Probability: Infinite marbles placed in, and selected from an urn

[IniquiTrance]

Homework Statement

I have a countably infinite set of marbles numbered; 1, 2, 3,..., n.

I also have an urn that can hold an infinite amount of marbles.

I then place marbles 1 and 2 into the urn, and remove one of them with the following probabilities:

The probability of removing a marble is proportional to its number.

So, the probability that I remove marble 1 is \stackrel{1}{3}, and that I remove marble 2 is \stackrel{2}{3}.

Once a marble is removed, I then place marbles 3 and 4 into the urn. I

Now if marble 2 was removed earlier, then marbles 1,3,4 are in the urn. The probability of removing any of them are now respectively, \stackrel{1}{8}, \stackrel{3}{8}, and \stackrel{4}{8}

I keep adding and removing marbles as above, in order of their number.

I am asked to show, that there is a positive probability that marble 1 remains in the urn forever.

Homework Equations

The Attempt at a Solution

Not quite sure how to pin this down. Any help is much appreciated!

 

[Dick]

Ok, so the probability marble 1 survives the first pick is 2/3. The probability that it survives the second is 7/8. At this point you have a choice which one to pick which is not 1. Pick the one which has the least probability to be picked which is not 1. That would be a lower bound for the probability that 1 will never be picked, right? I haven't tried to show the resulting infinite product is positive. Can you? That should get you started.

 

[IniquiTrance]

Thanks for the lead!

Ok, I'm trying to trace it out now...

Wouldn't the 2nd round have to be conditioned on the 1st round, so P(surviving 2nd round) would be:

\stackrel{2}{3}*\stackrel{7}{8}, no?

edit: n.m., I see your point, going to set it up as:

\prod_{n}^{}\mathbb{P}\left\{A_{n}|A_{n-1}\right\}