# Expected number of unique items when drawing with replacement [duplicate]

Having a set of size M, I'm drawing M items with replacement. What is expected number of unique items that got picked?

Thanks, Jarek

• Tell us if you have tried something. May 26, 2011 at 15:51

Let $Y$ be the number of unique items that get picked. Let $X_i$ be $1$ if item $i$ gets picked and $0$ if not. We have $E[X_i] = 1 P(X_i = 1) + 0 P(X_i = 0) = P(X_i = 1) = 1 - P(X_i = 0)$. Since $P(X_i = 0)$ is the probability that item $i$ is not picked, this is the probability that $i$ is not picked with any of the first, second, third, etc., choices. Thus $P(X_i = 0) = \left(\frac{M-1}{M}\right)^M$.

Therefore, \begin{align}E[Y] &= E\left[\sum_{i=1}^M X_i\right] = \sum_{i=1}^M E[X_i] = \sum_{i=1}^M P(X_i = 1) = \sum_{i=1}^M \left(1 - \left(\frac{M-1}{M}\right)^M\right) \\ &= M - \frac{(M-1)^M}{M^{M-1}}.\end{align}

In general, using indicator functions to calculated expected values can be very helpful. For more examples, see the answers to Find the expectation or Another balls and bins question or Expected number of neighbors.

• I don't understand why the expected number of unique elements in the sample is equal to the sum of the probabilities that item $i$ gets picked or sum of the expectations of item $i$ being picked. Jan 21, 2023 at 8:41

I just would like to add that if someone is interested in percent of picked items, then $$\frac{M - \frac{(M-1)^M}{M^{M-1}}}{M} = \frac{M - M\left(\frac{M-1}{M} \right)^M}{M} = 1 - \left(1 - \frac{1}{M} \right )^M$$
For large M it is approximately $$1 - \frac{1}{e} \simeq 0.632120559 \dots$$
$$\lim_{M\to\infty}\left(1 - \frac{1}{M} \right )^M = \frac{1}{e}$$