In a lottery, there are $K$ distinct prizes. The probability of winning any one of the prizes is the same for every ticket and is equal to $1/K$. In case the player wins all $K$ prizes, he will get a super prize. Find the expected value of tickets that the player should buy, in order to get a super prize.
My attempt: Let $X$ be the random variable that denotes the number of tickets bought. And I thought it is a binomial distribution, where $X\ge K$ and $P(X=k) = {n \choose k}p^kq^{n-k}$, $k=K$ is number of successes out of $n$ trials. But I am stuck.
Actually, I don't know how to do it right. Any help is appreciated.
Let us define the random variable $X$ which counts the number of trials needed to win $K$ prices.
Let's notice that $X \in \{K,K+1,\cdots \}$.
$\mathbb{P}(X=n)$ is simply the binomial distribution, which we know equals to
$$\mathbb{P}(X=n) = \left(1-\frac{1}{k}\right)^{n-k}$$
What we are looking for in order to solve the problem is by defition $\mathbb{E}[X] = \sum\limits n\cdot p_{X}(n)$
Where $p_{X}(n)= \mathbb{P}(X=n)$
Otherwise said : $$ \mathbb{E}[X] = \sum\limits_{n=K}^{\infty} n \binom{n}{K}\left(\frac{1}{K}\right)^{K}\left(1-\frac{1}{K}\right)^{n-K}$$
Which unfortunately I was unable to evaluate, but according to Wolframalpha $$\sum\limits_{n=K}^{\infty} n \binom{n}{K}\left(\frac{1}{K}\right)^{K}\left(1-\frac{1}{K}\right)^{n-K} = K^{3}+K^{2}-K$$
Because $K^{3}+K^{2}-K \geq 0$ $\forall \hspace{0.1cm} K \in \mathbb{N}$ it seems a reasonable result due to the fact that the number of prizes is a positive integer.