Probability question?

[manku]

5 card Omaha 8OB...home game.

I'm dealt AAA34.

I have the nut low after the river.

What are the chances of someone else having an A3 to quarter me?

I kept pumping the pot three handed and, of course, got quartered.

Inquiring minds want to know...

manku

[curmudgeon]

don't worry about being quartered unless the stakes are high

maybe i'll run a simulation w Wilson software and get you some real numbers......
its very good for that

[elitegimp]

[ QUOTE ]
5 card Omaha 8OB...home game.

I'm dealt AAA34.

I have the nut low after the river.

What are the chances of someone else having an A3 to quarter me?

I kept pumping the pot three handed and, of course, got quartered.

Inquiring minds want to know...

manku

[/ QUOTE ]

If you get dealt AAA3 and then see 5 cards on the flop, you know 9 out of the 52 cards out there. So the probability 4 _random_ cards include the other A and exactly one of the other three 3s is 1*3*[number of ways to deal out remaining 41 cards]. This number is 41 choose 2 = 820, so there are 3*820 = 2460 possible hands containing the last A and one three. There are also 1*3*41 = 123 hands with the A and two of the last threes, plus 1 way to deal A333 = 2584 possible hands you quarter with.

For the denominator, you need to count how many ways there are to deal 4 out of 43 cards you haven't seen: 43 choose 4 = 123410.

So the probability that you quarter with a random hand is 2584/123410 = 0.0209 or ~2.1% However, your opponent would likely play a random hand and a hand with A3 differently...

edit: D'oh! Missed the "5 card" part! hold on for revised numbers

[elitegimp]

okay, you know about 10 cards (5 in your hand, and 5 on the board), so
* A3xxx = 1*3*[40 choose 3] = 29640
* A33xx = 1*3*[39 choose 2] = 2223
* A333x = 1*1*38 = 38

there are 29640+2223+38 = 31901 hands w/ A3 in them
there are 42 choose 5 = 850668 hands possible to deal

P = 31901/850668 = 0.0375 = 3.75%

same disclaimers about how your opponent plays as above

[MKR]

[ QUOTE ]
So the probability that you quarter with a random hand is 2584/123410 = 0.0209 or ~2.1% However, your opponent would likely play a random hand and a hand with A3 differently...

edit: D'oh! Missed the "5 card" part! hold on for revised n

[/ QUOTE ]

I believe you've calculated the probability that another person holds an A3?? preflop when there is only one opponent. What you need to answer the question is the probabilty that there is at least one A3?? out against you. That probability is 1 - (1 - 0.0209)^N, where N is the number of opponents you have.

MKR

[elitegimp]

[ QUOTE ]

I believe you've calculated the probability that another person holds an A3?? preflop when there is only one opponent.

MKR

[/ QUOTE ]

Ooh, good catch! I was definitely doing this for heads-up, not three-handed. However, one opponent having A3?? is definitely dependent on the other opponent's hand (since if Opp1 has Axyz then Opp2 _can't_ have A3xx), so I'm not sure the (1-(1-p)^N) works. It's definitely a good ballpark figure, though.

[donkeyradish]

[ QUOTE ]
don't worry about being quartered unless the stakes are high

maybe i'll run a simulation w Wilson software and get you some real numbers......
its very good for that

[/ QUOTE ]


Is that a general rule or just apt for this example?

ie. At a limit table should one always completely forget about the possibility of being quartered, as it is adquately compensated by all the times you scoop at least half the pot?

[curmudgeon]

overall you will come out ok..........