#### Shoestringx

##### Guest

- Joined
- Nov 27, 2006

- Total posts
- 222

I'll start with somthing very simple, the probability of drawing a specific pocket pair (for this example I'll show with A's).

p(A,A) = p(A1) * p(A2) (where p(A1) and p(A2) are probabilities of getting an ace on the first and second card respectively.

P(A1) is very simple, there are 52 cards in the deck and 4 aces therefore 4/52.

P(A2) is a also very simple, there are 51 remaining cards in the deck and 3 aces, therefore 3/51

p(A,A) = (4/52)*(3/51) = 0.45%

That calculation was easy as there is only one possible outcome of cards, and they both need to be Aces.

However lets say you wanted to find the probability of finding any A,K then you need to do:

p(A,K) = p(A1)*p(K2) + p(K1)*p(A2)

= (4/52)*(3/51) + (4/52)*(3/51)

= 0.90%

Now lets say you are dealt those two A,A to start, what are the odds of flopping a set (or maybe better, but no quads)?

p(Aset) = p(A3)*p(X1)*p(X2) + p(X1)*p(A3)*p(X2) + p(X1)*p(X2)*p(A3)

=[(2/50)*(48/49)*(47/48)] + [(48/50)*(2/49)*(47/48)] +

[(48/50)*(47/49)*(2/48)]

= 11.51%

I'll take some time later tonight and work out a couple more complicated examples if there is any interest in that (I think Zebranky mentioned an example, I'll probably try that one later).