What are the limiting probabilities for this silly dice game?

Question

The game works as follows: Two $n$-sided dice are rolled. If the dice show unequal numbers, then the higher number is rerolled. If the dice show equal numbers, then that is the final result of the silly dice game. As $n \rightarrow \infty$, what is the probability of getting $m$ for some finite $m$? I know that the probability for $m=1$ is $1/2$ + $1/2n$ because when the dice are different and neither is $1$, the probability of getting $1$ as the result is $1/2$ because the chance of rolling the number on the other die and the chance of rolling $1$ are equal.

Answer 1

For simplicity, we call ending up with $m$ as a win. There are only three states of the game that result in win or at least a chance to win:

1. Both dice show $m$.
2. One die show $m$ and the other show a larger number.
3. Both dice show different number larger than $m$.

Probability to Win from State #2

Say the probability is $p_{2}$. If we roll an $m$ we win, probability $\frac{1}{n}$. If we roll a number lower than $m$ we lose, probability $\frac{m-1}{n}$. If we roll a number larger than $m$ the game remains in state #2 , probability $\frac{n-m}{n}$.

$$ \frac{1}{n}+\frac{n-m}{n}\cdot p_{2}=p_{2}\implies p_{2}=\frac{1}{m} $$

Probability to Win from State #3

Say the probability is $p_{3}$. If the two dice becomes equal we lose, probability $\frac{1}{n}$. If we roll lower number than $m$ we lose, probability $\frac{m-1}{n}$. If we roll $m$ the games move to state #2, probability $\frac{1}{n}$. If we roll larger number than $m$ and not equal to the other die the game remains in state #3, probability $\frac{n-m-1}{n}$.

$$ \frac{1}{n}\cdot\frac{1}{m}+\frac{n-m-1}{n}\cdot p_{3}=p_{3}\implies p_{3}=\frac{1}{m(m+1)} $$

Remaining Things to Do

Simply calculate the probability of ending in any of the three states above. It's trivial to see that as $n$ becomes larger and larger, the probability to end in state #3 approaches unity.