A full ring table 10 players, let's say everyone is at least 50 BB Deep. Let's also say that anyone with a pocket pair always calls A bet pre flop. How often will someone flop trips in lets say 50 hands? Or or will it take more than 50 hands for that to happen, then how often?

Please replace the word " trips " in this post and in the title with the word " set " TY sorry

OP messaged this problem to me. I provided a quickie solution of 5.4% for at least one of ten players having a set or better (ignoring flushes and straights).

I used the general equation:

Sum (n=0 to 10) {Pr(n players have a pair)*(1-Pr(no player sets+ on flop|n)}

However, it was calculated assuming all events were independent, which is certainly not the case.

An exact analytic solution appears to me to be very complex so I wrote a simulation program. For one player the exact probability is Pr(pair)*Pr(flop set+|pair) = 1/17*0.118 = 0.0694. The simulation gave 0.0697 so the simulation seems right for n=1.

For n= 10 the simulation showed for 1 million trials, 5.37%, so I’m fairly confident the 5.4% is a pretty good approximation.

An independent analysis by another poster is welcomed.

I coded up a simple simulation as well.

For a 10-player table, in 10 million trials I found that at least one player flopped a set (or quads) in 6.74927% of the deals.

I double-checked my code but could not see any errors (notoriously difficult to find one's own errors). So not sure why my simulation result does not agree with statmanhal's.


Edit: in case anyone wonders my code requires a player to have a pocket pair to flop a set (or quads).

Okay. Found an error in my simulation – forgot to reset a Boolean variable when the number of players > 1. Now get 6.942% for 2 million trials. Should have suspected something when the simulation showed such close results to the formula result assuming independence.

We know prob of one player flopping a set (or quads), requiring a pocket pair, is given by:

= (1/17) * [C(2,2)*C(48,1) + C(2,1)*C(48,2)]/C(50,3)

= (1/17) * (2,304/19,600)

= 2,304 / 333,200

= 0.006914766 (or 0.6914766%)

Call the above number P1.

Let P10 be the prob that at least one player on a 10-player table flops a set (or quads), requiring a pocket pair.

Then, for this problem we know by PIE that P10 < 10*P1 = 6.914766%.

I have run 50 million trials of the 10-player case and get an estimate for P10 to be:

= 6.756%

(The reason I am posting this is because I don't think a simulation of 2 million trials of the 10-player case should be giving a result as high as 6.942%.)

Your'e confirming what a crappy programmer I am. I had counted flops where 2 (3) players with pairs flopped a set as 2 (3) successes. When I corrected this got 6.7547%.

Glad to hear it (if you know what I mean).

I hope you didn't take offense to my questioning your simulation results. I was also wondering if my simulation results were a bit goofy.

It reminds me of a thread we had years ago when several people were contributing a mix of analytical and simulation results. I spent several days programming and running a simulation of the darned problem and could never replicate what by then were known and confirmed answers. A very frustrating time for me (boo hoo).

Here's my attempt at an analytical solution.

First let's calculate the distribution of the number of ranks appearing on the flop (pretend that the flop is dealt first to make things easy for us here).

Scenario 1: Prob(1 rank appears) = C(13,1)*C(4,3) / C(52,3) = 52 / 22,100

Scenario 2: Prob(2 ranks appear) = P(13,2)*C(4,2)*C(4,1) / C(52,3) = 3,744 / 22,100

Scenario 3: Prob(3 ranks appear) = C(13,3)*C(4,1)*C(4,1)*C(4,1) / C(52,3) = 18,304 / 22,100

Next, for each of these scenarios, calculate the prob of at least one player flopping a set (or quads), requiring a pocket pair.


Scenario 1: One Rank appears on Flop

Clearly, prob of anyone flopping a set (or quads), requiring a pocket pair = 0


Scenario 2: Two Ranks appear on Flop (call it XXY)

Case 1: one player has pocket pair XX

= C(2,2)*C(47,18)*(1*3*5*7*9*11*13*15*17)/[C(49,20)*(1*3*5*7*9*11*13*15*17*19)]

= 5/588

Case 2: one player has pocket pair YY

= C(3,2)*C(47,18)*(1*3*5*7*9*11*13*15*17)/[C(49,20)*(1*3*5*7*9*11*13*15*17*19)]

= 15/588

Case 3: one player has pocket pair XX, another player has pocket pair YY

= C(2,2)*C(3,2)*C(45,16)*(1*3*5*7*9*11*13*15)/[C(49,20)*(1*3*5*7*9*11*13*15*17*19)]

= 45/211,876

So via PIE, the prob for this scenario is:

= 5/588 + 15/588 - 45/211,876

= 85,940 / 2,542,512


Scenario 3: Three Ranks appear on Flop (call it XYZ)

