Ok, so from what I have found, the probability of getting dealt a pocket pair and hitting quads by the river is 1/2083 for just one person. What I want to know is what is the probability of any of the 9 players at the table getting a pocket pair and hitting quads by the river. Simply, what are the chances at a 9 handed table that someone will hit quads with a pocket pair. We are assuming that no one ever folds and every hand goes to the river.


I'm sure there is a simple equation for this but I'm out of my comfort zone here and want to be certain.

Correct these if I'm wrong:

To be dealt a pocket pair - 1/17 (.0588)
To hit quads if you have a pocket pair - 1/123 (.00816)
To be dealt a pocket pair AND hit quads - 1/2083 (.00048)

My guess is for any player at a 9 handed table to get a pocket pair AND hit quads is 1/293... Please help me out.

With a very small probability such as getting quads, the chance of at least one success from n opportunities is very close to n*p. So, if p=0.00048, the chance of at least one quads in a 9 player showdown table is about 9*0.00048 = 0.00432 or 1 in 231.

Of course the 9 player showdown is quite unrealistic.

Also, it would not be too difficult (which is my polite way of saying that I am not going to do it) to determine how much "double-counting" the n*p formula entails.

Clearly, if we are talking NLHE, there is only one way for two players to make quads on the same deal. Each player gets dealt a pocket pair and the board also contains both of those pairs.

If one were to calculate this probability, the exact probability of a deal in 9-handed NLHE having at least one quads (assuming all go to showdown) is simply the n*p from the above posts minus the prob of two quads on the same deal.

I think this is something that is best done with simulations. I think you want this:

In a nine handed game: By the river, at least one player has a pocket pair and quads, but isn't making the quads due to four of the same rank on the board.

I ran the below code for this for 50,000,000 hands. Came up with 0.43%.

Welcome to the Probability Forum!

And thank you for your simulation results. The only caveat I would add is that I would favor a direct probability calculation over simulation results. Of course, ideally they should give the same information. The situation OP asked about is specific enough to be amenable to direct calculation.

Agree 110%!

Agree - but feel that is a pretty complex problem - would be very impressed to see someone do it, with the problem of not having quad from playing the board.

IMHO this is not a complex problem. See the above posts.

Consider one player.

Prob of one player making quads with a pocket pair = Prob of one player getting dealt a pocket pair * Prob of board having other two cards of his pair rank

= [C(13,1)*C(4,2)/C(52,2)] * [C(2,2)*C(48,3)/C(50,5)]

= 17296 / 36018920

= 0.000480192 or 1 out of every 2082.5 deals

The posts above show that multiplying this by 9 is very very close to the correct answer OP asked about (and post #3 above shows how to make this calculation exact).

Are you saying your answer to the posted question is 0.000480192?

I honestly have no idea how anyone can read my post and draw that conclusion.

Edit: I will post more below.

Alright I said I wasn't going to do it, but it seems like it might be of some "value" to the thread in light of other posts.

Here is what I get for the prob of exactly two people each making quads with different pocket pairs (using their hole cards) on a random deal of 9-handed NLHE:

[C(13,2)*C(4,2)*C(4,2)*C(11,1)*C(4,1)*C(9,2)*C(2,1) *C(2,2)*C(2,2)*C(43,14)*C(14,2)*C(12,2)*C(10,2)*C( 8,2)*C(6,2)*C(4,2)*C(2,2)] / [C(52,5)*C(47,18)*C(18,2)*C(16,2)*C(14,2)*C(12,2)*C (10,2)*C(8,2)*C(6,2)*C(4,2)*C(2,2)]

which I think simplifies to:

= 72 / 22511825

= 0.00000320

Recall from above our slightly high estimate of any person making quads with a pocket pair (using their hole cards) on a 9-handed NLHE table was:

= 9 * 17296 / 36018920

= 0.00432173

Removing the double-counting prob found above yields the overall exact prob of at least one player making quads with a pocket pair (using their hole cards) on a random 9-handed NLHE deal as:

= 0.00432173 - 0.00000320

= 0.00431853

which in exact form (I think) is: 97218 / 22511825

The exact calculation from start to finish:

9 * 13 * C(7,4) * C(4,2)/C(7,2) / C(52,4) -
C(9,2) * C(13,2) * 9 * C(4,2)*C(5,2)/C(9,4) * 3/5 * 1/3 / C(52,8)

= 0.43185304 %
= about 1 in 231.56

... which is the exact answer I found in post #10 above!

I just wanted to take this opportunity to say:

(1) Thanks heehaww for your awesome post and awesome posting

(2) These types of problems are indeed amenable to direct calculation

(3) There may be different paths to grandma's house but we all get there in the end (compare my slothful derivation to heehaww's gazelle-like derivation)

(4) Simulations are useful in confirming/refuting direct calculations and providing estimates when direct calculations are not tractable (I am not knocking simulations at all, I have and will continue to do tons of them)

Over and out.

I'm impressed you where able to do that.

Any thoughts on the UTH question I posted? It is a simple bet:

You play against a blind hand. If you make trips or better you win at the odds posted below. But you must beat the blind hand to collect. If you tie the blind hand, it is a push.

• Royal: 50-1
• Str8 Flush: 40-1
• Quads: 30-1
• Boat: 10-1
• Flush: 8-1
• Str8: 6-1
• Trips: 3-1

You do collect if you have AK on a board of 22289, when the blind hand is AQ. You collect on the trips bet.

This was my first guess. I really just thought that was too easy and couldn't be right. Thank you all for the replies. This was a huge help!!