There are $32$ balls in total, equally split into red, green, blue, and white colors ($8$ balls each). I also have $4$ bins. What is the probability that in at least one of the bins, all the balls are of the same color?
My process: I am trying to use complementary probability, which in this case is $$1 - P(\text{no bin has all balls of same color})$$. I started this off by doing for the first bin, where the number of options is $$\frac{32 * 24 * 31 * 30 * 29 * 28 * 27 * 26)}{8!}$$, but I feel like I am going in the wrong direction. I have thought about PIE as well but am not sure how to do it as it seems quite messy. Any help would be appreciated.
Let's use PIE.
To begin, the total number of ways to assign balls to bins is $\binom{32}{8} \binom{24}{8} \binom{16}{8}$.
Case 1: At least one bin has all balls the same color. There are $4$ ways to choose a bin, and $4$ ways to choose a color. Following this, there are $\binom{24}{8} \binom{16}{8}$ ways to fill the remaining bins.
Case 2: At least two bins have all balls the same color. There are $\binom{4}{2} = 6$ ways to choose the bins, $4$ ways to choose the color for the first bin, and $3$ ways to choose the color for the second bin. Following this, there are $\binom{16}{8}$ ways to fill the remaining bins.
Case 3: All bins have all balls the same color. For this case, our only choice is in assigning colors to bins. There are $4$ choices for the first bin, $3$ choices for the second, and $2$ choices for the third.
Now let's account for overcounting. Each arrangement in Case 2 was counted twice in Case 1. Therefore, the number of arrangements in Case 2 needs to be subtracted (once) from the total. Each arrangement in Case 3 was counted four times in Case 1, but was discounted $\binom{4}{2} = 6$ times in Case 2. Therefore, the number of arrangements in Case 3 needs to be added three times to the total.
Bringing everything together, we have
$$P = \frac{(4) (4) \binom{24}{8} \binom{16}{8} - (6) (4) (3) \binom{16}{8} + (3) (4) (3) (2)}{\binom{32}{8} \binom{24}{8} \binom{16}{8}}$$
$$= \frac{123731423}{81340761805875} \approx 1.5211 \times 10^{-6}$$
As expected, not very likely at all.