Calculating % X-HKO

Say monster has 70000 HP, your effective damage range is 30000 to 50000.

The formula gives, for 2 hits, (100k - 70k) / (100k - 60k) = 30k / 40k = 75%. For 3 hits, it's fairly evident that it'll be 100% because the minimum you can hit in 3 hits is 90k which is > HP.

The actual scenario is that the probability for the 2nd hit to kill depends on how much the first hit dealt, and how much left is needed.

Say that the first hit was for 50k (the max). Then, the monster only has 20k left, which is less than the minimum (30k) so it's an ensured 2HKO.

Now the question to be asked is, what's the least amount the first attack can hit for in order to make an ensured 2HKO? 70k - 30k = 40k. So if you hit anywhere in the upper 50% of the range (40k - 50k), it'll be an ensured 2HKO.

So, what if the first hit is under that? Say you hit the minimum, 30k, on the first hit. That leaves 40k left. By your formula, the chance to 1-hit a monster with 40k HP left, with the same damage range, is 50%.

It turns out that this chance increases linearly from 50% to 100% as your first hit goes from 30k to 40k. So the average over the lower half of the range would be 75%. Combined with the upper half, the total probability for 2HKO is 87.5%.

The reason for all this is, assuming the damage for 1 hit is uniform distributed, the total damage for 2 hits is a triangular distribution. Think of it in terms of dice sums, as JoeTang mentioned.

Okay that makes a lot more sense (I just bolded the sections that summed up the post for me), as yea, I hadn't even taken that into consideration. Silly me! Anyways, thanks a lot for that reminder!