31 replies to this topic

[Mekkah]

These are the kind of calculations you don't find on the site, but might want to use in your debates anyway to hype your own advantages or debunk your opponent's. Logic is fun and all, but there's few things more foolproof than massive numbers displaying how much your unit wins. And in recent debates, both tier lists and official ones, I sometimes see way too few numerical support, even when enemy stats are available.

The Formulae

Death Chances

Chance of a unit to die in x hits is (imo) the best way to display who is winning durability, and by how much.

If you want to show chance of death after x hits at y% hit, where x is also the amount of hits your unit survives, the formula is:

CoD = y/100 ^ x

For example, if Mia is getting 3HKO'ed at 20% real hit, her chance of death after 3 attacks is

20/100 ^ 3
= .2 ^ 3
= .008

This way only works if you want to calculate a unit's chance of death at the number of attacks they're getting killed in. If you have a really large durability advantage, that's usually sufficient. If not, however, you can use Reikken's awesome death chance calculator. This topic contains discussion on what formula is used for this, so if you're interested just look through it.

Chance of Critting on a double

Fairly simple, but listing it anyway.

If a unit is doubling with a certain crit% percentage, you can calculate his chance to crit at least once out of those two attacks as follows:

Chance not to crit on a single hit: a = 1 - crit%
Chance not to crit on a double hit: b = a ^ 2
Chance to crit at least once on a double hit = 1 - b

Chance of triggering something

For example, a Zihark has 20% crit, 30% Adept and 20% Astra.

Like above, you simply calculate the chance not to activate any of these, and substract that from 1.

.8 (no crit) * .7 (no Adept) * .8 (no Astra) = .112 (none triggered)
1-.112 = .888 = 88.8% chance of triggering one of these

Combining this and chance of critting on a double, I'm sure you can work out the chance of triggering any of these on a double.

Edited by Mekkah, 19 July 2009 - 04:13 PM.

[Ether]

I remember seeing a chart or something that gave true hit in relation to display hit,but I can't find it anymore,what is the formula for true hit? (Or if you happen to have the chart I'm talking about you could post that)

[Tino]

http://serenesforest...al/truehit.html

[Ether]

Wow...I feel kind of stupid now...

[Mekkah]

It can also be calculated by hand, but I see no reason for it. I actually have a (somewhat) complicated formula for it...one day either me or A22ZOMG might actually write that ultimate statulator that takes this into account.

[kirsche]

There's one major flaw with your chances of death formula: It only works for the number of rounds that unit is KO'ed in. If Mia was 3RKO'ed at 20% hit and i wanted to know her chances of death in 4 rounds that formula would be useless.

This is the formula, unless I've made a mistake:

P = (Hit^R) + N-R(R+X(Hit^R(Miss^(X+1))))

P = Probability of death
N = Number of rounds of combat
R = Number of rounds to kill
X = Number of bracketed equations to it's left -1. (Does not include the bracketed equation it is in or bracketed equations in other brackets.)

So the chance for Mia to die in 4 rounds if she's 3RKO'ed at 20% hit is:

P = 0.008 + (3(0.008 * 0.8))
P = 0.008 + 0.0192
P = 0.0272
P = 2.72%

This formula is basically just this condensed:

Hit * Hit * Hit = 0.008
Miss * Hit * Hit * Hit = 0.008 * 0.8 = 0.0064
Hit * Miss * Hit * Hit = 0.008 * 0.8 = 0.0064
Hit * Hit * Miss * Hit = 0.008 * 0.8 = 0.0064

P = 0.0064 + 0.0064 + 0.0064 + 0.008 = 0.0272 = 2.72%

Edit: i think this can be pretty much applied to anything. With adept it's number of adept's wanted for R, and number of attacks for N. Same with criticals.

Edited by kirsche, 02 July 2009 - 10:14 AM.

[Tino]

Wait, she has a 0.8% chance of dying in three rounds, but a 35% chance of dying in four? Ridiculous. I get 2.72%.

