can someone please help me calculate the likelihood of being up against AAxx when holding KKAx?

Thanks very much,
5th

edit: sorry for the titular mistake, maybe a mod could fix it

There are 3 aces left out of 48 cards.

Pr(AAxx|KKAx) = C(3,2)*C( 45,2)/C(48,4) = 1.5%

thanks very much for the reply, normally i wouldn't ask for clarification but seeing as that we're in the probability forum, would you mind depicting what's going on in the formula?

Given a random deal, the probability of event (hand) E is equal to

Pr(E) = (Number of deals where E occurs ) / (Total number of possible deals)

I used combinations to count the number of ways.

C(n,r) = number of combinations of r events occurring out of n trials

= n! / (r! * (n-r)!) where x! = x * (x-1) * (x-2) * … *3 *2 *1 (e.g., 4! = 4*3*2*1 = 24)

For the given problem, there are 48 cards left and four are to be dealt to villain. The total number of ways this can occur is the C(48,4), which would be the denominator of the probability equation.

For villain to have exactly two aces, since there are 3 aces left, there are C(3,2) ways for villain getting two aces. That means there are C(44,2) ways for the other 2 cards, none being an ace. (I was sloppy in using C(45,2) but it doesn’t change the result to 3 decimals. In fact, if you allow 3 aces it still is 1.5%).

That's only the probability that a single random hand is AAxx. Keep in mind that the chance is higher vs. multiple opponents and as ranges narrow preflop. Vs a nitty player's 3-bet it's about 100% chance he has AA.

thanks guys, very informative