Find expected value of discrete trials same probability.

Question

I am struggling with a question. I have found the probability of success to be $\frac{k!}{k^k}$. How do I find the expected number of trials before a success? For example, this is similar to if I rolled a dice with a probability of $\frac{1}{6}$ what is the expected number of trials I would have to take to get lets say the side 2? Thanks

Answer 1

Let's say that $p$ is the probability of success and that $X$ denotes the number of failures that precede the first success. If $S$ is the event that the first trial is a success, and $F$ the event that it is a failure then:

$\mathbb{E}X=\mathbb{E}\left(X\mid S\right)P\left(S\right)+\mathbb{E}\left(X\mid F\right)P\left(F\right)=0.p+\left(1+\mathbb{E}X\right)\left(1-p\right)=\left(1+\mathbb{E}X\right)\left(1-p\right)$

This equation enables you to find $\mathbb EX$.

Essential is here that $\mathbb{E}\left(X\mid F\right)=1+\mathbb{E}X$.

After the failure the process "starts over" with $1$ failure in our pocket.