Now, it's been a bit since I took probability in College, but I believe the Math in the first post isn't correct.
The best way to determine the odds of a success/fail roll is using the Binomial Distribution Formula. If you look up the formula on Wikipedia, it seems pretty complicated but really isn't. The first part is read as N choose K, and the second part is pretty self explanatory. N is how many trials we will be performing, and K is how many successes we need.
Thus, since we have 1 roll plus two rerolls, and we only need 1 success, we would say 3 choose 1. 3 choose 1 doesn't seem like math, but it is. You can plug 3 choose 1 exactly into google and it will spit out your answer. Since the probability of a double hit occurring is 1/8, or .125, we would say: 3 choose 1 * (.125)^1 * (1-.875)^2. That would equal 28.71% chance of having a double hit. Of course this assumes that you are rerolling everything until that sweet double.
Now you can use that new probability to plug in the same exact formula for each dice you need. For instance, in your example you go with 4 Red Dice. Thus, for exactly 1 double hit: 4 choose 1 * (.2871)^1 * (1-.2871)^3 or 41.61%. You can do this for as many successes that you want. Example for two successes: 4 choose 2 * (.2871)^2 * (1-.2871)^2 = 25.13%.
This formula, along with expected value can be used to solve almost any damage related question for upgrades and dice in this game.
