ORKO chance / Risk of Death calculations

[Espinosa]

A large part of my life has passed since I last studied math (though I occasionally tutor it to my siblings when they're in trouble and I pretty much need to teach it to myself from zero before that since nobody else can do that; luckily that takes little time) so I'm no longer sure if I ever knew the means to calculate some of the key issues we tend to discuss from time to time here.

On a basic level, let's say Rutger has 56 true hit on an enemy whom he 2HKOs with 0 crit. This one I can count easily enough, 0.56 x 0.56 is 0.3136 or roughly 31% chance to ORKO.

Then we have the same situation, but with Rutger having 7 crit. He still has his 31% chance to ORKO but if we suppose one of the hits whiffs, the other could crit and so he wouldn't need both the hits to land in order to remove the boss off the field. I could be doing this wrong but let me have a go: a hit needs to connect before it crits, so 0.56 x 0.07 = 0.0392 or roughly 4% chance to OHKO the boss on any of the hits. That's 96% of not critkilling. 0.96^2 = 0.9216, so there's roughly an 8% chance that at least one of the hits connects and crits. Not sure what to do with this 8%, add it to 31%? Or find the chance of failure and then subtract the 8 from it?

A related issue - there's a 30% chance of A happening, and a 20% chance of B taking place. 6% of both A and B occurring. But let's say we also have C that occurs with the probability of 40%, and we want to know the probability of _at least_ ANY two of the three taking place. How do I do something like that? I've asked this question to friends who are former economists and college teachers (without the Rutger bit of course, not to confuse them) and they had no answer.

Finally, the RoD issue. Let's say Vanessa has 50 HP. We have 5 guys who hit her for 15 damage with a 20 true hit and 10 guys who hit her for 5 damage with 50 hit. How likely is she to die? No crits possible. How do we count this?

[Quintessence]

Really interesting, I'd like to know the answer too.

Btw, does this have to do with Bayes' Theorem?

[charlie_]

A related issue - there's a 30% chance of A happening, and a 20% chance of B taking place. 6% of both A and B occurring. But let's say we also have C that occurs with the probability of 40%, and we want to know the probability of _at least_ ANY two of the three taking place. How do I do something like that? I've asked this question to friends who are former economists and college teachers (without the Rutger bit of course, not to confuse them) and they had no answer.

Too lazy to do the other ones but here.

Those are some really bad college teachers.

[shadykid]

Those are some really bad college teachers.

yeah, no kidding....

[Espinosa]

Too lazy to do the other ones but here.

Those are some really bad college teachers.

Can you briefly explain what you're doing there? I fail to identify two of the numbers and that also gets in the way.

[charlie_]

Basically, at the first probability is 3/10 of success (Y) and 7/10 of failure (N). From each of those you branch off the second, which has a 2/10 chance of succeeding and a 8/10 chance of failing. You do the same thing with the third.

Then you multiply all of the fractions in the path (3x2x4=24 for example) and work out which ones you need (the ones with 2 or 3 Ys, in this case) and add them all together (24+36+96+56=212), which would make it 212/1000 and 21.2%.

Alternatively you decide I suck at explaining and google Bayes' Theorem.

[Redwall]

Then we have the same situation, but with Rutger having 7 crit. He still has his 31% chance to ORKO but if we suppose one of the hits whiffs, the other could crit and so he wouldn't need both the hits to land in order to remove the boss off the field. I could be doing this wrong but let me have a go: a hit needs to connect before it crits, so 0.56 x 0.07 = 0.0392 or roughly 4% chance to OHKO the boss on any of the hits. That's 96% of not critkilling. 0.96^2 = 0.9216, so there's roughly an 8% chance that at least one of the hits connects and crits. Not sure what to do with this 8%, add it to 31%? Or find the chance of failure and then subtract the 8 from it?

There are several ways to kill.

Rutger hits twice but doesn't crit: (0.56*0.93)*(0.56*0.93)

Rutger crits on the first hit: 0.56*0.07

Rutger misses the first hit, but crits on the second hit: 0.44*0.56*0.07

Rutger hits the first time (no crit) and then hits the second time (yes crit): (0.56*0.93)*(0.56*0.07)

All four of these events are non-overlapping; hence, you just add them up to get the probability that Rutger ORKOs. You can also find the odds that Rutger *doesn't* kill and subtract this from 1.

