As in title I am looking for the answer to the following problem:
We have an urn with 1 black and 1 white ball in it. If you pick a black ball, then you put in another black ball each time until you pick the white ball. Once the white ball is picked the game stops. What is the expected number of balls to be drawn in order for the game to end?
Here is what I did:
I denoted the probability of stopping at the $k$-th draw with $P(X=k)$. Then,
$$P(X=1) = \frac 12, $$ $$P(X=2) = \frac 12*\frac13, $$ $$P(X=3) = \frac12*\frac23*\frac14,$$ $$P(X=4) = \frac12*\frac23*\frac34*\frac15$$
and continuing in this manner I find
$$P(X=k) = \frac1{k(k+1)}$$
Now, to calculate the expected value, I multiply with $k$ and sum over all $k$.
$$E[X] = \sum_{k=1}^\infty \frac k{k*(k+1)} = \sum_{k=1}^\infty \frac 1{(k+1)}$$
which as we know from p-test, diverges.
What am I doing wrong? Any suggestions?
Thanks.
The answer to the OP's question about whether the game will continue indefinitely is no. The probability that the game eventually stops is $\ P\left(\bigcup_\limits{k=1}^\infty\left\{X=k\right\}\right)=\sum_\limits{k=1}^\infty\frac{1}{k\left(k+1\right)}=1\ $. Thus, with probability $1$, the game will eventually stop.