Probability that my set of dice 'wins' My question is about probabilities. Given are two sets of regular, 6-sided, fair dice of at least one die (but no upper bound on the number of dice). Now, 1 of the sets is considered to be 'the winner'. This is determined with the following method:
Sort both sets on the number of eyes. Compare the dice of both sets 1-by-1 starting at the highest by pairing up the dice from both sets (compare 1st with 1st, 2nd with 2nd etc.). As soon as 1 die is higher than the other, then the respective set is the winner and the other dice are ignored. If both sets have an equal number of dice and all dice are equal, then it's a tie. If it's a tie except that one set still has dice left (is a bigger set), then that set is the winner.
Examples:
The range of results $n$ dies produce is $n$ to $6n$. This means there are $6n - (n-1) = 5n + 1$ possible results for each die, meaning the total number of possibilities is $(5a + 1)(5b + 1)$. The number of times the smaller party wins is the sum of numbers from 1 to its number of dies - 1:
(showing ties)

We can see that wins for 'column' party are $1 + 2 + 3 + 4 + 5$,  or
$$\sum\limits^{6c-1}\limits_{n=1} n$$
where $c$ is actually the size of the smallest party.
Dividing this, by the number of possibilities should yield the probability (unless I am an ignorant or have done something wrong).
So, let $c$ be the smallest groups size, $a$ be the first group's size and $b$ be the second group's size, following the logic previously stated(which might be incorrect) we get:
$$f(a, b) = \frac{\sum\limits^{6c-1}\limits_{n=1} n}{(5a + 1)(5b + 1)}$$
I tried it with the simple test case of a = b = 1 and it works, giving the answer $\frac{15}{36} = \frac{5}{12}$.
