Join Date: Mar 2009
Posts: 6,729
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.