I will take a first stab at one of the required calculations. I will assume full-ring (Hero plus eight Villains). We will assume that Hero flops a straight which obviously means that Hero has two unpaired cards and Flop has three unpaired cards (none of which pair Hero).
For the purposes of these "pure" probability calculations, I will follow convention and assume all players go to showdown on every deal.
Let's first calculate prob of Hero losing to quads versus exactly one Villain. I will first tally the villain-makes-quads cases (numerators) and at the bottom derive total number of cases (denominator) and then divide the sum of the numerators by the denominator to arrive at the desired probability.
Case 1: Turn and river are a pair and pair one of the flop cards (so there is trips on board)
= C(3,1)*C(3,2)*C(1,1)*C(44,1)
= 396
Case 2: Turn and river are a pair but do not pair one of the flop cards (so there is one pair on board)
= C(8,1)*C(4,2)*C(2,2)
= 48
Case 3: Turn and river both pair one of the flop cards, but not each other (so there is two pair on board)
= C(3,2)*C(3,1)*C(3,1)*C(2,1)*C(2,2)
= 54
Case 4: Turn or river pair one of the flop cards, but not both (so there is one pair on board)
= C(3,1)*C(3,1)*C(38,1)*C(2,2)
= 342
TOTAL:
= 396 + 48 + 54 + 342
= 840
Denominator (total number of possible "runouts" with exactly one villain):
= C(47,4)*C(4,2)
= 1,070,190
-> Prob of Hero losing to quads versus exactly one villain
= 840 / 1,070,190
= 0.078490735%
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Now consider two villains. Is it possible for both villains to make quads on the same deal? Let's go through the same cases we used above.
Case 1: Turn and river are a pair and pair one of the flop cards (so there is trips on board)
= 0 (impossible)
Case 2: Turn and river are a pair but do not pair one of the flop cards (so there is one pair on board)
= 0 (impossible)
Case 3: Turn and river both pair one of the flop cards, but not each other (so there is two pair on board)
= C(3,2)*C(3,1)*C(3,1)*C(2,2)*C(2,2)
= 27
Case 4: Turn or river pair one of the flop cards, but not both (so there is one pair on board)
= 0 (impossible)
TOTAL:
= 0 + 0 + 27 + 0
= 27
Denominator (total number of possible "runouts" with exactly two villains):
= C(47,6)*C(6,2)*C(4,2)/2
= 483,190,785
-> Prob of Hero losing to both villains making quads
= 27 / 483,190,785
= 0.000005588%
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It is clear that no matter how many villains Hero faces, no more than two can ever make quads on the same deal. Thus, our friend the Principle of Inclusion-Exclusion tell us:
Prob of Hero losing to quads facing N villains:
= N*P(1) - C(N,2)*P(Both)
where P(1) is the first prob we calculated, the prob of Hero losing to quads facing exactly one villain, and P(Both) is the second prob we calculated, the prob of Hero losing to quads facing exactly two villains when both villains make quads.
Thus, for N=8 we have the prob:
= 8*(840)/(1,070,190) - C(8,2)*(27)/(483,190,785)
= 48,148 / 7,669,695
= 0.627769422%.
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Others should feel free to confirm/refute the above results (I make more than my share of mistakes).
Of course, the assumption that all players go to showdown on every deal is unrealistic and would probably need to be dialed back.
Above I assumed that Hero flopped a straight. Someone can calculate that probability if they wish (it shouldn't be too difficult).
Others can also address the "twice in 3 hours" phenomenon if they wish.
Last edited by whosnext; 08-05-2019 at 12:25 AM.