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Old 05-21-2020, 05:09 AM   #1
Im Nacho Friend
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Odds of this happening

Hi! Played probably the funniest hand ever last night:

PokerStars, $0.23 + $0.02 - Hold'em No Limit - 60/120 (15 ante) - 8 players
Hand delivered by Upswing Poker

UTG: 1,592 (13 bb)
UTG+1: 4,168 (35 bb)
MP: 1,448 (12 bb)
MP+1: 4,545 (38 bb)
CO: 1,591 (13 bb)
BU: 6,542 (55 bb)
SB: 5,298 (44 bb)
BB (Hero): 4,062 (34 bb)

Pre-Flop: (300) Hero is BB with K K
UTG raises to 240, UTG+1 3-bets to 4,153 (all-in), 5 players fold, Hero calls 3,927 (all-in), UTG calls 1,337 (all-in)

Flop: (9,851) K 8 K (3 players, 3 all-in)

Turn: (9,851) 4 (3 players, 3 all-in)

River: (9,851) 2 (3 players, 3 all-in)

Total pot: 9,851

Showdown:
UTG+1 shows A A (two pair, Aces and Kings)
(Equity - Pre-Flop: 67%, Flop: <1%, Turn: 0%, River: 0%)

UTG shows J J (two pair, Kings and Jacks)
(Equity - Pre-Flop: 15%, Flop: <1%, Turn: 0%, River: 0%)

BB (Hero) shows K K (four of a kind, Kings)
(Equity - Pre-Flop: 18%, Flop: >99%, Turn: 100%, River: 100%)

BB (Hero) wins 9,851

Now I was wondering what are the odds of this happening?
So P(AA v KK v JJ u flopping quads)? Now I've had some probability courses last year, but it's been a long time since I've worked out a problem like this one. So tell me if I'm doing something wrong.

So the chance that we actually get dealt KK is: 6/C(52,2)
Then given we have KK the chance one opponent holds aces: 8*(6/C(50,2)) - C(8,2)/C(50,4)
Then given AA and KK the chance one opponent holds JJ: 7*(6/C(48,2)) - C(7,2)/C(48,4)
Then the amount of times we'll flop quads is: 44/C(46,3)

Multiplying the probabilities: =~ 1 in 52.574.641

Could be correct I guess. Any comments?
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Old 05-21-2020, 12:47 PM   #2
heehaww
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Re: Odds of this happening

Quote:
So the chance that we actually get dealt KK is: 6/C(52,2)
Then given we have KK the chance one opponent holds aces: 8*(6/C(50,2)) - C(8,2)/C(50,4)
Then given AA and KK the chance one opponent holds JJ: 7*(6/C(48,2)) - C(7,2)/C(48,4)
Then the amount of times we'll flop quads is: 44/C(46,3)
Close but a few issues:
• There are 7 villains instead of 8.
• P(at least one JJ | at least one AA among 7) isn't the same as P(at least one JJ among 6)
• P(KKx flop | hero KK & at least one AA & one JJ among 7) isn't the same as P(KKx flop | 2 kings & 44 non-kings left)

I'd do it like this:

P(KKx flop ) = 6*48/C(52,3)
P(hero KK | KKx flop) = 6/C(49,2)

P(>0 AA and >0 JJ | above) =
C(7,2)*6^2 / 3 / C(47,4) -
C(7,3)*2*6 / 5 / C(47,6) +
C(7,4) * 3/(7*5) / C(47,8)

product = 1 in 64,228,040
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Old 05-23-2020, 05:40 AM   #3
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Re: Odds of this happening

Thanks for your input, I have some questions though.

Why do you calculate the odds of flopping quads first? Aren't the odds more accurate if we take the hole cards of our opponents in consideration?

Also I want to know 1 opponent with AA and 1 with JJ so P(AA =1 u JJ =1)

And lol these were 9 handed tables so I went with that in my calculations -_-
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Old 05-23-2020, 01:33 PM   #4
heehaww
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Re: Odds of this happening

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Originally Posted by Im Nacho Friend View Post
Why do you calculate the odds of flopping quads first? Aren't the odds more accurate if we take the hole cards of our opponents in consideration?
It can be done in either direction, but I thought doing it in chronological order would be a little more work due to the other players. If you were only dealing yourself cards and a flop, then you can see that the order of the calculation doesn't matter:
(1/221)*48/C(50,3) = 6*48/C(52,3) / C(49,2)
For that matter, it also doess it matter if you deal your hole cards first or the flop first.

Better yet, we could do it without splitting it up into preflop and flop: 2 of the 4 kings have to be within a specific region of 3 cards and the other 2 have to be in a specific region of 2. We can treat the kings as indistinguishable and just divide the binary permutations. Suppose the deck is a string of 48 0's and four 1's. You need two 1's in the first two spots, then two 1's within the next 3 spots, so the chance of Hero getting KK and flopping quads is
C(3,2) / C(52,4)

Quote:
Also I want to know 1 opponent with AA and 1 with JJ so P(AA =1 u JJ =1)
Ok then all I need to do is tweak a couple coefficients. It also occurs to me that my earlier calculation ignored the possibility of the 3rd flop card being an A or J.

Start with 3/C(52,4)
P(3rd flop card ≠ A or K or J) = 40/48

P(exactly one AA & exactly one JJ, 8-handed | above) =
C(7,2)*6^2 / 3 / C(47,4) -
2*C(7,3)*2*6 / 5 / C(47,6) +
4*C(7,4) * 3/(7*5) / C(47,8)

The first line double-counts AA|JJ|JJ and JJ|AA|AA, and quadruple-counts AA|AA|JJ|JJ. We want those counted zero times, so we subtract them twice.
The 2nd line subtracts AA|AA|JJ|JJ eight times (twice when subtracting AA|JJ|JJ, twice vice versa, and then the whole line is multiplied by 2). So now that possibility is counted -4 times, which is why in the 3rd line we add it back 4x.

But actually, there's a slicker way to find the AA/JJ probability: take that 1st line and multiply by P(no other players get AA/JJ), which is 1 - P(at least one other player gets AA/JJ).

[C(7,2)*6^2 / 3 / C(47,4)] * [1 - 5*2/C(43,2) + C(5,2)/3/C(43,4)]

All told, it's 1 in 77,503,601 if neither villain is allowed to boat on the flop.
If they can, then to that we'll add: 3/C(52,4) * 8/48 * [C(7,2)*6/C(47,4) - 2*C(7,3)*3/5/C(47,6)]
bringing it to 1 in 70,422,149 (8-handed).

For 9-handed:

3/C(52,4) * {
(40/48)[C(8,2)*36/3/C(47,4)][1 - 6*2/C(43,2) + C(6,2)/3/C(43,4)] +
(8/48)[C(8,2)*6/C(47,4) - 2*C(8,3)*3/5/C(47,6)]
}
= 1 in 52,929,049

which is approximately C(8,2)/C(7,2) or 4/3 of the 8-handed answer.
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