BRUTAL- What are the odds?

[sickofants]

My gut tells me that these ARE gonna be pretty long odds.

Player 1: Pr(Pocket pair)=13*(4C2/52C2)=1/17
Player 2: Pr(Different pocket pair)=12*(4C2/50C2)=72/1225

So odds of both starting with a pocket pair are 72/20825

Now we need both players to hit their set and for the final flop card to be a blank, that is:

(2/48)*(2/47)*(44/46)=11/6468

Multiplying these probabilities together gives: 6/1020425 or about 1:170072 (which is somewhere close to 256*64)

You can be rest assured that if you have QQ in the hole against one opponent, the probability that opponent has AA or KK are:

2*(4C2/50C2)=12/1225

so the odds are about 1:104

[T50_Omaha8]

If you have QQ and the flop comes AQx and...say...8 other players are present, there are c(48,16) different sets of cards that could be distributed among your opponents.

There are c(3,2)*c(45,14) ways that exactly two aces can be distributed among your opponents, and c(45,13) ways that exactly three aces can be disributed among your opponents.

Given that exactly two aces exist among your opponents, there are 2!*8*14! different ways that the two aces belong to the same opponent. Given that exactly three aces exist among your opponents, the probability that nobody has pocket aces is (14/15)*(12/14) or 12/15 or 4/5, so the probability someone has pocket aces is 1/5.

Now we sum.

The probability two aces are distributed among your opponents is

c(3,2)c(45,14)/c(48,16)
= 3*32*16*15/2*48*47*46
= 23040/207552
= .1110083

while the probability three aces are distributed among your opponents is

c(45,13)/c(48,16)
= 3360/207552
= .0161887.

Given that two aces exist among your opponents, there are 16! different permutations of those two aces, and 2!*8*14! different ways they can be dealt to the same person. Then we have

2!*8*14!/16!
= 16*14!/16!
= 1/15, which matches intuition.

It has already been determined that if three aces exist among our opponents, there is a 1/5 chance that someone has aces. Thus the probability of someone else having a set of aces in this situation is

.1110083/15 + .0161887*.2
= .0074055 + .0032377
= .0105432
Or 1 in 95.

[KoH]

Thank you both for the help.

As far as taking the flop with 8 other people, actually only two players limped, and the guy with aces raised. I reraised, forcing the limpers to fold and he and I took the flop heads up.

[T50_Omaha8]

Ah. When I say present I only meant dealt in. It is assumed that any person with aces will see the flop. Given that someone raised to you, the conditional probability that aces are present increases.

[KoH]

Yea, especially since he wasn't really the aggressive or tricky type, I had to put him on AK, AQ, or a big pair. AK or AQ would have been more likely, and it could have been any other big pair...*sigh* I'll shake this beating off by Christmas I'm sure.

Given his tight play, I might almost be inclined to just call but that would have been a weak play and in any event, I needed those two other players out of the pot...and of course with no draws or boat possibilities...I'm just not going to get away from a set. I don't know why though, but when he called my post-flop bet, I just knew he had AAs in the pocket...I don't know why but I did. I'm just glad this was a limit game...

Anyway, thanks again.