How many boxes remain empty when n balls are distributed in 4 containers where each container has 13 unlabeled slots?

Question

If each container has n unlabeled slots, the problem is same as the one "How many ways can b balls be distributed in c containers with no more than n balls in any given container?" The answer is known as

N(b,c,n)=∑i(−1)i(ci)(b+c−1−i(n+1)c−1).

I want to know what the expected number of empty boxes is for n balls. Obviously if there is one ball, 3 boxes will be empty. And if there are 40 or more balls, no boxes will be empty because the extreme case is that 3 boxes have 13 balls each and 1 ball left over is in the 4th box.

What is the expected number of empty boxes for 1 < n < 40?

I'm not sure if my wording is correct for the question I'm trying to ask. But my initial question that I'm trying to solve is:

r cards are selected from a standard deck of 52 playing cards to create a new deck. Select one card from the new deck. What is the probability of the selected card being the highest card of its suit in the new deck?

Which boiled down to the expected number of suits in the r card selection, over r. Hence the ball and box question. So I guess I'm really asking, what is the expected number of suits in a selection of r cards from a standard deck?

I've tried thinking about how many ways there are to order a deck but a whole suit or 2 or 3, doesn't appear in the first r cards of the deck. As this would be similar.

Answer 1

Your title question is not obviously the same as your initial question, but let's find the expected number of suits see in the selection of $r$ cards without replacement.

With $0 \le r \le 39$ cards draw from a standard deck:

• the probability there are no spades is $\dfrac{39 \choose r}{52 \choose r}=\dfrac{{\frac{39!}{(39-r)!}}}{{\frac{52!}{(52-r)!}}}= \frac{39!(52-r)!}{(39-r)!52!}$ and similarly for the other three suits
• the probability there are some spades is $1-\dfrac{39 \choose r}{52 \choose r}$ and similarly for the other three suits
• the expected number of suits seen is then $4\left(1-\dfrac{39 \choose r}{52 \choose r}\right)$ by linearity of expectation.

If $39 \lt r \le 52$ then these are instead $0$, $1$ and $4$ respectively.