Plus Magazine

A collector's piece: solution

In last issue's outer space we worked out how many pruchases you would expect to have to make in order to obtain every one of a set of N distinct cards. The puzzle was to show that the standard deviation of this result is close to 1.3N for large N. To work out the expectation, we considered N random variables X₀, X₁, etc, up to X_N-1. Each random variable X_i counts how many purchases you have to make to get a new card if you already have i distinct cards. The overall expectation is then just the sum of the individual expectations of the N variables.

You can use the same technique to work out the standard deviation (which is of course the square root of the variance): since the N random variables are independent of each other, the overall variance is simply the sum of the individual values.

For each individual variable X_i, the probability that you have to make j purchases to get a new card is $$\frac{N-i}{N} \times \left(1 - \frac{N-i}{N}\right)^{j-1}.$$ So each individual X_i has the well-known geometric probability distribution with parameter $$p = \frac{N-i}{N}.$$

This distribution has variance $$\frac{1-p}{p^2} = \frac{iN}{(N-i)^2}.$$ Now if X is the sum of the N random variables X₀, X₁, ... , X_N-1, then the variance of X is $$variance(X) = \sum_{i=0}^{N-1} \frac{iN}{(N-i)^2} = N \sum_{i=0}^{N-1} \frac{i}{(N-i)^2}.$$ A little thought shows that this sum can be expressed as $$N \sum_{i=1}^{N} \frac{N-i}{i^2} = N\sum_{i=1}^N \frac{N}{i^2} - N\sum_{i=1}^N \frac{i}{i^2},$$ which is equal to $$N^2 \sum_{i=1}^N \frac{1}{i^2} - N \sum_{i=1}^N \frac{1}{i}.$$

Dividing through by N² gives $$ \frac{variance(X)}{N^2} = \sum_{i=1}^N \frac{1}{i^2} - 1/N \sum_{i=1}^N \frac{1}{i}.$$ As N gets large, the first term tends to $\pi^2/6$, as was stated in the hint, while the second term tends to zero. This proves that the standard deviation for large N is close to $\sqrt{\pi^2 N^2/6},$ which is approximately 1.3 N — QED.