Card game probability 3 players

Question

I don’t know how to handle this 3 players case, may anyone give a hand? Thanks!

When playing card game, there are 3 players playing 32 cards, each receive 10 cards, 2 remind on the table. Overall are 4 aces and 4 kings in the game. How likely is P (E | D)? D = "The table remains 2 aces" E = "I (at least) get one king".

I guess I should have steps as follows.

1. P(E) = P(1King)+P(2K)+P(3K)+P(4K)

2. P(D) = (4/32) * (4/32) = 1/256

3. P (E | D) = P(E) + P(D)

Equations may be used:

Counting combinations

    The number of combinations of n objects taken r at a time is
    nCr = n(n - 1)(n - 2) . . . (n - r + 1)/r! = n! / r!(n - r)!

               Number of favorable outcomes
P(I) =     Total number of possible outcome

Answer 1

Guide:

The condition $D$ implies that a person gets $10$ cards from a stack that contains $4$ kings in total.

Find the probability $p$ that in that situation the person does not get any kings at all.

Then the answer is $1-p$.