Calculate the winning probability when using two dice with a different number of sides

Question

I have two players with one die.

The first player throws a die between 1 and 125.

The second player throws a die between 1 and 350.

The winner is the person who gets the higher roll. How do I make an average for this, aka the average times player one or player two will win. I seem to be having a brain fart here. I can make a simulator to generate this, but that seems rather silly in favor of a calculation.

Answer 1

Let A represent the distribution of rolls that the player with the $125$ sided die can achieve and B represent the other. The probability that the player $A$ with the $125$ sided die wins is $$\sum_{k=1}^{k=125}P(A=k)\cdot P(B \lt k)$$ $$=\sum_{k=1}^{k=125}\frac{1}{125}\cdot \frac{k-1}{350}$$ $$=\frac{1}{43750}\sum_{k=2}^{k=125}k$$ $$=\frac{1}{43750}\cdot\Big(\frac{125}{2}(125+1)-1\Big)$$ $$=\frac{3937}{21875} \approx 0.1799771428571428571...$$

Answer 2

$225$ times out of $350$ the second player rolls a number higher than $125$ and wins no matter what the first player rolls. The other $125$ times he rolls a number between $1$ and $125$ so it's a fair game between the two players. You just have to account for the possibility of a tie.

Answer 3

I assume that ties play again.

Let $p$ be the chance that the second player ($350$ roller) wins. On $\frac{225}{350}$ of the rolls, the second player wins, on the other $\frac{125}{350}$ of the rolls, they have $\frac{124}{125}\frac12+\frac1{125}p$ chance of winning. That is, $$ p=\overbrace{\ \ \ \frac{225}{350}\ \ \ }^{\substack{\text{second player}\\\text{rolls $\gt125$}}}+\overbrace{\ \ \ \frac{125}{350}\ \ \ }^{\substack{\text{second player}\\\text{rolls $\le125$}}}\left(\vphantom{\frac12}\right.\overbrace{\ \ \frac{124}{125}\,\frac12\ \ }^{\substack{\text{first player}\\\text{doesn't tie}}}+\overbrace{\ \ \frac1{125}\,p\ \ }^{\substack{\text{first player}\\\text{ties}}}\left.\vphantom{\frac12}\right) $$