Birthday problem with limited bucket size

Question

Let's say there are 1095 balls, 23 of them are black. Randomly assign each ball into 365 buckets, each has a maximum capacity of 3 balls. What is the probability of at least one bucket having at least 2 black balls?

Initially I thought the order does not matter, so we can imagine assigning the black balls to the buckets first. This will give probability of 0.5, just like the birthday problem. Then I think this is incorrect, as the bucket size is limited, the order of assignment actually matters. We cannot assign black balls first and the other balls later.

Update:
Could the probability simply be:

Probability of at least one unlimited bucket with at least 2 black balls - probability of at least one unlimited bucket with at least 4 black balls?

(By unlimited bucket I mean bucket with no capacity limitations.)

Answer 1

Let $p(n,b)$ be the probability that with $3n$ balls of which $b$ are black and with $n$ buckets, you have no buckets with $2$ or $3$ black balls after putting $3$ balls in each bucket drawn at random without replacement from the $3n$:

Then you have $p(n,0)=p(n,1)=1$ and $p(n,b)=0$ when $b \gt n$. Otherwise I think you have $$p(n,b) = \frac{(3n-b)(3n-b-1)(3n-b-2) }{3n(3n-1)(3n-2)} p(n-1,b) +3 \frac{b(3n-b)(3n-b-1) }{3n(3n-1)(3n-2)} p(n-1,b-1)$$

so for example you get $p(2,2)=\frac35$ and $p(3,2)=\frac34$ and $p(3,3)=\frac{9}{28}$

I think you get $p(365,23) \approx 0.62176$ and so the probability you get at least one bucket with at least two black balls is about $0.37824$

Since it may be of interest, I also think $p(365,27) \approx 0.51549$ and $p(365,28) \approx 0.48943$

Answer 2

You can just draw groups of $3$ and ask the chance that at least one group of $3$ has at least two black balls. The chance the first group has at least two black balls is $3\cdot \frac {23}{1095}\cdot \frac {22}{1094}\cdot \frac {1072}{1093}+\frac {23}{1095}\cdot \frac {22}{1094}\cdot \frac {21}{1093}\approx 0.00124$. The chance any given draw has at least two black balls is the same by symmetry. By the linearity of expectation, the expected number of buckets with at least two black balls is $365$ times this, or about $0.4524$. The chance of having at least one bucket with at least two black balls is a little less than this because of the chance you have more than one bucket with at least two black balls.