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We have an urn with two marbles numbered $1$ and $2.$ We pick a marble randomly, write down its number and return it to the urn. Then we add a marble with the number 3 to the urn, choose one of the three marbles randomly, record its number and return it to the urn. We repeat this over and over: before the $(k+ 1)$st pick we add a marble with with the number $k + 2$ in the urn (so that it contains the numbers $1, \dots , k + 2)$, choose a marble randomly, record its number (that's the $(k + 1)$st pick) and return it to the urn. At the end of the experiment we have an infinite sequence of integers. Show that with probability one the marble with the number $1$ will be picked at some point.

It was recommended to break up the event into disjoint pieces. But I have no idea how to do this. Also, does this mean that all drawings must give a different number, with say $1$ in the $n$th draw?


Any hints are much appreciated.

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    $\begingroup$ Hint: after $n$ turns, what is the probability that the marble with $1$ was not picked? $\endgroup$ – Wojowu Feb 8 at 12:08
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Let $E_n$ be the event "you haven't drawn the $1$ in the first $n$ trials."

Then $$P(E_n)=\prod_{i=2}^n \frac {i-1}i=\frac 1n $$

Where, for the last equality we used the fact that the product telescopes.

It is clear, therefore, that $$\lim_{n\to \infty} P(E_n)=0$$ and we are done.

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$$\Pr(\text{1 is picked up in some time}){=1-\Pr(\text{1 is not picked up in any time})\\=1-{1\over 2}\times{2\over 3}\times{3\over 4}\times\cdots\\=1-\lim_{n\to \infty}{1\over n+1}\\=1}$$

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The probability that marble $1$ is picked for the first time at round $k$ is $$ \frac{1}2\cdot \frac23\cdots \frac{k-1}{k}\cdot \frac{1}{k+1}=\frac1{k(k+1)} $$ Therefore, the probability $1$ is picked up during some round is the sum of the probabilities of these disjoint events: $$ \sum_{k=1}^\infty \frac1{k(k+1)}=\sum_{k=1}^\infty \frac1{k}-\frac1{k+1}=1 $$ since the sum telescopes.

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