Kids and Prizes

Title:
SGU 495

Have $n$ prizes,$m$ individuals, each has$\frac{1}{n}$The probability of selecting a prize, if the prize has not been selected before, the prize will be won. Ask the expectation of the number of people who won the prize (in fact, it is the probability).

Idea 1:
Let $f\left[i\right]$ display numberiiThe probability that$i$ individual gets a gift, then
$$f[i]=(1-f[i-1])f[i-1]+f[i-1](f[i-1]-\frac{1}{n})$$
$(1-dp[i-1])\ast dp[i-1]$ means$i-1$ person did not get the prize,iiThe probability that individual$i\; will$ receive a prize (equal to$i-$The probability of$1$ person winning a prize),$dp[i-1]\ast (dp[i-1]-1/n)$ means the$i-1$ person got the prize and theiiThe probability that individual$i\; will$ receive a prize, because the$i-1$ person got the prize, so theiiThe probability of individual$i$ getting a prize is$dp[i-1]-1/n$ , that is, the probability of winning the prize is reduced by$1/n$。

Idea 2:
The probability that each prize will not be selected $=(\frac{n-1}{n}{)}^m$
Expectation of each prize being selected$=1-(\frac{n-1}{n}{)}^{m\; the}$
answer isnnExpected sum of$n$ prizes$=n(1-(\frac{n-1}{n}{)}^m)$