# Probability of x pocket pairs at a table of n people (NLHE)?

With n people at a table, what is the probability that x of them are dealt pocket pairs? There are several easy ways to approximate this but I was wondering there was an elegant solution. Any takers?

• I have an issue. Doesn’t the fact that one person was dealt a pocket pair slightly in crease the likelihood that someone else will get a pocket pair. Doesn’t that logic lend to increasing the probability the more people who get a pocket pair? Obviously it wouldn’t be a dramatic increase, but it shouldn’t be ignored. Think of it this way, if 52 players are at a table and 51 of them have a pocket pair, there is a 100% chance that the last person will get a pocket pair as well. Just a thought – Patton Feb 19 at 0:27

I assume the value `n` is reasonable for a poker table (n = [2, 10]). In this case, it won't be relevant for the result.

You give the first player a card. The probability that the second card you give him has the same rank so that he'll get a pocket pair is: `3 / 51` (the 3 cards of the same rank that remained out of the remaining total of 51 cards).

After this, you can repeat the same rationale for the 2nd player: first cards and `3 / 49` chance that the second one will be a matching rank.

Extrapolating to `x` players, you get the following formula (in pseudo-code):

``````probability = 1
for i = 1 to x
probability = probability * (3 / (52 - 2 * i + 1))
``````

Like I said: `n` is irrelevant in this calculation. And this makes sense, if you think about it.

The problem with this formula is that it doesn't take into account the situation when the matching card of the one you get was already dealt to another player. But there's just no way to know that... (or maybe there is, but it's way too complicated).

PS: Applying the formula to 3 players gives you `1 / 4350`. It's not that improbable. Actually, it's sufficiently probable that it made this actually happen.

The calculations probably get out of hand, so that a Monte Carlo estimate would probably be the easiest way to get the answer. One way to approximate the answer is to just assume there is a 1/17 chance of anyone getting a pair. (if you consider 14 people in the game, you can see why this isn't accurate.)

The binomial distribution would apply, with n = number of players, x = number with a pair. p = probability of a pair = 1/17. q = probability not a pair = 1-p = 16/17.

probability x with pair = n choose x * p^x 8 q^(n-x); where n choose x = n!/(x!*(n-x)!)

and further x! = x factorial = x*(x-1)*(x-2)...3*2*1.

I will use n= 10.

No pairs = (16/17)^10 = 0.545 Cumulative = 0.545

1 pair = 10*16^9/17^10 = 0.341 Cumulative = 0.886

I did the rest with a spreadsheet.

2 pair = 0.096 cumulative = 0.982

3 pair = 0.016 cumulative = 0.998

4 pair = 0.0017 cumulative = 0.9999

5 pair = 0.00013 cumulative = 0.99999

6 pair or more = not very likely at all; cumulative approx. 100%.

The "cumulative" is the cumulative probability of that many pairs or less.

n = 9

0 = 0.549

1 = 0.326

2 = 0.081

3 = 0.012 cum 0.9988

n = 6

0 = 0.695

1 = 0.261

2 = 0.041

3 = 0.003 cum = 0.9998

n = 4

0 = 0.875

1 = 0.196

2 = 0.018

3 = 0.0008 cum = 0.99999

Just from the top of my head I would say: I think Radu Murea's calculations are pretty ok, however I think that it really is mandatory to calculate that someone else can receive the 2nd card that could make your pair. You could do it in 2 steps:

• What is the chance to receive a pocket if you were dealt 2 cards after each other:

First we receive any card (let's say a 6). Then, what is the chance to receive another 6 from that deck? 3/51 since there are 3 sixes left and 51 cards in the deck. If you were playing with n players, that would be n times 3/51 (1/17). Let's say, you play with 7 people, that would be 7/17 chance. However, we need to elliminate the fact that other people could get your cards as well...

• Chance that someone else gets your card:

If you play with n players (let's say 5), there would be a chance first there would be 3/51, then 3/50, then 3/49 and then 3/48 that one of the players would get your card. Once you receive your second card, you would have 3/47 chance (if your card wasn't gone yet) that you would receive a pocket.

So, calculating both together, I would say that the chance of getting a pocket pair with n players is around: n/17 - n * 3/(51 - n/2 + 1)

I am of course not sure if this is correct. It's just something I came up with in my head so...

EDIT: I am 100% sure my calculations are incorrect. Maybe someone can finetune this. I will also think about it with pen and paper.

This just a copy of guest geek with more detail. I tested it out myself. Give the check to guest geek.

Turns out this a binomial distribution

Binomial distribution is the distribution of a total number of successes in a given number of Bernoulli trials. The common notation is b(k; n, p), where k is the number of successes, n is the number of trials, p is the probability of success.
We know that b(k; n, p) = C(n, k) x p^k x (1 - p)^(n - k)

C(n, k) x p^k x (1 - p)^(n - k) C is combination
x is multiply
^ is power
p is probability of pocket pair = 1/17
n is number of hands = 10 k is number of hands with pocket pair

``````k   probability     cumulative
0   0.5453943226371 0.5453943226371
1   0.3408714516482 0.8862657742854
2   0.0958700957761 0.9821358700614
3   0.0159783492960 0.9981142193574
4   0.0017476319543 0.9998618513117
5   0.0001310723966 0.9999929237083
6   0.0000068266873 0.9999997503956
7   0.0000002438103 0.9999999942058
8   0.0000000057143 0.9999999999201
9   0.0000000000794 0.9999999999995
10  0.0000000000005 1.0000000000000
``````

The cumulative converging to 1 indicates it is correct. Well if it did not total 1.0 it would indicate it is wrong.

I am amazed how quickly it drops off.