A box contains 4 piece of papers, each paper marked with A,B,C, and D respectively. A person draws a paper and observes its letter and puts it pack.
Papers are now drawn repeatedly without replacement until the first paper drawn is selected again.
What is the mean number of draws required?
I first thought that the mean was $\frac 52$
Since for any letter I picked in the beginning, the probability of getting it on the first draw is $\frac 14$. On the second, $(\frac 34)(\frac 13)= \frac 14$ and so on.
I get the distribution $1$ draw-$\frac 14$ , $2$ draws- $\frac 14$ $3$ draws $\frac 14$ $4$ draws-$\frac 14$
So I get expected value $\frac {10}4$
In fact, the answer is $\frac 57$.
Is there something wrong with my approach?
Thanks in advance.
The answer $\frac 57$ is wrong. Your answer $\frac {10}4$ I would say is correct. Your approach is correct as well. For one, as a comment mentioned, we would need at least one draw. Secondly, since all draws are equally probable, it's just the mean of $1,2,3,4 = 2.5$