Coupon Collector Variance

Problem

Variance of the number of draws needed to collect all n coupon types:

Var = sum from j = 0 to n - 1 of j * n / (n - j)^2.

This follows from writing the total time as a sum of independent geometric waiting times. Return 0.0 for n <= 0.

Hint

With j coupons collected, the waiting time is geometric with success probability (n - j) / n.

case 1

in: [1]

expected: 0

case 2

in: [2]

expected: 2

case 3

in: [3]

expected: 6.75

case 4

in: [5]

expected: 25.17361111111111

Starter codePython

def coupon_collector_variance(n):
    # Your code here
    pass

4 test cases · run your code against them in the editor