I deal a deck into $2$ equal piles of $26$, and what is the probability all $4$ ace cards are in one pile?

Question

I worked it out as follows:

I lay out the $52$ cards in a row. The first $26$ are in the first pile and the next $26$ in the second pile. I want all Aces to be in the first pile. So I choose $4$ out of $26$ places for the aces to go $\binom{26}{4}$. There are $4!$ ways to arrange the $4$ aces. There are $48!$ ways to arrange the rest of the $48$ cards.

This gives the total arrangements where all $4$ aces are in the first $26$ cards to be: $\binom{26}{4} \times 4! \times 2 \times 48!$. However, the aces could also be in the second pile so to get the total arrangements where all $4$ aces are in one pile, I multiply this by $2$.

Further, there are $52!$ arrangements of all the cards.

So, the answer is:

$$P \ = \ \frac{\binom{26}{4} \times 4! \times 2 \times 48!}{52!}$$ which is approximately $0.11$.

I'm not sure if this is correct though. Can you please comment on if this seems right. Thanks!

Answer 1

Your logic is fine. A much simpler approach is to let the ace of spades be in one pile. The chance the ace of hearts is in the same pile is $\frac {25}{51}$. Given that they are both in the same pile, the chance the ace of diamonds is in the same pile is $\frac {24}{50}$ and the ace of clubs $\frac {23}{49}$. The total probability is $$\frac {25 \cdot 24 \cdot 23}{51 \cdot 50 \cdot 49}=\frac {92}{833}\approx 0.11$$ You should be able to cancel things algebraically to make your answer match this.

Answer 2

Here a solution that seems to me a bit more "intuitive" and uses directly binomial coefficients as ordering of the cards does not play a role.

You have $4$ aces and $48$ other cards. So, for one pile you have either

• $\binom{4}{4}$ ways for the aces and $\binom{48}{22}$ for the other cards or
• $\binom{4}{0}$ ways for the aces and $\binom{48}{26}$ for the other cards

All in all there are $\binom{52}{26}$ ways of choosing $26$ cards for one pile.

$$\frac{\binom{4}{4}\binom{48}{22}+\binom{4}{0}\binom{48}{26}}{\binom{52}{26}}=\frac{92}{833}$$