Note that events 1 and 4 can be combined into a single event, that of Rutger hitting twice.

A related issue - there's a 30% chance of A happening, and a 20% chance of B taking place. 6% of both A and B occurring. But let's say we also have C that occurs with the probability of 40%, and we want to know the probability of _at least_ ANY two of the three taking place. How do I do something like that? I've asked this question to friends who are former economists and college teachers (without the Rutger bit of course, not to confuse them) and they had no answer.

I'm confused by your presentation of the problem. Are A, B, and C each independent events?

Finally, the RoD issue. Let's say Vanessa has 50 HP. We have 5 guys who hit her for 15 damage with a 20 true hit and 10 guys who hit her for 5 damage with 50 hit. How likely is she to die? No crits possible. How do we count this?

For this specific problem, you need a joint binomial distribution (read about the binomial distribution on wikipedia for a more detailed explanation). Joint binomial calculations are annoying to do by hand--I think you generally will need to write a short computer code--but regular binomial calcs can be done easily in Excel.

I'll do a simplified version for you. Suppose Vanessa has 50 HP, and there exist five enemies who hit her for 15 damage each, and 20 true hit each. Vanessa can survive one hit, two hits, and three hits. In Excel, simply enter the command '=BINOMDIST(3,5,0.2,1)'. This command basically does the following.

The prob of seeing one hit connect is C(5,1)*0.2*0.8^4

The prob of seeing two hits connect is C(5,2)*0.2^2 * 0.8^3

The prob of seeing three hits connect is C(5,3)*0.2^3 * 0.8^2

The final argument in BINOMDIST is set to 1 in order to add up these three possibilities. You can also use the binomial distribution to calculate the odds of reaching stat benchmarks.

Edited September 23, 2013 by Nessie

[Espinosa]

Bayes' Theorem, joint binomial distribution, BINOMDIST... My generation was not taught any of this lol. What age group/major (if any at all) learns this nowadays?

I got the explanations and should be able to reproduce this with different numbers unless complexity takes a leap (like with the joint thing that needs a bit more research). Many thanks everyone.

[Quintessence]

Well, I have 20 years and was taught this (Bayes' and Probability Distributions) last semester (1st sem of 2013) in Statistics I. Now, in Statistics II, I'm learning Normal Distribution/Chi-Squared, etc.

[shadykid]

I, uh....did the above examples by hand

[dondon151]

a lot of this stuff just involves a bit of brute forcing and thinking around various outcomes. i think nowadays, most people learn probability in their stats class in high school or college. i didn't take stats in high school, so i had to learn probability theory independently for math team problems.

[BBM]

Bayes' Theorem, joint binomial distribution, BINOMDIST... My generation was not taught any of this lol. What age group/major (if any at all) learns this nowadays?

I got the explanations and should be able to reproduce this with different numbers unless complexity takes a leap (like with the joint thing that needs a bit more research). Many thanks everyone.

Didn't learn joint binomial distribution, but I learnt Bayes' Theorem and regular binomial distribution in high school, and it should all be taught in any stats class in university.

[Meteor]

There are several ways to kill.

Rutger hits twice but doesn't crit: (0.56*0.93)*(0.56*0.93)

Rutger crits on the first hit: 0.56*0.07

Rutger misses the first hit, but crits on the second hit: 0.44*0.56*0.07

Rutger hits the first time (no crit) and then hits the second time (yes crit): (0.56*0.93)*(0.56*0.07)

All four of these events are non-overlapping; hence, you just add them up to get the probability that Rutger ORKOs. You can also find the odds that Rutger *doesn't* kill and subtract this from 1.

Note that events 1 and 4 can be combined into a single event, that of Rutger hitting twice.

That is correct, although as a programer, I like to give the variables names. Makes it easier for me anyway. Like so:

hit  = 0.56;
miss = 1 - hit;
crit = .07;
norm = 1 - crit;

lethal = (hit * crit) +          # first hit crits
         (hit * norm * hit) +    # two hits
         (miss * hit * crit);    # miss then crit

#=> .348096

# also...

CritHit = (hit * crit);
NormHit = (hit * norm);
AnyHit  = hit;

lethal = CritHit + (NormHit * AnyHit) + (miss * CritHit);

Edited September 26, 2013 by Meteor