I've been trying to calculate the probability of drawing exactly two aces and two kings, or exactly one ace and three kings when the player already has an ace in hand. The player draws 24 more cards, and the deck is full aside from the ace in hand.
I've calculated the probability of getting exactly two aces and two kings like so:
$\dfrac{{3\choose2}.{4\choose2}.{44\choose20}}{{51\choose24}} \approx 13.81\%$
Which seems a little high to me. However, moving on with the same equation for drawing exactly one ace and three kings:
$\dfrac{{3\choose1}.{4\choose3}.{44\choose20}}{{51\choose24}} \approx 9.20\%$
And so, the probability of getting one or the other is $13.81\% + 9.20\% = 23.01\%$.
Can someone tell me where I'm going wrong? Because I have trouble believing there's a $23.01\%$ chance of the described scenario arising.
Best Answer
For exactly two more aces and exactly two kings in 24 draws from a deck of 51 cards (missing an ace), I also get
$$ \frac{{3 \choose 2}{4 \choose 2}{44 \choose 20}}{{51 \choose 24}} = 0.138,$$
(to three places), computed in R as:
Here is a simulation of a million such draws with probabilities correct to 2 or 3 places.
The approximate probability of $P(A = 2, K = 2)$ is found in cell $(2,2)$ of the table. The approximate probability $P(A = 1, K = 3)$ is in cell $(1, 3).$
Related probabilities can also be approximated from the table. For example, the total probability $P(A = 2) \approx 0.358$ is found separately in the printout above and as the total of row 2 of the table.
Its exact probability (to three places) is
$$ \frac{{3 \choose 2}{48 \choose 23}}{{51 \choose 24}} = 0.358,$$
However, to get the probability of either 'two aces and two kings' OR 'one ace and three kings', you should add only two entries in the table $(2,2)$ and $(1,3).$
Addendum: 'Expanded' R code, demonstrating method of counting aces (cards 2 through 4) in 24 draws:
The count of the number of aces in a 'draw' does not depend on which aces or their order. This is done a million times.