Probability of 2 PP's at 9 handed table?

timmay28 · #1 10-09-2007, 05:19 PM

I tried googling and searching this forum to no avail.. not sure how to go about figuring it out mathematically.

Gonso · #2 10-10-2007, 04:54 AM

Do you mean if you know you look down and see you have a PP, what are the chances that "at least one" other player has one also? I suspect this is what you're looking for.

Or do you mean, 'if you dealt nine hands out, what are the chances that exactly two players have pairs and the other seven do not'?

Each player has about a 5.88 percent chance of being dealt a pair individually, but it's easier to answer if you clarify that.

tshort · #3 10-10-2007, 07:17 AM

If you assume any given player has a 3/51 chance of being dealt a PP:

C(9,2) * (3/51)^2 * (48/51)^7 = 0.0815

Binomial Distribution

This should be close enough to the correct answer, but isn't exactly correct because the events are not independent.

I see no way of working around the numerous cases of conditional probabilities to get the exact solution.

timmay28 · #4 10-10-2007, 11:26 AM

[ QUOTE ]
Do you mean if you know you look down and see you have a PP, what are the chances that "at least one" other player has one also?

[/ QUOTE ]

Sorry.. yeah that one.

jay_shark · #5 10-10-2007, 01:08 PM

1 - [(1225-73)/1225]^8 ~ 38.83%

This is a good approximation to solve these types of problems . Notice that I subtract 73 hands from 1225 since there are 73 pairs remaining which may also include the one you're holding . If you're interested in distinct pairs from your own then the answer is 1- [(1225-72)/1225]^8 ~ 38.4%

timmay28 · #6 10-10-2007, 11:22 PM

Thanks guys.

EasilyFound · #7 10-11-2007, 07:55 PM

[ QUOTE ]
C(9,2) * (3/51)^2 * (48/51)^7 = 0.0815

[/ QUOTE ]

I guess I should know the answer to this, but I don't.

What does the "," in "9,2" stand for?

roggles · #8 10-11-2007, 08:06 PM

You serious? C(9,2) is the number of combinations of two elements out of 9 distinct elements to choose from.

So C(3,2) = 3 since if the elements are a,b,c the combinations are (a,b), (a,c) and (b,c)

EasilyFound · #9 10-11-2007, 08:55 PM

yeah, i'm dead serious. i'm not a mathematician, so i don't know all of the symbols used when you people speak to each other. thanks for the answer, though. now i know.

rrrorrim · #10 10-12-2007, 04:33 PM

I'll just piggyback on your thread with my question!

Few days ago in micro NLHE I had KK. I pushed moderately on PF and two normally passive players showed aggression. They kept reraising in small amounts. I thought, What the hell? It's like they both have Aces or something. I was kind of joking with myself, but I ended up folding before I was committed.

Turns out they both had pocket Aces. What are the chances of KK, AA, AA on the flop in a 10-player game?