Probability that an $n$ sided dice reaches a sum of at least $x$ before an $m$ sided dice.

Question

What is the probability that an $n$-sided dice reaches a sum of at least $x$ before an $m$-sided dice?

Assume:

1. $n$ and $m$ are not equal values;
2. Dice are fair;
3. Dice can roll $0$.

Answer 1

Let $X_k$ be the sum of the $m$-sided die after $k$ rolls, $Y_k$ the sum of the $n$-sided die after $k$ rolls.

Then $$E(X_k - Y_k) = \frac k2(m - 1) - \frac k2(n - 1) = \frac k2(m - n)$$ and $$Var(X_k - Y_k) = \frac k{12}(m^2 - 2m - 2) + \frac k{12}(n^2 - 2n - 2) = \frac k{12}(m^2 + n^2 - 2m - 2n - 4).$$ The standard deviation of $X_k - Y_k$ is therefore $\sqrt{\frac k{12}(m^2 + n^2 - 2m - 2n - 4)}.$ Because $k$ grows faster than $\sqrt k,$ if $m > n$ then there is some value of $k$ such that $0$ is more than three standard deviations below the mean of $X_k - Y_k,$ that is, $X_k > Y_k$ with high probability. For larger values of $k,$ the number of standard deviations between $0$ and the mean of $X_k > Y_k$ is even greater.

So if $m > n,$ and if we set $x$ to a large enough number, we can make it so that by the time the dice have been rolled enough times for there to be any chance that the sum of the rolls of the $n$-sided die is $x,$ the sum of the rolls of the $m$-sided die is almost certain to be greater and hence it is almost certain that the sum of those rolls reached $x$ first. This is despite the fact that the expected number of rolls for either die to reach $x$ is almost linear; for example, if $m = 6$ and $n = 5$ then we expect it take approximately $\frac65$ times as many rolls for the $n$-sided die to reach $x$ as for the $m$ sided die to reach $x,$ no matter how large $x$ is.

So the ratio of the expected number of rolls is not the correct answer.

I don't know if there is any simple formula that answers this question. For specific values of $m,$ $n,$ and $x$ you could calculate an exact answer by brute force (with the help of a computer, preferably). It may also be possible to find a formula for any $m$ and $n$ that asymptotically approximates the probability as $x$ goes to infinity. (The formula should approach either $0$ or $1$ depending on whether $n < m$ or $n > m.$)