Case 1: one player has pocket pair XX

= C(3,2)*C(47,18)*(1*3*5*7*9*11*13*15*17)/[C(49,20)*(1*3*5*7*9*11*13*15*17*19)]

= 15/588

Case 2: one player has pocket pair XX, another player has pocket pair YY

= C(3,2)*C(3,2)*C(45,16)*(1*3*5*7*9*11*13*15)/[C(49,20)*(1*3*5*7*9*11*13*15*17*19)]

= 135/211,876

Case 3: one player has pocket pair XX, another player has pocket pair YY, another player has pocket ZZ

= C(3,2)*C(3,2)*C(3,2)*C(43,14)*(1*3*5*7*9*11*13)/[C(49,20)*(1*3*5*7*9*11*13*15*17*19)]

= 9/582,659

So, via PIE and the symmetry of XYZ, the prob for this scenario is:

= 3*(15/588) - 3*(135/211,876) + 1*(9/582,659)

= 86,973 / 1,165,318


Combining the Three Scenarios

Therefore, the overall probability of at least one player at a 10-handed table flopping a set (or quads), requiring a pocket pair, is seen to be:

= (52/22,100)*0 + (3,744/22,100)*(85,940/2,542,512) + (18,304/22,100)*(86,973/1,165,318)

= 1,520,478 / 22,511,825

= 0.067541303292825...


Comments welcome (identification of mistakes encouraged).

COOL!

Quoting myself - that's one thing I got right.

Wow that seems like a lot, learn something new every day thank-you thank-you both very much. Shaking my head in amazement

While we are here, let's post the results for all standard number of players at a poker table. Here is the probability of anyone flopping a set (or quads), pocket pair required, with N players at the table (NLHE assumed).

N=1

This has been derived many times (including in this thread above) and in many different ways. This prob is:

= 179,712 / 25,989,600

= 144 / 20,825

= 0.006914766

= 0.6914766%

N=2

Derivation is very straightforward (can use the approach given above in this thread). Prob is:

= 193,763,232 / 14,047,378,800

= 310,518 / 22,511,825

= 0.013793551

= 1.3793551%

N=3

Derivation is very straightforward (can use the approach given above in this thread). Prob is:

= 95,663,101,248 / 4,635,635,004,000

= 2,322,822 / 112,559,125

= 0.0206364611

= 2.06364611%

N=4 thru 10

The derivation can follow the approach given above. The algebra becomes a bit messy, and the resulting formula is a bit clunky (and requires a "simplification process" to evaluate on some calculators), but here it is. [Note, the notation X!! refers to the double-factorial expression and for X an odd integer is the product of the odd integers from 1 to X.]

Prob of anyone flopping a set (or quads), pocket pair required, for N = 4 to 10 is:

= [155,664*(2N-3)!!*C(2N,2) - 2,430*(2N-5)!!*C(2N,4) + 36*(2N-7)!!*C(2N,6)] / [22,511,825*(2N-1)!!]


Summary Table

Here is a table that summarizes our results.

Number of Players	Prob of Anyone Flopping a Set*
1	0.6915%
2	1.3794%
3	2.0636%
4	2.7444%
5	3.4215%
6	4.0951%
7	4.7651%
8	5.4317%
9	6.0947%
10	6.7541%

* Prob is of at least one player flopping a set (or quads), pocket pair required.

P(unpaired flop) = C(13,3) * 4^3 / C(52,3) = f1
P(paired flop) = 13*12 * 4 * C(4,2) / C(52,3) = f2

P(at least one set | unpaired flop) =
10*3*3 / C(49,2) -
C(10,2) * 3 * 3^2 / C(49,4) / 3 +
C(10,3) * 3^3 / C(49,6) / 5!!
= p1

P(boat or quads | paired flop) = [10*(3+1) / C(49,2)] - [C(10,2) * 3 / C(49,4) / 3] = p2

Overall P(flop set+, not counting single-rank flops) = f1*p1 + f2*p2 = 6.7541%
If only sets count then P(flop set) = f1*p1 = 6.1815%

So I agree with whosnext.

Edit: @OP, those percentages are for a single dealt hand, not 50.

If you wanted to know how many times they'll occur on avg in 50 hands, multiply by 50.
If you were asking how often it will happen at least once in 50 hands, that's: 1 - P(not)^50

So if only sets count then it's 1 - .061815^50 = 95.88% and on avg it will happen 3.09 times.

Thanks heehaww for the confirmation/contribution. PIE and you are truly amazing.

By the way, heehaww's post points out that what I/we have been calling "flopping a set (or quads), requiring a pocket pair" includes the possibility of flopping a full house where the flop has one of the rank of my pocket pair and the other two cards being a different pair.