Calculating % X-HKO

I was thinking about this the other day as I was playing around with the magic damage formula, as people are constantly wondering just how often they're able to XHKO. After pondering just how to calculate it for a little bit (thinking of calculus theory even though it's not even related XD), I think I came up with an extremely simple answer. Mathematically it works (as far as I'm aware), so without further ado, here is the formula:

MAGIC DAMAGE FORMULA

 Spoiler

MAX MAGIC DAMAGE = ((Magic²/1000 + Magic)/30 + INT/200) * Spell Attack * Amp * Elemental Staff
MIN MAGIC DAMAGE = ((Magic²/1000 + Magic * Mastery * 0.9)/30 + INT/200) * Spell Attack * Amp * Elemental Staff

Once you have calculated your Minimum and Maximum damages with your respective ultimate (after factoring in amp, Elemental bonuses, etc.), plug those values and a few others you'll need to research into this formula:

(MAXper - MAXmon)
(MAXper - MINper)

x 100% = your % XHKO.

Terminology:

• MAXmon - Maximum HP of the monster. Input full HP to calculate % OHKO, 1/2 full HP to calculate % 2HKO, and so on and so forth.
  
• MAXper - Maximum personal damage. Calculated using the magic damage formula.
  
• MINper - Minimum personal damage. Calculated using the magic damage formula.
  

Common Monster Reference HPs:

 Spoiler

• Skelegons - *85000 HP to OHKO, 42500 HP to 2HKO
  
• Newt Golems - 80000 HP to OHKO, 40000 HP to 2HKO
  
• Newties - 68000 HP to OHKO, 34000 HP to 2HKO
  
• Blue Wyverns - 57000 HP to OHKO, 28500 HP to 2HKO
  
• Red Wyverns - 53000 HP to OHKO, 26500 HP to 2HKO
  
  *Values in order to OHKO/2HKO the whole map (Skelosaurus HP given above). To OHKO/2HKO Skelegons specifically, the values are 80000/40000, respectively.
  

Alright, time to explain what the heck this all means. The logic behind the theory is relatively simple: the game uses the magic damage formula with your specific stats in order to create a damage range for whatever skill you may be using. As a player, you're capable of hitting any number damage within that range, and so assuming you would eventually see every single number at least once within the range during training, you can calculate just how often you're able to XHKO. Simply put, the formula subtracts the minimum HP of the monster you're trying to hit from your personal maximum damage in order to see how many numbers within your damage range allow you to either kill the monster or "count" as a whatever-HKO. Then, it sets that value over the TOTAL number of values you can possibly hit (aka, your damage range), and multiplies it by 100% in order to give you a percentage "hittability," we'll say =D.

Now this is just a THEORY, of course, something I came up with on a whim. There are several factors that could completely ruin it, the major one being whether or not the game calculates the amount of damages you hit on a bell curve, or something closer to "the average." In this case, while your damage range will be what the formula gives you, the game may instead allow you to hit most often within the middle 50% or so, completely changing how often you can XHKO, as you'll hit towards the extremes a lot less. However, when I tested this out with my personal data I came up with a 90.911% OHKO ability at Skeles, which seems to be exactly correct just by eyeballing survivors while training.

In any case I wanted to throw this out for people to see; please go ahead and debunk it if you know more than I do, or factor your knowledge into the formula, WHATEVER! I'm curious to see what people have to say, and if it looks like this works out, then good luck with your calculations!

That's only 1HKO, though.

I have a (slightly iffy) formula that works (assuming damage is uniform), that I've implemented here: http://krakia.110mb.com/testing.htm

Edit: FML. It's kinda buggy. I was tinkering with it previously and apparently didn't finish. Now I have no idea what the heck I was doing.
It should be working as long as it isn't obviously not-working, though.

It's fine for 1hko % but I don't think it'll work right at higher numbers. The actual way to calculate for larger amounts is difficult though. Eg. if you hit maximum on the first shot, you may only need to be in the top 80% of your range to kill in the 2nd shot. But if you hit average on the first, you need to make it into top 40%. And so on. It just gets more complicated for extra hits.


edit: Oh, I see Russt posted his calculator

I've always wondered about the distribution in damage range. Is it flat or gaussian (or something else)? Has anybody tested this yet?

I'm guessing flat though =)


PS: question: For the elemental bonus, just add *1.5 at the end or somewhere else?

The basic formula is the same as a probability to dice roll a certain sum with variable number of dice, with variable faces and minimum to maximum values, I believe.

In my past attempts, I found formula which used large permutations and combinations (due to the large range within a damage range; i.e. one value per face on the dice, which could have been reduced by using fractions and rounding but still) to find with exact precision the probability of something like
%chance to kill a monster with 5000 HP using a damage range of 100~200 or something, where the formula held strong for most examples but crumbled after certain lengthy calculations; I'm still unsure if it was because the formula was unsound or it was possibly due to rounding issues in the machine, but it took me a lot of work to correctly implement it in java only to have it require absurd amounts of processing power and not work for any practical example.

How does this not work with 2HKO, etc? I'm sure someone can prove it to me, but as far as I'm aware it works....you'll just 2HKO over 100% (calculating like 160%) of the time when your damage minimum exceeds the damage necessary to 2HKO.

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.

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!

Looks like I was a lot more verbose than I needed to be. Oh well, the point got across. lol