(4 choose 3) * (0.2)^3 * (0.8)^1 + (4 choose 4) * (0.2)^4 * (0.8)^0 = 0.0272 = 2.72%

[kirsche]

Wait, she has a 0.8% chance of dying in three rounds, but a 35% chance of dying in four? Ridiculous. I get 2.72%.

It's 0.8%, and I got ~4%. But you're right, my formula was based on a 2RKO, instead of N-2 it should've been N-R.

My bad.

Edited by kirsche, 02 July 2009 - 10:13 AM.

[Mekkah]

I know this, kirsche. The reason why that formula isn't included is because I found I actually didn't grasp the mechanics of it as well as I used to, so as I said in the OP:

Now at the moment I'm getting weird ass results with those, proving that I would fail my high school maths tests if I did them right now, so I'll leave that to someone more savvy at that for now.

that's what I was referring to.

[Ether]

There's one major flaw with your chances of death formula: It only works for the number of rounds that unit is KO'ed in. If Mia was 3RKO'ed at 20% hit and i wanted to know her chances of death in 4 rounds that formula would be useless.

This is the formula, unless I've made a mistake:

P = (Hit^R) + N-R(R+X(Hit^R(Miss^(X+1))))

P = Probability of death
N = Number of rounds of combat
R = Number of rounds to kill
X = Number of bracketed equations to it's left -1. (Does not include the bracketed equation it is in or bracketed equations in other brackets.)

Kirsche,with this particular formula,wouldn't X always equal 0?If so,why not just change the equation accordingly?

[kirsche]

Kirsche,with this particular formula,wouldn't X always equal 0?If so,why not just change the equation accordingly?

It doesn't. Basically, when you multiply something, you just add the number over and over again. Example:

2 x 2 = 2 + 2 = 4
3 x 3 = 3 + 3 + 3 = 9
3 x 4 = 3 + 3 + 3 + 3 = 12
etcetera.

So the formula does this. Say N-R was 2:

(R+X(Hit^R(Miss^(X+1)))) + (R+X(Hit^R(Miss^(X+1))))

Now, the X on the right equals 1. If N-R = 3 then:

(R+X(Hit^R(Miss^(X+1)))) + (R+X(Hit^R(Miss^(X+1)))) + (R+X(Hit^R(Miss^(X+1))))

The fathest X is 2, the middle X is 1 and the X to the left is 0.

And so on.

My maths is only so advanced.

I know this, kirsche. The reason why that formula isn't included is because I found I actually didn't grasp the mechanics of it as well as I used to, so as I said in the OP:

Now at the moment I'm getting weird ass results with those, proving that I would fail my high school maths tests if I did them right now, so I'll leave that to someone more savvy at that for now.

that's what I was referring to.

That's fair enougth, though Binominal stuff is much harder than algebra. So less woudl under stand that than the maount of people who would understand the formula I presented.

Edited by kirsche, 02 July 2009 - 01:18 PM.

[Vykan12]

Let

n= number of attacks a character faces
r= number of attacks needed to land for the character to die
p= probability of each hit landing

Then the probability of a character dying after n hits is:

\sum_{i=r}^{n} (n C i) (p^i) (1-p)^(n-i) (look here)

Example:

Oscar is 3HKOed at 30% true hit. What are his odds of dying in 5?

n= 5
r= 3
p= 0.3 (or 30%)

Therefore Oscar’s chances of dying in 5 hits are:

(5 C 3) (0.3)^3 (0.7)^2 + (5 C 4) (0.3)^4 (0.7)^1 + (5 C 5) (0.3)^5 (0.7)^0 = 16.308%

Of course explaining the combinatorics or even the use of the summation symbol is another story.

Edited by Vykan12, 02 July 2009 - 02:17 PM.

