48
In percent,
P(0) = 95.1%
P(1) = 3.9%
P(2) = 0.988%
P(3) = 0.021%
I won't do it for more players. :-)
Please notice how the probability of a second player forming a pair if the first one has made one is increased by about 1.5%.
When you have a pair, there is MORE chance that someone else also has one. (The same in Caribbean Stud, etc.)

Probability Of Flopping A Set With Pocket Pair

kubikulann

How Many Combinations Of Pocket Pairs

There is definitely something wrong in my computations. If the P of geting a pair is 1/17 = 6%, the prob of one pair should be closer to 3 times that..
Sorry. I do the check.
EDIT: ok! got it. It's the ordering that omits some possible cases (namely the one where 2 does not get a pair but 3 does).

Probability Of Being Dealt Pocket Pair

mathman9
a pocket pair is a pair being dealt to you in texas hold em poker pre-flop. each player is dealt two cards pre-flop. so a pocket pair means your two cards are a pair