Flopping a set heads up

all_in_lam · #1 07-20-2006, 03:05 AM

I understand a similar situation was posted before, but the situation im looking for is a bit different.

How do you calculate the probability that player A and player B both flop a set, assuming it's heads up and they both have different pocket pairs.

is it as simple as just 2*2*48/48C3??

all_in_lam · #2 07-20-2006, 03:13 AM

i know that P(player a floping a set) = (2* 48C2/48C3) but then i get dizzy cuz i dont think that player B also flopping a set is independent of player A flopping a set.
i.e., if player A flops one player B only has 2 of the 3 flop cards to match with his pocket pair.

EDIT: Sorry, I edited this by mistake, but I changed it back to what you had originally. -BZ

BruceZ · #3 07-20-2006, 03:44 AM

[ QUOTE ]
I understand a similar situation was posted before, but the situation im looking for is a bit different.

How do you calculate the probability that player A and player B both flop a set, assuming it's heads up and they both have different pocket pairs.

is it as simple as just 2*2*48/48C3??

[/ QUOTE ]

2*2*44/C(48,3) for both flopping exactly a set.

BruceZ · #4 07-20-2006, 04:19 AM

[ QUOTE ]
i know that P(player a floping a set) = (2* 48C2/48C3).

[/ QUOTE ]

That should be 2*C(46,2)/C(48,3) including full houses.

For exactly a set it should be (2*44*40/2 + 2*2*44)/C(48,3), where the 2*2*44 corresponds to both players flopping a set.

For a set, full house, or quads, it would be [2*C(46,2) + 48]/C(48,3), where the 48 corresponds to the quads.

If we don't know player B's cards, and if we also include full houses, then it would be 2* C(48,2)/C(50,3).

[ QUOTE ]
but then i get dizzy cuz i dont think that player B also flopping a set is independent of player A flopping a set.
i.e., if player A flops one player B only has 2 of the 3 flop cards to match with his pocket pair.

[/ QUOTE ]

They are not independent. That's why the answer is not [P(player A flopping a set)]^2.