It originates from this problem:
- We have a box with $n$ balls that are evenly mixed, all are $\color{blue}{blue}$ except for one that is $\color{red}{red}$.
- If we pick blindly one ball per try and after each try we don't put it back in the box, what is the average number of tries it will take us to take get the $\color{red}{red}$ ball $?$.
- If I'm correct the probability of picking the $\color{red}{red}$ ball at the try number $x$ is: $$ P[X=x]=\frac{(n-x)!}{n!},\quad X = \mbox{number of tries.} $$
- Thus the expected value of $X$ is: $$ E(X)=\sum_{x=1}^{n}x\frac{(n-x)!}{n!} $$ whose 'clean' analytical expression I can't find.
Could you help me advance in the problem please $?$. Unfortunately the book where I found the problem doesn't have the solution.
The red ball is equally likely to come in any position. There are $(n+1)!$ orders is which the balls may be drawn, and in $n!$ of them the red ball is in position $k$, for any $1\leq k\leq n+1$ so the probability, that the red ball is in position $k$ is $\frac1{n+1}$
Now I'm sure you'll find it easy to finish.