[kirsche]

Hit * Hit * Hit = 0.027
Miss * Hit * Hit * Hit = 0.0189
Hit * Miss * Hit * Hit = 0.0189
Hit * Hit * Miss * Hit = 0.0189
Miss * Miss * Hit * Hit * Hit = 0.01323
Miss * Hit * Miss * Hit * Hit = 0.01323
Miss * Hit * Hit * Miss * Hit = 0.01323
Hit * Miss * Miss * Hit * Hit = 0.01323
Hit * Miss * Hit * Miss * Hit = 0.01323
Hit * Hit * Miss * Miss * Hit = 0.01323

0.07938 + 0.027 + 0.0567 = 0.16308

P = 16.308%

I give up, don't use my formula.

[Vykan12]

Your method is correct, I just made a calculation error somehow.

I don't understand the formula you gave earlier, but I have something similar programmed into my TI calculator.

\sum_{i=1}^{r} (n-i C r-1 ) (p^r) (1-p)^(r-i) (see here, second post)

The idea is to set the hit that kills the character and then determine the amount of combinations where that's possible. By placing that hit, we have 1 less slot and 1 less hit to work with when working out our combinations.

Oh, and maybe a sample calculation is in order.

N=5
R=3
P= 0.3

Answer: (4 C 2) (0.3)^3 (0.7)^2 + (3 C 2) (0.3)^3 (0.7)^1 + (2 C 2) (0.3)^3 (0.7)^0 = 16.308%

Edited by Vykan12, 02 July 2009 - 02:49 PM.

[kirsche]

Your method is correct, I just made a calculation error somehow.

No, I made a mistake with the formula. What's bugging me is the R+X part of teh equation, which just doesn't like fitting. See, after 2 extra rounds after the RK rate (4 rounds if the unit is 2RKO'ed), the term X becomes very annoying.

Basically, the formula only works if the unit is 2RKO'ed or if N-R =1

Edited by kirsche, 02 July 2009 - 02:45 PM.

[Vykan12]

I was referring to this part:

Hit * Hit * Hit = 0.027
Miss * Hit * Hit * Hit = 0.0189
Hit * Miss * Hit * Hit = 0.0189
Hit * Hit * Miss * Hit = 0.0189
Miss * Miss * Hit * Hit * Hit = 0.01323
Miss * Hit * Miss * Hit * Hit = 0.01323
Miss * Hit * Hit * Miss * Hit = 0.01323
Hit * Miss * Miss * Hit * Hit = 0.01323
Hit * Miss * Hit * Miss * Hit = 0.01323
Hit * Hit * Miss * Miss * Hit = 0.01323

That yields the correct answer, though I'm pretty sure your formula doesn't generate that.

P = (Hit^R) + N-R(R+X(Hit^R(Miss^(X+1))))

X = Number of bracketed equations to it's left -1. (Does not include the bracketed equation it is in or bracketed equations in other brackets.)

I do not get how you define X, but it seems you're trying to use that to replace a binomial coefficient.

[kirsche]

I do not get how you define X, but it seems you're trying to use that to replace a binomial coefficient.

Meh, defining X was tricky, but i couldn't think of any other way to do it.

I'm no AS level mathematician.

That yields the correct answer, though I'm pretty sure your formula doesn't generate that.

It doesn't, it only generates the answer if the unit is 2RKO'ed or N-R= <2

Which is why I said don't use the formula.

Edited by kirsche, 02 July 2009 - 02:57 PM.

[Skyward]

Should I use it even though I can't understand it? (except for the hit part.)

[Vykan12]

Do you use computers while fully understanding how they work?

[Tino]

Not quite the same thing. You also don't laugh seriously about a joke you don't understand at all; perhaps you laugh because everyone else laughs, but that doesn't mean you actually get it. A computer doesn't require full understanding of all its functionality; even the pre-installed programs can suffice for some people and they don't need to understand anything more about it.

If he uses all those formulae without having the slightest idea what he's talking about... I can understand if he doesn't use them. It may give him a disadvantage in a debate, perhaps (depends, I guess), but at least he won't break his head over some formula while debating here is still mainly supposed to be fun.

Edited by Tino, 03 July 2009 - 03:45 AM.