What is the probability of two people rolling a die and one person rolling higher if one of them adds 1 or 2 to their die's total? Two players roll a 6-sided die, it is a 50/50 chance that one will roll higher, but what would the probability be if one player gets to add 1 or 2 to their die's total? I'm developing a tabletop ancient/medieval battle strategy game. Say a light infantry unit is engaged with a heavy infantry unit, the light unit has no modifications to combat rolls but the heavy infantry gets +2 to combat rolls. What is the probability that the heavy infantry unit will have a higher roll? Also, since tieing would be a 1/6 chance (which results in no damage being dealt), would that affect the probability? And how would the probability change if rolling a "1" cannot count as more than 1 regardless of modifications?

I will handle the case where the die of the high infantry gets an increase of 2. The same approach can then be used to determine the probabilities when adding any other number.

If the light infantry throws a 1 or 2, then the high infantry always wins. If the light infantry throws a 3, the probability of a victory for the high infantry equals $\frac{5}{6}$ (throw higher than 1). Throwing a 4, we find a probability of $\frac{1}{6}$ for a tie (throw 2) and $\frac{4}{6}$ for a victory (throw higher than 2). Applying the same to rolls of 5 and 6, and we find, for the events $V$ (high infantry wins), $T$ (a tie) and $L$ (high infantry loses):

$$P(V) = \frac{1}{6} \left( 1 + 1 + \frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} \right) = \frac{26}{36} = \frac{13}{18}$$

$$P(T) = \frac{1}{6} \left( 0 + 0 + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \right) = \frac{4}{36} = \frac{1}{9}$$

$$P(L) = 1 - P(V) - P(T) = \frac{6}{36} = \frac{1}{6}$$