Suppose there is an urn with $n\in\mathbb{N}$ balls, of which one is marked. If we draw the marked ball from the urn it was a 'good' draw, after each draw we put the drawn ball back into the urn with an additional unmarked ball. The number of balls is equal to the number of turns. What is the probability that we have infinite 'good' draws if we play forever?
With $n\in\mathbb{N}$ are $X_n \sim B(\frac{1}{n})$ independent Bernoulli distributed random variables with values in $\{0,1\}$. We can model the given problem as the sum $S_n = \sum_{i=1}^{n} X_i$, than $\mathbb{P}(S_n \leq t)$ is the probability that we draw the marked ball less than or equal to $t\in\mathbb{N}$ times.
Based on the expected value going to $\infty$ I think that the probability to have infinite 'good' draws should be positive.
What I got so far:
$$ \begin{aligned} \mathbb{E}(X_n) &= 1 \cdot\frac{1}{n} + 0 \cdot(1-\frac{1}{n}) = \frac{1}{n} \xrightarrow{n\rightarrow\infty} 0 \\ \mathbb{E}(X_n^2) &= 1^2 \cdot\frac{1}{n} + 0^2 \cdot(1-\frac{1}{n}) = \frac{1}{n} \xrightarrow{n\rightarrow\infty} 0\\ \mathbb{E}(S_n) &= \mathbb{E}(\sum_{i=1}^{n} X_i) = \sum_{i=1}^{n} \mathbb{E}(X_i) = \sum_{i=1}^{n} \frac{1}{i} \xrightarrow{n\rightarrow\infty} \infty\\ \mathbb{V}(S_n) &= \mathbb{E}(S_n^2) - \mathbb{E}(S_n)^2 \\&= \mathbb{E}\left( (\sum_{i=1}^{n} X_i )^2\right) - \left(\sum_{i=1}^{n} \frac{1}{i} \right)^2 \\&= \mathbb{E}\left( \sum_{i=1}^{n} X_i^2 + 2\sum_{1\leq i < j \leq n} X_iX_j ) - (\sum_{i=1}^{n} \frac{1}{i^2} + 2\sum_{1\leq i < j \leq n} \frac{1}{i}\frac{1}{j}\right) \\&= \sum_{i=1}^{n} \mathbb{E}(X_i^2) + 2\sum_{1\leq i < j \leq n} \mathbb{E}(X_i)\mathbb{E}(X_j ) - \sum_{i=1}^{n} \frac{1}{i^2} - 2\sum_{1\leq i < j \leq n} \frac{1}{i}\frac{1}{j} \\&= \sum_{i=1}^{n} \frac{1}{i} + 2\sum_{1\leq i < j \leq n} \frac{1}{i}\frac{1}{j} - \sum_{i=1}^{n} \frac{1}{i^2} - 2\sum_{1\leq i < j \leq n} \frac{1}{i}\frac{1}{j} \\ &= \sum_{i=1}^{n} \frac{1}{i} - \sum_{i=1}^{n} \frac{1}{i^2} \\ &= \sum_{i=1}^{n} \frac{1}{i} \left( 1- \frac{1}{i} \right) \xrightarrow{n\rightarrow\infty} \infty - \frac{\pi^2}{6} = \infty\\ \end{aligned}$$
This seems to be Poisson distributed, with parameter $\alpha = \lim_{n\to \infty} n \cdot p_n =\lim_{n\to \infty} n \cdot \frac{1}{n} =\lim_{n\to \infty} 1 = 1 $.
Therefor we get
$$\begin{aligned} \lim_{n\to \infty} \mathbb{P}(S_n \leq t) &= \sum_{k=1}^{t}\left( e^{-1}\frac{1^k}{k!} \right) \\ &= e^{-1} \sum_{k=1}^{t} \frac{1^k}{k!} \xrightarrow{t\rightarrow\infty} e^{-1} e = 1 \end{aligned}$$
which makes sense, that's a characteristic of the cumulative distribution function.
And this is where I'm stuck, how do I calculate the probability to have infinite 'good' draws?
The expected values and variance is not needed to solve this problem. With @QiaochuYuan comment that I should apply the Borel-Cantelli-Lemma I was able to solve it.
The event we are looking for is describes that no matter how finitely often we draw the marked ball from the urn and put it back with another unmarked ball, what is the probability to draw the marked ball again.
With the events $A_n=\{X_n=1\}$ which are independent because the random variables $X_n$ are independent, we get the event $\limsup_{n\rightarrow\infty} A_n={\bigcap_{n=1}^\infty}\left({\bigcup_{m=n}^\infty}A_m\right)$, that describes the subset of all elements that are in all but finitely many events of the family $\left({\bigcup_{m=n}^\infty}A_m\right)_{n\in\mathbb{N}}$, so the probability to have infinite 'good' draws.
We observe that $\sum_{n\in\mathbb{N}}(A_n)=\sum_{n\in\mathbb{N}} \frac{1}{n}=\infty$, which means according to the 2. Borel-Cantelli-Lemma $\mathbb{P}(\limsup_{n\rightarrow\infty} A_n)=1$, so the probability to have infinite 'good' draws is 1.