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The question is the same as the Collecting Problem. However, when I tried another approach, I have met some difficulites.

Let $A_i$ be the event that the $i$th toy we collected is a new type and let indicator variable $I_i$ to indicate whether $A_i$ happens. Let $X$ be the number of different types we collected after $t$ toys are collected, then $X=I_1+I_2+\cdots+I_t$, then the expectation of $X$ is $$E(X)=\sum_{i=0}^{t}{E(I_i)}=\sum_{i=0}^{t}{P(A_i)}$$ I find out that $P(A_i)$ should be $(\frac{n-1}{n})^{i-1}$, but how to prove it?

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Let your $i$th toy be $k$. Then, for it to be a new toy, none of the previous toys could have been $k$ for $k$ to be new. Thus, there are $n-1$ choices out of $n$ for the past $i-1$ picks.

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