Probability to have at least two kings when drawing 13 cards from a standard deck. What is wrong with reasoning?

Question

I know there are $\binom{52}{13}$ ways to pick 13 cards from a standard deck.

There are $\binom{4}{2}$ was to pick the two kings. Then I reasoned there are 50 cards remaining (52 cards minus the two kings) to pick 11 cards. This results in $\frac{\binom{4}{2}\binom{50}{11}}{\binom{52}{13}}$. However, this answer is wrong according to my notes. But what is wrong with my reasoning?

The correct answer should be $\sum_{i=2}^{4} \frac{\binom{4}{i} \binom{48}{13-i}}{\binom{52}{13}}$. I understand the reasoning behind this solution, i.e. the probability to have at least 2 kings is equal to the sum of the probability of having 2, 3 or 4 kings.

Thanks in advance.

Edit: I also simulated this problem and the simulation agreed with the answer in the notes.

Answer 1

The product $\binom 42 \binom{50}{11}$ counts the outcomes in two stages: first, you pick the two kings, and then, you pick the other cards. This would be a correct way to count if every outcome was obtained exactly once when we do this.

But an outcome with three or more kings is obtained multiple times. If you draw the king of hearts, king of diamonds, and king of spades (and ten more cards), then we can obtain that by:

• picking the king of hearts and king of diamonds in the first stage, then the other cards in the second stage;
• picking the king of hearts and king of spades in the first stage, then the other cards in the second stage;
• picking the king of diamonds and king of spades in the first stage, then the other cards in the second stage.

So it is triple-counted. An outcome with all four kings is counted six times.