Probability of 3 Aces, Last Picked is an Ace

Question

"Aaron picks an integer $k \in [1,52]$. Then, he draws the first $k$ cards from a standard, shuffled 52-card deck. Aaron wins a prize if the last card he draws is an ace and if there exists exactly one ace in the remaining cards. What $k$ should Aaron pick?"

1. I am struggling to understand how to get to the probability that Aaron wins with this approach:

We need to have 2 aces in the first $k-1$ cards, ${k-1 \choose 2}$, then the $k^{th}$ card needs to be an ace. Then we need to count the ways of positioning the last remaining ace, which is ${52-k \choose 1}$. Finally, there are $4!$ ways of ordering the 4 aces.

This aims to count all combinations of the 52 cards in the deck that lead to a win. I would then simply divide this by $52!$ the number of possible combinations.

But the solution I found takes a different calculation, $\frac{{k-1 \choose 2}{52-k \choose 1}}{{52 \choose 4}}$. Could you please help me understand this?

Then I understand the optimisation step to get to the optimal $k$, no issue there.

2. Is there a way to approach this problem ignoring cards values, and thinking of it in terms of 52 balls, 48 black and 4 white?

Thank you!

Answer 1

The second question is actually the approach of the solution that you found. Since you only care if a card is an ace or not, you interprete any cards, which are not an ace, as black balls, which cannot be distinguished from each other, and all aces as white balls, which also cannot be distinguished from each other. Then $\binom{52}{4}$ is the number of possibilities to order the 52 balls. The numerator gives you the number of possibilities that the first $k-1$ balls contain exactly two white balls and the $k$-th ball is white: You have $\binom{k-1}{2}$ possibilities where you can put two white balls in the first $k-1$ slots. The $k$-th ball must be white, so there is only one possibility. Finally there are $52-k=\binom{52-k}{1}$ possibilities where you can put the last white ball. Since every possible ordering has the same probability, we can divide the number of possibilities, where we succeed by the total number of possible orderings. This leads to a probability of $\frac{\binom{k-1}{2}\binom{52-k}{1}}{\binom{52}{4}}$. In your approach you made the mistakte to divide by the number of possible orderings of the cards, which can be distinguished from each other, while in the numerator you have the number of possible orderings of the balls, where black balls and white balls cannot be distinguished from each other.

Answer 2

You are right in thinking of it in terms of $52$ balls, $48$ black and $4$ white. In fact, if you don't differentiate between aces and non-aces, you can't solve the problem.

And you should be able to now understand that the book formula of $$\frac{{k-1 \choose 2}{52-k \choose 1}}{{52 \choose 4}}$$ does exactly that, although for even more clarity, I'd prefer to write it as $$\frac{\binom{k-1}2\binom11\binom{52-k} 1}{{52 \choose 4}}$$

Note that in order to compute probability correctly, there has to be consistency between numerator and denominator ( either both combinations, or both permutations )

Also note that using combinations here is by far the easiest way. If you want to use permutations, the formula becomes a fearsome

$$\frac{(k-1)!}{2!(k-3)!}\cdot\frac{(52-k)!}{(51-k)!1!}\bigg/\frac{52!}{4!48!}$$

where both the numerator and denominator are, so to say, "double stacked" !