3
$\begingroup$

An ice-cream truck stops at a park and randomly distributes 10 ice-cream cones among 20 children. What is the probability that a randomly selected child receiving no ice-cream cone? Exactly one ice-cream cone? Two or more ice-cream cones?

My current approach is to find all the integer compositions of 10 ice-cream cones using 20 children, where (obviously) zeros are allowed. So, for example, one of the integer compositions might be:

1+1+1+1+2+4+0+0+0+0+0+0+0+0+0+0+0+0+0+0=20

In the above integer composition, 14 out of 20 children received no ice-cream, 4/20 children received one ice-cream cone, and 2/20 children received 2 or more ice-cream cones. I can perform a Monte Carlo simulation with many random integer compositions to get probabilities.

p(x=0) = 0.66 p(x=1) = 0.24 p(x=2) = 0.08 p(x=3) = 0.01 p(x=4) through p(x=10) = less than 0.01 --- p(x>=0 & x <= 10) = 1

It would seem that this pattern is a discrete probability distribution. Can someone please point out the name of the distribution? For some reason I am having trouble seeing how one of the common distributions (Poisson, binomial, etc.) fits this problem!

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I might call it the coupon-collector distribution. There are previous related questions here but briefly, the probability of $x$ children getting at least one ice cream is $S_2(10,x) \, 20! / ((20-x)!\, 20^{10})$ where $S_2(,)$ is a Stirling number of the second kind, the expected number is $20(1-(1-1/20)^{10})$ and the variance is $20(1-1/20)^{10} + 20^2(1-1/20)*(1-2/20)^{10} - 20^2 (1-1/20)^{2\times10}$ $\endgroup$ Commented Jan 7 at 22:49
  • $\begingroup$ The probability of a particular child getting $k$ ice creams is simpler: that distribution has a binomial distribution with parameters $10$ and $\frac1{20}$. But what happens to a particular child is not independent of what happens to the others. $\endgroup$ Commented Jan 7 at 22:54
  • $\begingroup$ @Henry: The coupon-collector distribution is a slightly different distribution which forms the solution to an inverse problem relating to this situation (see e.g., O'Neill 2022). $\endgroup$ Commented Jan 8 at 2:52

1 Answer 1

Reset to default
4
$\begingroup$

Taking your generalised problem, suppose there are $n$ ice-creams and $m$ children and let $X_1,...,X_m$ denote the number of ice-creams distributed to each child. For a randomly selected child the number of ice-creams follows a binomial distribution:

$$X_i \sim \text{Bin}(n, \tfrac{1}{m}).$$

If you are instead interested in the number of children who receive a stipulated number of ice-creams then this is a variant of the occupancy problem. The number of children who receive at least one ice-cream follows the classical occupancy distribution (see e.g., O'Neill 2019). Taking $K_n \equiv \sum_{i=1}^m \mathbb{I}(X_i > 0)$ to be the number of children who get at least one ice-cream (called the "occupancy number") we have:

$$K_n \sim \text{Occ}(n,m).$$

This result allows you to find the distribution of the number of children who get no ice-cream (which is $n-K_n$). For higher numbers of ice-creams, you can use a variant of the occupancy distribution which involves some recursive application of the classical occupancy distribution. For now, I will let you have a read of the linked paper and leave the latter result as an exercise. (You might also be interested in some other distributions relating to occupancy problems in this type of situation --- see e.g., O'Neill 2021, O'Neill 2022, O'Neill 2023).

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ +1 Weirdly it didn't occur to me that the number of occupied cells in the occupancy problem would just be called the (classical) occupancy distribution, but it makes perfect sense. Johnson and Kotz is plenty authoritative / standard enough for me on that terminology. $\endgroup$ Commented Jan 8 at 2:38
  • $\begingroup$ Interestingly enough, it is also called the "Arfwedson distribution" after a Swedish mathematician that examined it in the 1950s. The "classical occupancy distribution" is a much better name! $\endgroup$ Commented Jan 8 at 2:49
  • $\begingroup$ Yeah, I agree (I took a look at your 2019 article as soon as I saw your answer). $\endgroup$ Commented Jan 8 at 2:52
  • $\begingroup$ Thank you, the classical occupancy distribution and spillage distribution with $\theta=1$ (no falling through bins) are what I was looking for. Unfortunately, in my case where there are more bins than balls, it looks like there is no shortcut for recursively computing the Stirling numbers. $\endgroup$ Commented Jan 8 at 15:43

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.