[Pre-BB] MapleStory Formula Compilation

it would be interesting to make tests on the damage distribution, like if your range is 10~20 how many twenties you hit, how many fifteens, etc.

if it's a normal distribution or something else... i was testing this someday with my pure dex islander on snails, but i forgot that my pole arm slash/stab ratio influenced... should be done with a sword...

btw, pure dex islanders are better for this as their damage range is around 10 units.

how many hits do you think it would be enough for testing?

It's uniform distributed. I did research on it.

Yay. Now I don't have to feel like I'm using dubious methods to find probabilities of OHKOing flyeyes and the like.

Yay. Now I don't have to feel like I'm using dubious methods to find probabilities of OHKOing flyeyes and the like.

You still have to use normal distribution for finding 2hkos and the like.

Not normal, not exactly. 2 numbers make a triangle, as you add more it approaches normal. I did take that into account.

Critical with Strafe is hellish to compute.

Not normal, not exactly. 2 numbers make a triangle, as you add more it approaches normal. I did take that into account.

Critical with Strafe is hellish to compute.

Ahh, thought I wrote this already, seems like I didn't post it, meh, lol:

2 numbers do make a triangle, a binomal distribution. However, at higher differences (for instance, a 43995 difference) the values will be extremely hard to calculate. You "got to" use normal distribution, as nCr (for e.g. 43995 C 21997), 419301357155661. This is a number way higher than what integers operate with, and is the exact accuracy doubles work with, Anything higher will deal out wrong answer as it's not 15 counting digits and, the probability for one way of getting to that damage is rather low. Taking this into account, I'd say that floating-point will happen somewhere, and normal distribution will either give out a more exact, or the same result on higher differences.

And uh, what's the problem with calculating damage with strafe+critical? Maybe I can help out.

Use a long?

I see nothing wrong with working with the enormous numbers, there's many ways to still deal with them and get an accurate result.

I agree though, however confusing a normal distribution would be more accurate, and probably easier.

strafe+crit = 16 possiblities, all with different probabilities and results. It just flat out sucks.

Then again, you could always just do it arrow-by-arrow, then round up to the nearest 4th arrow, so if a 1 hit was 5 arrows, you'd know it would be a 2 hit with strafe.

Normal distribution is ironically an approximation of xhko.

Binomial distribution is actually not relevant to this purpose. It's useful when there are two possibilities (e.g. flipping a coin or passing a scroll), and spits out all the possibilities given the number of trials and probability of each outcome (heads/tails, pass/fail). But when you've got a damage range that isn't 1-2 or 9-10 or 32767-32768, it becomes useless.

What would really be useful for calculating this would be a function that spits out the distribution of a sum of dice rolls, given the number of sides on the dice and the number of dice rolled.

I'm currently trying to work something out with nested arrays in Mathematica to see if it could spit out some numbers for me, but even if I manage to get it to work (and I'm doubtful) my computer might not appreciate doing all that number bashing.

Use a long?

Long isn't big enough. I think.

Use a long?

I see nothing wrong with working with the enormous numbers, there's many ways to still deal with them and get an accurate result.

I agree though, however confusing a normal distribution would be more accurate, and probably easier.

strafe+crit = 16 possiblities, all with different probabilities and results. It just flat out sucks.

Then again, you could always just do it arrow-by-arrow, then round up to the nearest 4th arrow, so if a 1 hit was 5 arrows, you'd know it would be a 2 hit with strafe.

A long must be converted to a double to get a right answer when multiplied with a double, else it'll only display long*(rounded down to integer double) which would be in probability "always" 0.

It's actually funny, how 0.1 + 0.1 != 0.2 in doubles. Calculation with doubles is a pain in the ass.

Binomial distribution is relevant for 2hkos. Edit: I'll add in this when I got time. I don't atm, in 1 hr or less. School.

What's a long?

Binomial distribution is basically Pascal's triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

etc

2hkos, or in other words rolling 2 dice with n sides, is

1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1

etc.

And yes, you'd think that's even simpler. But the trouble happens when you have 3hkos and beyond. Essentially to find 2hko for strafe you have to find 8hko, since there are 8 independent arrows to consider... plus critical/noncritical for all 8 of them. The nested arrays might do the trick, I don't know at this point.

What's a long?

Binomial distribution is basically Pascal's triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

etc

2hkos, or in other words rolling 2 dice with n sides, is

1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1

etc.

And yes, you'd think that's even simpler. But the trouble happens when you have 3hkos and beyond. Essentially to find 2hko for strafe you have to find 8hko, since there are 8 independent arrows to consider... plus critical/noncritical for all 8 of them. The nested arrays might do the trick, I don't know at this point.

A long is a 64 bit value.

Yeah, I know how binomial distribution works. Pascal's triangle shows how you deal with the 2hko binomial distribution. For 3hko and further I'd "recommend" normal distribution (I'd go normal all the way for programming, but that's just me) just to simplify it. It's a difference of basically nothing.

I meant that binomial distribution was different than '2hko distribution'.

But about normal, yeah, I suppose so. It's like the difference between pi and 355/113. No one would notice if they weren't looking for it.

I'm not exactly sure how to set it up, though, without it returning some inane chance of 3hkoing Zakum or something.

Anyone here know how much experience you get from Heal per HP? D:

I looked that up on SW and Neko cited it as 1 exp per ~25 healed. Looking at his data, it seems pretty linear.

I noticed that the Lucky Seven formula is in blue (not recently verified). That's the correct formula and you may want to note that it applies to Triple Throw as well.

I noticed that the Lucky Seven formula is in blue (not recently verified). That's the correct formula and you may want to note that it applies to Triple Throw as well.

I see no verification.

It's an old formula, and the fact that it's not a "damage range" formula says that it would've had to be tested, and back then hardly anything was tested.

It's almost certainly the right one, but I've never ever seen any actual proof.

586 luk, 64 watk.

It projects a range of 1406-2812 regular, 2344-4688 crit.

I spent some 400 seconds on a sparsely populated map (Mushroom Garden, for those interested) and observed the following:

Min reg: 1421
Max reg: 2798
Min crit: 2362
Max crit: 4672

In short, if that's not it, I will eat my hat (considering that that's a disproportionate amount of luk/watk).

Sounds like a confirmation to me. You were using TT no?

*Tries something*
Damage Range 30~50, probability to 2 hit a monster with 80 HP

Let me try to do this via the way you do it with dice.

21^2 = 441 possibilities.

Chances to do 80 damage+ added together:
30/50
31/50
32/50
...
50/50
=20

31/49
32/49
...
50/49
=19

etc etc

49/31
50/31
=2

50/30
=1
= 210 chances, 210/441 = 47.6% chance to 2 hit.
Rawr, that can't be right.

That is right. Assuming you calculated it correctly.

Now factor in critical and fractional damage ranges. (We know that 30~50 could be 30.23~50.97, or 30.76~50.01, etc etc, and the probability of hitting a number depends on those fractional parts.)

I tried writing a program to do it awhile ago, but I lost interest.