Hi

My big playing friend flopped two straights in 3 hours live play. His opponent(s) made quads twice. The stakes were 1/2. I told him this will never happen to him again in life. I.e. this was a "special" bad luck.

My question: Is this true for this kind of loss? He did take this to heart and left for the night. Can he basically not expect this loss to ever happen to him again: flopping straights and them making quads twice in one session? I would really like to confirm this with some basic stats. Can anybody help?

Regards
Francois

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.

Never is a very long time.

Okay, here is an easy piece of the puzzle: What is prob of flopping a straight in NLHE?

From a probability perspective it the same probability of having a straight in 5-card stud; that is, five cards make a straight. We assume, as before, that all deals see a flop and all hands go to showdown.

Clearly there are 10 different straights (high card Ace thru Five) and any suits will work for each of the five cards, except we want to exclude flushes (straight flushes).

Thus, straight tally is:

= 10*(4^5 - 4)

= 10,200

Total number of 5-card hands is:

= C(52,5)

= 2,598,960

So prob of flopping a straight is:

= 10,200 / 2,598,960

= 0.392464678%

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Now let's combine the two results posted so far:

Prob of flopping a straight and losing to quads versus eight villains (all hands go to showdown on every deal):

= (Prob of flopping a straight) * (Prob of losing to quads versus eight villains | Hero flopped a straight)

= (10,200 / 2,598,960) * (48,148 / 7,669,695)

= 24,074 / 977,119,143

= 0.002463773%

which is approx once every 40,588 deals.

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There are several aspects of "converting" this probability of occurrence into the associated probability of seeing this occur twice in X hands.


Edit to add: I guess I should add that the above derivation ignores the possibility that Hero may wind up with a better hand than a straight and the possibility that the winning hand is a royal/straight flush. I think the above derivation gets to the heart of OP's question.

It looks like others have done the math, so I'll just add...

Flopped straights lose all the time. The interesting part is that they both lost to quads. I don't think your friend would feel any better if they had lost to flushes though.