Texas Hold'em Poker Probability - Paired Flop cont.

With your two hole cards, the other players might have...

Possible starting hands: 1081=47*46/2=C(47,2)

Assuming the flop is 99K,

a) if you flop a three of a kind with 9x

Four of Kind: 0

Three of a Kind or Full House: 1*46+3=49  type 1: 9y, type 2: KK

Two Pair: 3*43+10*6+3=192  type 1: Ky, type 2: yy, type 3: xx

Others: 43*42/2-(10*6+3)=840  any combination without 9 and K without one x card(43*42/2), minus yy and xx pairs.

There is about 22.3% (1 in 4.5) having a two pair or better.

b) if you have a paired starting hand with QQ

Four of Kind: 1

Three of a Kind or Full House: 2*45+3=93  type 1: 9y, type 2: KK

Two Pair: 3*42+10*6+1=187  type 1: Ky (y is any card <> K9), type 2: yy (y<>Q), type 3: QQ

Others: 42*41/2-(10*6+1)=800  any combination without 9s and Ks and two Q cards(42*41/2), minus pairs.

There is about 26% (1 in 4) having a two pair or better.  It's slight better than the original 25% that doesn't consider your hole cards.

c) if you flop a two pair with Kx

Four of Kind: 1

Three of a Kind or Full House: 2*45+1=91  type 1: 9y, type 2: KK

Two Pair: 2*43+10*6+3=149  type 1: Ky, type 2: yy ( y not in (9,K,x)), type 3: xx

Others: 43*42/2-(10*6+3)=840

There is about 22.3% (1 in 4.5) having a two pair or better.

d) if your starting hand is XY, X<>Y and not in (9,K)

Four of Kind: 1

Three of a Kind or Full House: 2*45+3=93  type 1: 9y, type 2: KK

Two Pair: 3*42+9*6+2*3=186  type 1: Kz (z is any card not in (9,K)), type 2: zz (z not in (X,Y)), type 3: xx ( x is X or Y)

Others: 42*41/2-(9*6+2*3)=801

There is about 26% (1 in 4.5) having a two pair or better.