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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.

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    $\begingroup$ I think your answer is correct. $E(X)$ diverges to $+\infty$. $\endgroup$ – Yuta May 9 at 10:26
  • $\begingroup$ Does this mean that the game may not eventually stop and go on forever? If so, is there a way we can calculate the percentage of games that will stop? $\endgroup$ – Quannos May 9 at 13:07
  • $\begingroup$ when you say "If you pick a black ball, then you put in another black ball each time until you pick the white ball.", do you mean that the same black ball is put back into the urn, or the same black ball along with an extra black ball is put back in? From your solution, it is obvious that you've considered the latter, but is that correct? I'm asking because this is confusing me. $\endgroup$ – ExtremeRaider May 9 at 14:31
  • $\begingroup$ in response to my above comment- turns out that it does not matter whether an extra ball is added or not (atleast in this specific case); the answer will turn out to be 1 either way. Also, @Yuta, no, the answer is incorrect, the last step where $P(x=k)$ is multiplied by $k$ does not seem right. $\endgroup$ – ExtremeRaider May 9 at 14:36
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    $\begingroup$ As I noted below, the fact that the length of the game is finite with probability $1$ does not contradict that the expected length is infinite. $\endgroup$ – Ned May 9 at 16:39
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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.

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    $\begingroup$ The fact that the game has finite length with probability $1$ does not imply that the expected length is finite, as the game is in this problem shows. For another simple example, flip a fair coin until you have flipped more heads than tails. This game is also finite with probability $1$ and yet its expected length is infinite. $\endgroup$ – Ned May 9 at 16:37

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