I'm confronted with a probability problem that may have already been solved.

I'm considering random sequences of m events. The number of possible events is n, and each type of event has the same fixed probability of occurrence 1/n. I would like to know the probability distribution of the number of different events in the sequence (comprised between 1 and min(n,m))


Indeed, the problem was already solved. It's a generalization of Birthday paradox. The formula for probability mass function is

The probability of getting $k$ unique values from $[0, n)$ when choosing $m$ times is given by:

$$P(V = k) = \binom{n}{k}\displaystyle\sum_{i=0}^k (-1)^i \binom{k}{i} \left(\frac{k-i}{n}\right)^m $$

where $V$ is a random variable giving the number of unique outcomes and $\binom{\cdot}{\cdot}$ is the binomial coefficient.

as quoted by George V. Williams on math.stackexchange.com who refers to http://www.math.uah.edu/stat/urn/Birthday.html#General


m <- 100
n <- 40

f <- function(x, m, n) {
  vapply(x, function(k) {
    choose(n, k) * sum( ((-1)^(0:k) * choose(k, (0:k)) * ((k - (0:k))/n)^m) )
  }, numeric(1))

out <- replicate(1e6, {
  length(unique(sample.int(n, m, replace = TRUE))) 

plot(prop.table(table(out)), xlab = "k", ylab = "P(V=k)")
lines(1:m, f(1:m, m, n), col = "red")

Simulated values vs true pmf

  • $\begingroup$ +1 Is it possible that the use of $n$ and $m$ is swapped in the OP with respect to the answer? $\endgroup$ – Antoni Parellada Oct 31 '16 at 10:54

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