BASIC: Probability of a box to contain n-keys Consider the situation in which 4 boxes exist, as well as 4 keys. Each key is stored inside of a randomly chosen box. Now suppose we chose a box at random. What is the probability of the chosen box to contain a given amount of keys?
My thoughts were as follows. The odds of any given key to end up in any given box is P = 0.25. Similarly, the odds of containing precisely all keys is P = 0.25^4 = 0.0039.
Thus, I have considered the solution being




# of keys
P




0
1-cumsum = 0.667


1
0.25^1 = 0.25


2
0.25^2 = 0.0625


3
0.25^3 = 0.0156


4
0.25^4 = 0.0039




But this appears wrong for a multitude of reasons.
I have created a simple simulation of the problem using Excel (https://docs.google.com/spreadsheets/d/1q9EZZvNzgSrq0iF82DjOZfpNKO4qwcJSJLtkPmjJq2s/edit#gid=962295911), and the actual values (for any given box) are more close to




# of keys
P




0
0.32


1
0.42


2
0.21


3
0.045


4
0.005




I am looking forward to seeing a correct theoretical analysis of the problem.
EDIT: This problem has been solved. Please see frank's answer for the proper of way of computing the probability.
 A: Let's say, without loss of generality, that we have chosen box 1.
What is the probability for box 1 to contain zero keys?
It is the probability to distribute all keys over only the boxes 2, 3, and 4: Each key has the probability of $\frac{3}{4}$ to end up in those last three boxes, and we have 4 keys, and their positioning is independent, so we get $(\frac{3}{4})^4$.
What is the probability for box 1 to contain one key?
There are four possibilities for box 1 (for each of the four keys) and each time the probability of the other three to be distributed over the last boxes is $(\frac{3}{4})^3$, i.e. we get: $4\cdot\frac{1}{4}\cdot(\frac{3}{4})^3 = (\frac{3}{4})^3$.
What is the probability for box 1 to contain two keys?
The probability that two keys end up in box 1 is $(\frac{1}{4})^2$. Then the probability of the other two to be in the last boxes is $(\frac{3}{4})^2$. And there are $\binom{4}{2}$ choices of two keys, so the total probability for this situation is $\binom{4}{2}(\frac{1}{4})^2(\frac{3}{4})^2$.
What is the probability for box 1 to contain three keys?
Similar arguments as the ones above lead to: $\binom{4}{3}(\frac{1}{4})^3\frac{3}{4}$.
What is the probability for box 1 to contain four keys?
Well, that is just $(\frac{1}{4})^4$.
I am afraid, your simulations are wrong. But this type of problem is notorious for being tricky in getting right. It is so easy to make a mistake. I hope my solution is correct. You also might want to check whether those probabilities add up to one...
