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March 2006

## 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 XN-1. Each random variable Xi 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 Xi, the probability that you have to make j purchases to get a new card is

So each individual Xi has the well-known geometric probability distribution with parameter

This distribution has variance

Now if X is the sum of the N random variables X0, X1, ... , XN-1, then the variance of X is A little thought shows that this sum can be expressed as which is equal to

Dividing through by N2 gives

As N gets large, the first term tends to , 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 which is approximately 1.3 N — QED.

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