Difference between two sample space setups

Question

An ordinary deck of cards is dealt randomly to four players so that each player receives 13 cards. Find the probability that each player is dealt exactly one ace.

I was wondering if I could compare this scenario with the following: Given 4 indistinguishable balls and 4 distinguishable boxes, if we distribute the balls randomly, what is the probability that each box has exactly one ball? The answer to that is $\frac{1}{4+4-1\choose 4}=\frac{1}{7\choose 4}=\frac{1}{35}$. So is that also the answer to the question above? If not then why?

Answer 1

The situations are not the same. It is simplest to consider two cases:

• Case 1: $16$ cards, including four aces, dealt four cards to each player
• Case 2: just four aces assigned (in possible multiples) to four players.

In Case 1: The chance a give player gets all four aces is $\frac{4}{16} \frac{3}{15} \frac{2}{14} \frac{1}{13}$

In Case 2: $\frac{1}{4} \frac{1}{4} \frac{1}{4} \frac{1}{4}$

(The case of $16$ cards is simpler to analyze than $52$, but the principle remains the same.)

The two cases are not the same.

Answer 2

There are $$\binom{52}{13,13,13,13}=\frac{52!}{(13!)^4}$$ ways to distribute the cards.

We can give each player one Ace in $4!=24$ ways, and then give them each $12$ more cards in $$\binom{48}{12,12,12,12}=\frac{48!}{(12!)^4}$$

Divding gives a probability of $$\frac{24\cdot13^4}{52\cdot51\cdot50\cdot49}\approx.105498$$