Probability of die threshold

Question

Suppose there is a fair die with $n$ faces labeled from $1$ to $n$, and a number $1\leq k \leq n$.

How many times does one need to throw the die (on average) to get a number $j$ with $j\geq k$?

Answer 1

Let $X$ be the number of times the die is thrown until a number $j$ with $j\geqslant k$ is obtained. Then $X$ is geometrically distributed with parameter $(n-k+1)/n$. So the probability mass function of $X$ is given by $$ \mathbb P(X=m) = (1-(n-k+1)/n)^{m-1}(n-k+1)/n. $$ We compute the expectation by \begin{align} \mathbb E[X] &= \sum_{n=1}^\infty m\cdot\mathbb P(X=m)\\ &= \sum_{n=1}^\infty m(1-(n-k+1)/n)^{m-1}(n-k+1)/n\\ &= \frac n{n-k+1}. \end{align}