Suppose raindrops are falling on your head, randomly and independently, at an average rate of one per minute. What's the average of the 𝑐𝑢𝑏𝑒 of the number of raindrops that fall on your head in one minute?
The probability that (k) raindrops fall on your head in a minute is given by the Poisson distribution of mean 1, so it's
[ \frac{1}{ek!} ]
I could explain this but let's move on. The puzzle asks us to compute the expected value of (k^3) for this probability distribution, which is
[ \sum_{k=0}^\infty \frac{k^3}{ek!} ]
The heart of the puzzle is to figure out this sum. It turns out that
[ \sum_{k = 0}^\infty \frac{k^n}{k!} = B_n e ]
where (B_n) is the (n)th 'Bell number': the number of partitions of an (n)-element set into nonempty subsets. This is called 'Dobiński's formula'. I'll prove it in my next post. Now let's just use it!
We're interested in the case (n = 3). There are 5 partitions of a 3-element set
[ {{1,2,3}}, ]
[ {{1,2}, {3}}, ; {{2,3}, {1}}, ; {{3,1}, {2}}, ]
[ {{1}, {2}, {3}} ]
so (B_3 = 5).
So, the average of the cube of the number of raindrops that fall on your head in one minute is 𝟓.
Wild, huh? From probability theory to combinatorics.
