You roll a die infinitely many times. After rolling, you get the number on the die in dollars. For example, if you roll a 6, you'll win $\$6$. You start with no money. Let $p_n$ be the probability that at some point you have $\$n$. What is the limit of $p_n$ as $n \rightarrow \infty$?
I believe this must be a standard problem, but I wasn't able to find anything on it. Here is my current progress.
First, each step has an average size of $7/2$, so we expect the "time we spend" on each destination to be $2/7$. So we expect $p_n \rightarrow 2/7$. I've confirmed this by writing a program using the recurrence relation $p_n = 1/6(p_{n-1}+...+p_{n-6})$.
Here's how I'm approaching proving this. Consider the more general problem where you have a possibly weighted dice that has positive integers written on it, instead of a fair dice with numbers $1...6$. Say the expected value of a roll is $\mu$. So $\mu = 7/2$ in our example.
Let $X_n$ be the money I have after $n$ rolls. By the strong law of large numbers, $X_n/n \rightarrow \mu$ almost surely (a.s). Therefore $n/X_n \rightarrow 1/\mu$ (a.s). Let $S_n$ be the number of destinations in $1, 2, ..., n$ that are reached at some point. For each $n$, one of $n, n+1, ..., n+5$ are hit. So $S_n/n \rightarrow 1/\mu$ (a.s). Therefore $E[S_n/n] \rightarrow 1/\mu$ (I proved this using weak the law of large numbers, but maybe it is a standard result). Therefore $(p_1+...+p_n)/n \rightarrow 1/\mu$. Therefore IF $p_n \rightarrow p$ for some $p$, then $p = 1/\mu$.
The condition that $p_n \rightarrow p$ is important (for example, consider a dice that only has even rolls. Then you will never hit an odd number). I'm guessing that if the set of rolls share no common divisor, then it is true that $p_n$ converges. I'm not sure how to prove this.
I also have one related question. Let's say that I stop the game if I get $M$ dollars or more. What is the distribution of $X_M$, the total money I make, as $M \rightarrow \infty$? I'm guessing that $P(X_M=M) : P(X_M=M+1) : ... : P(X_M=m+5) = 6 : 5 : ... : 1$, since $M$ can be reached from 6 places ($M-1, ..., M-6$), $M+1$ can be reached from 5, ..., $M+5$ can only be reached from 1. And so the distribution is $6/21, 5/21, ..., 1/21$.
In the scenario above, let $Y_M$ be the money I had immediately before my last roll. What's the distribution of $Y_M$? Is it related to the distribution of $X_M$ in any nice way (e.g., by reflection)?
I've currently only taken a first year probability course so please keep your answers simple!
Thanks!
