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What exactly are you asking?

How many players at table see flop?

The probability of flopping an "under" two pair varies based upon your hole cards. It is easier to flop an under two pair when holding 32 than when holding AQ.

So do you want the probability over all possible hole card holdings or a formula that gives you the probability based upon your specific hole cards?

Let's see if I can do a simple case. Suppose we are interested in answering the following question.

Assume 1 hero and 1 villain who each see the flop. Given that hero has hole cards xy, how many flop/villain hand combos would give hero a lower two-pair than villain.

We will exclude the case where there is a pair on the flop and each make two pair such as Q7 vs. 97 and flop is KK7.

Thus, we will only be interested in the case where flop is unpaired and both hero's hole cards and villain's hole cards are unpaired. An example is Hero has J8, Villain has J9 and flop comes J98.

There are three cases to consider (remember hero has xy). It is straightforward to tally how many flop/villain combos are possible in each case.

Case 1. Flop is wxy and villain has wx (w>x>y)
C(3,1)*C(3,1)*C(14-x,1)*C(4,1)*C(3,1)*C(2,1)

Case 2. Flop is wxy and villain has wy (w>x>y)
C(3,1)*C(3,1)*C(14-x,1)*C(4,1)*C(3,1)*C(2,1)

Case 3. Flop is xwy and villain has xw (x>w>y)
C(3,1)*C(3,1)*C(x-y-1,1)*C(4,1)*C(3,1)*C(2,1)

If I did this correctly, when you combine terms you get:

TOTAL = 216*[27-(x+y)]

Of course, the usual numerical ranks apply (Ace=14, King=13, Queen=12, Jack=11, Ten=10, ..., Deuce=2).

Given hero has xy, there are C(50,3)*C(47,2) = 19,600*1,081 = 21,187,600 flop/villain combos in this case.

If we are interested in knowing the probability of hero flopping an under-pair in two-handed NLHE (before we know what hero holds), then we would need to use the formula above to tally all possibilities.

If we are interested in knowing the probability of hero flopping an under-pair in N-handed NLHE (given hero is dealt xy), then the formula above would need to be adjusted for how many villains are at the table (see the flop).

Okay, let's tackle one of these questions this morning.

Suppose you want to know what is the probability that Hero will flop an under-two-pair in a 2-handed NLHE game in which Hero and Villain always see a flop. That is, before any cards are dealt.

Based upon the formula given in the previous post, we can simply tally across all [x,y] possibilities. Dusting off some high school algebra for summing the first N integers and the first N squares, I find that there are a grand total of:

GRAND TOTAL = 2,965,248

deals of 2-handed NLHE thru the flop which give Hero an under-two-pair (remember that we are excluding deals such as Q7 vs. 97 on a KK7 flop).

There are C(52,7)*C(7,2)*C(5,3)*C(2,2) = 28,094,757,600 such deals of 2-handed NLHE thru the flop.

2,965,248 / 28,094,757,600 = 0.00010554453 or around 1 in every 9475 deals.

Of course, if we are interested in either Hero or Villain flopping an under-two-pair (i.e., if we seek the probability of anyone at the table flopping an under-two-pair) we would need to double this probability.

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Now that I think of it, there is surely a direct way to calculate the 2,965,248 figure. Flop has to come with 3 different ranks, and the hero and villain cards must be such that hero flops an under-two-pair (see the three cases listed in the previous post). Essentially, we are interested in the three specific ways in which each has two ranks equal to the board and they (of course) share one rank.

Thus, total number of deals thru the flop which satisfy this is:

3*C(13,3)*C(4,1)*C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(3,1 )*C(2,1) = 2,965,248

which, thankfully, agrees with what we found above.

Seems too low. > around 1 in every 9475 deals.

Here is the easiest way to do this.

Answer the question; "what is the probability that I flop top two pair and you flop bottom two". Then multiply your result by three since you could also have top and bottom or I could have top and bottom and you have bottom two.

To answer that first question you multiply the chances of the flop being unpaired (52/52 x 48/51 x 44/50) times the chances I hold the top two of that flop (6/49 x 3/48) times the chances that you (given my top two) hold the middle and bottom card. 2/47 x 3/46 gives you the chance that the middle card was the dealt to him first so you double that number since they can be dealt in either order.

In other words its 52/52 x 48/51 x 44/50 x 6/49 x 3/48 x 2/47 x 3/46 x 2 times three.