# Another marble and urn problem, this time to $\infty$

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.

• Hint: after $n$ turns, what is the probability that the marble with $1$ was not picked? – Wojowu Feb 8 at 12:08

Let $$E_n$$ be the event "you haven't drawn the $$1$$ in the first $$n$$ trials."
Then $$P(E_n)=\prod_{i=2}^n \frac {i-1}i=\frac 1n$$
It is clear, therefore, that $$\lim_{n\to \infty} P(E_n)=0$$ and we are done.
$$\Pr(\text{1 is picked up in some time}){=1-\Pr(\text{1 is not picked up in any time})\\=1-{1\over 2}\times{2\over 3}\times{3\over 4}\times\cdots\\=1-\lim_{n\to \infty}{1\over n+1}\\=1}$$
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.