Poker Probabilities

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?

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.

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

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

Quote: mathman9

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.

Quote: mathman9

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

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.
It 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.
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.

Quote: kubikulann

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.

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.)

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).

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