Say for a moment two people each have a single die. One has $m$ sides while the other has $n$ sides. Assuming the dice are each fair (as in, each die has an equal chance to roll any side… obviously if $m > n$ it doesn't seem fair!) how can I calculate the chance of either one winning?
In my exact situation, my "dice" have many thousands of sides, so it would be prohibitive of me to make a "truth table" so to speak. I doubt that affects the formulas but figured I should add it for relevance.
I was thinking that assuming $m > n$, then when $m$ rolls a 1, then $n$ wins $n-1$ number of times (i.e. when they don't roll a 1). Same situtation where $m$ rolls a 2, $n$ has $n-2$ rolls that win. So we have this $n – 1 + n – 2 …$ which seems to be $(n *( n+1))/2$ for large n. Since there are $m*n$ number of possible combos, the chance of $n$ winning is $((n*n+1)/2)/(m*n)$ and cancelling $n$ yields $(n+1)/(2*m)$ but I could be completely off base with it and could use some guidance.
This result doesn't sit well with me because for very large n (say 5000 and 5001) they should be very, very close to eachother where this gives a massive bonus to the 5001. So clearly I messed something up!
For purposes of this, you may ignore ties. It is sufficient to select one player as the "interesting" player and simply determine if that player "wins" or not. For such large n, ties are effectively irrelevant anyway.
Best Answer
The number of outcomes where player $M$ beats (or ties) player $N$ is given by the sum $$\sum_{i=1}^{n}\sum_{j=i}^m1 = \sum_{i=1}^{n}(m-i) = mn - \sum_{i=1}^{n}i = mn-\frac{n(n+1)}{2}.$$ Dividing through by the total number of outcomes $mn$ yields $$P(\text{M does not lose}) = 1 - \frac{n+1}{2m},$$ which for large $n, m$ roughly equal is close to $\frac{1}{2}$ as expected.