Probability of having two 4 of a kinds....

[BDP]

what is the % of 2 people having 4 of a kind in Hold 'Em?

Thanks

[BruceZ]

[ QUOTE ]
what is the % of 2 people having 4 of a kind in Hold 'Em?

Thanks

[/ QUOTE ]

The most likely way for that to happen is to have quads on the board, but you probably don't want to count that. It also depends on the number of players. For 10 players, and NOT counting quads on the board, the probability is exactly:

C(10,2)*13*6/C(52,2)*12*6/C(50,2)*44/C(48,5)

=~ 0.0003997899% or 1 in 250,131.

And here's a second way, just as a check. We should get exactly the same thing, and we do:

C(10,2)*C(13,2)*6*6*44/C(52,5)*2/C(47,2)*1/C(45,2)

=~ 0.0003997899% or 1 in 250,131.

The probability of quads on the board is 48/C(52,5) = 0.0018468926% = 1 in 54,145.

Adding that to the first case of 2 different quads gives 0.0022466825% or 1 in 44,510.

[BruceZ]

[ QUOTE ]
The probability of quads on the board is 48/C(52,5) = 0.0018468926% = 1 in 54,145.


[/ QUOTE ]

That should be 13*48/C(52,5) = 0.0240096038% or 1 in 4165.

Adding that to the first case of 2 different quads gives 0.0244093937% or 1 in 4097.

[aloiz]

Also if you want to include the time the board boats and there are two people with quads then you'd have: C(10,2) * 13*6/C(52,2) * (12 * 4 * 11 * 4)/2/C(50,2) * 2/C(48,5) = .00000267 or 1 in 375197, and adding that to the other cases gives us .00247% or 1 in 4052.

aloiz

[BruceZ]

[ QUOTE ]
Also if you want to include the time the board boats and there are two people with quads then you'd have: C(10,2) * 13*6/C(52,2) * (12 * 4 * 11 * 4)/2/C(50,2) * 2/C(48,5) = .00000267 or 1 in 375197, and adding that to the other cases gives us .00247% or 1 in 4052.

aloiz

[/ QUOTE ]

I agree that this is another case, but you need to multiply your probability by 2. The reason is that C(52,2)*C(50,2) counts each pair of hands twice, for 2 different orders (permutations of hands rather than combinations) but your numerator only counts each pair of hands once (combinations of hands rather than permutations). This is a subtle point which must be thoroughly understood because it comes up all the time. Notice in my original calculation for the main case I have 13*6/C(52,2)*12*6/C(50,2). In that case, I counted all permutations in both the numerator and denominator, so there is no factor of 2 needed. I hope you can see the difference.

C(10,2) * 13*6/C(52,2) * (12 * 4 * 11 * 4)/2/C(50,2)*2 * 2/C(48,5)

=~ 0.000533% or 1 in 187,599.

Here is another way, considering the board before the players hole cards.

C(10,2) * C(13,2)*2*4*6/C(52,5) * 2/C(47,2)*44/C(45,2)*2

=~0.000533% or 1 in 187,599 same as before.

Note the multiply by 2 at the end for the same reason.

At least half of these cases should definitely be counted as "legitimate", because in half the cases the kicker plays when it is higher than the pair on the board, and then both players are using both of their hole cards.

Adding up all 3 cases gives 0.0249424469% or 1 in 4009. Adding up the cases where both players must use at least 1 hole card gives 0.0009328431% or 1 in 107,199. Adding up the cases where both players use both hole cards gives 0.000666316% or 1 in 150,079. The case where both players have pocket pairs gives 0.0003997899% or 1 in 250,131.

[aloiz]

Thanks, that makes sense. If I had used C(12,2)*2*4*4 as the second numerator in my equation then I'd be counting up all possible permutations.
BTW I'm assumming that last 2 in your second calculation:
C(10,2) * C(13,2)*2*4*6/C(52,5) * 2/C(47,2)*44/C(45,2)* 2
is a typo.

aloiz

[BruceZ]

[ QUOTE ]
Thanks, that makes sense. If I had used C(12,2)*2*4*4 as the second numerator in my equation then I'd be counting up all possible permutations.

[/ QUOTE ]

No, the problem isn't that you want permutations of 2 CARDS to make the second hand; the problem is you need to compute the permutations of 2 HANDS. The way you have it assumes the first player is the pair. When you multiply the number of pairs by the number of non-pairs, you get combinations of 2 hands, which counts each combination just once instead of twice for the 2 possible orderings. If you do this, you need to divide the C(52,2)*C(50,2) by 2 in the denominator to get combinations of hands to be consistent. This is the same as multiplying the numerator by 2 to get permutations of hands. C(52,2)*C(50,2) is permutations of hands, just as 52*51 would be permutations of cards, and you have to divide by 2 to get combinations.

[ QUOTE ]
BTW I'm assumming that last 2 in your second calculation:
C(10,2) * C(13,2)*2*4*6/C(52,5) * 2/C(47,2)*44/C(45,2)* 2
is a typo.

aloiz

[/ QUOTE ]

The last 2 is correct, and I explained that one. The one at the beginning of the last term is a typo. It should be 1/C(47,2).... The number is correct.

[aloiz]

hmm...Ok if we dumb down the example a bit and we wanted to know the odds that the board will be a full house, but that it is either a board of xxxyy or yyxxx. Doing 13 * 4/C(52,3) * 12 * 6/C(49,2) will only give us half of the probability because we count the number of combinations of trips followed by pairs, but not vise-versa. Multiply by two accounts for all permutations of trips and pairs, and thus gives us the correct probability. This sound ok?

aloiz