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Given is a sequence $\langle a_1,a_2,\ldots,a_n\rangle$ over the alphabet $\{1,2,\ldots,m\}$ chosen uniformly at random among the $m^n$ possibilities. What is the expected size of the set $\{a_1,a_2,\ldots,a_n\}$?

If $m=n$ it seems the answer tends to $(1-1/e)n$ as $n\to\infty$, but I don't know why.

I bumped into this while benchmarking some code for hashtables, so I wouldn't be surprised if it is a standard result in the hash world.

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2 Answers 2

up vote 21 down vote accepted

Define the indicator random variable $I_i$ for $1 \leq i \leq m$ as $1$ if alphabet $i$ is present in the set ${a_1,\dots,a_n}$. Then the size of the set is simply $\sum_{i=1}^m I_i$. The expectation of this can be easily computed by linearity of expectation. The probability that $I_i$ equals $1$ is given by $1-\left( \frac{m-1}{m} \right)^n$ and therefore the expected size of the set is $m \left[ 1- \left( 1 - \frac{1}{m} \right)^n \right]$. For $m=n$, the limiting value is indeed as you mentioned in the question.

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This is dealt with in depth at

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That link is broken for me right now (Feb 2012). Do you mean ? – mcherm Feb 28 '12 at 19:01
Yes, the site must have been updated. I fixed the URL. – Jeff Hussmann May 24 '12 at 23:20

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