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:
Here is an alternate, very rough approximation for large $a,b$.  Another answer establishes that the probability of ties converges to $0$.  Likewise I’d expect that the probability of a tie on just $6$s is also small.  In that case we can just model the difference in the number of sixes as a random variable normally distributed with mean $(a-b)/6$ and variance $5(a+b)/36$.
The probability that this is positive (i.e. that the first player wins by having more 6s) is the same as a standard normal variable being $< (a-b)/\sqrt{5(a+b)}$.  This cumulative distribution can be computed using the erf or erfc function in many programming languages.  But I wouldn’t trust it for single digit values of $a,b$.
