Probability puzzle involving crickets on a chess board I was given the following problem in a technical interview:

Suppose you have a normal 8x8 chessboard, and crickets are placed on
  every single square. The crickets begin to hop from square to square
  at random. Every cricket hops to every square on the board with equal
  probability, and there is no limit to how many crickets fit on a
  single square. 
Here is the question: We let them hop for a little bit, and then take
  a picture of the board. How many squares do we expect to find empty?

At first, I thought this would be easy. The crickets hop completely independent of one another, so it should just be some riff on the regular binomial distribution. When you try to work it out, though, you realize that the events we're logging are not independent, since whether a given cricket lands on an empty square depends on where the crickets before him have landed.
Every attempt I make to solve this ends up involving variants of the partition function. I haven't been able to get a good answer. I'm wondering if anyone sees an obvious solution, or a non-obvious one. I'm also interested in how this problem looks if you start making the chessboard get bigger, up to size infinity.
 A: Isn't this precisely a binomial distribution, because we have 64 crickets and place each randomly on one of 64 squares?  (The starting arrangement and the intermediate hops are irrelevant.)  The probability is thus $0.364987$.
A: Due to the large number of grasshoppers (trials) and low probability of a grasshopper being placed on a square (1/64), I would use poisson probability  to model the number of grasshoppers on one square
we have lamda = (1/64) and 64 'trials' so the probability of zero is given by
e^-lamda (lamda^x) / x!
therefore
e^-1 = .37..
so 37% empty

by the way, if you look at @aes comment below. for infinite chessboard with infinite grasshoppers, his formula (1 - 1/64)^64 tends to (1 - 1/n)^n = e^-1
it does then tend towards poisson, since poisson is the infinite limit of binomial 
A: The number of ways to arrange $n$ blank spaces on the board is $\binom{64}{n}$.  Given that every non-blank space has at least one cricket on, the number of distinct ways of then distributing the $n$ spare crickets among the remaining $64-n$ spaces is given by $\binom{64-n}{n}$
If $n$ is the number of empty squares and $f(n)$ enumerates the number of distinct ways of achieving $n$ then we have:
$$f(n)=\binom{64}{n}\times \binom{64-n}{n}$$
The expected value is given by $$\frac{\sum_{}f\times n}{\sum f}$$
$$=\frac{\sum\limits_{n=0}^{63} n\times\binom{64}{64-n}\times \binom{64-n}{n}}{\sum\limits_{n=0}^{63} \binom{64}{64-n}\times \binom{64-n}{n}}$$
$$=\frac{1481084144392088988865778942592}{69699678772087924716181154417}≈21.2495$$
And the generalisation you requested for a board of $k$ squares is given by:
$$\frac{\sum\limits_{n=0}^{k-1} n\times\binom{k}{k-n}\times \binom{k-n}{n}}{\sum\limits_{n=0}^{k-1} \binom{k}{k-n}\times \binom{k-n}{n}}$$
This can be generalised further to the case $m\neq k$ where $m$ is the number of crickets, but that is a little more complicated.
Furthermore, this answer makes the assumption not stated in the question that sufficient time has passed for every cricket to have jumped at least once.  If that has not happened then the answer is not calculable without more information but some estimation of the mean time before jumping would be sufficient to build a distribution of the rate at which the expected value of 21.249 would be approached and this would occur in accordance with a Poisson distribution.
