multiple % chances to inflict different statuses - Calculation

[murkly gurgles]

So I have a bow with:
20% chance to cause bleeding
20% chance to cause atrophy
20% chance to cause crippled

Of course all 3 of these are blocked by physical armor, so I removed some armor on my test subject so the bow's base damage would always clear more than enough to allow the status effects chance to land.

I spent about 30min shooting at my group member, testing this to see if I could figure out how these % chances worked together. I never once saw all 3 statuses land together. The odd time I'd see bleed+cripple or bleed+atrophy at the same time, but never all 3. Is it impossible for 1 weapon to land 3 separate status effects or is there some kind of mechanic with these particular statuses that would disallow them to stack all at once?

For the 20% chances, I don't think each shot has 3 separate instances of a 20% chance (which would basically kinda mean 60% that at least 1 of the 3 would proc), its more likely calculated differently. PLEASE correct me if I'm wrong, but I think its more like 20% chance for ONE of the 3 statuses to land, then if that happens, THEN there's a 20% an additional status will land, and in that rare case, there should be a further unlikely but possible chance for that extra 20% chance of a 3rd - but as stated, I've never seen that.

I really have no idea how the actual calculation works for this, please enlighten me if you know, thanks!

[Bunny Girl Senpai]

I think your right. But to me its more of a question of what is the probability of more than one status at a given time. The chance to land one status is 20%. Then to land 2 statuses is less than 20%. Then to land 3 statuses, it is even less of a chance.

What do you think?


BTW that bow sounds awesome

[Ashera]

I think it's more like, 1/5 attacks will land you with 1 of the three effects.
1/10 attacks will land you with 2 of the effects.
1/15 of the attacks will land you with all 3.

You just have bad luck and haven't seen the 3 yet.

having 3 effects doesn't mean it is 20% + 20% + 20% = 60% chance to land an effect.
It is still 20% chance to land an effect. 10% for 2. 5% for 3.

I'm bad with math... so probably safer to disregard my opinion.

[Avstah]

I think your chance of landing all three is basically 3 individual rolls at 20% or 1/5, which is 1 in 5^3, which is a 1/125 chance of all three happening. Maybe some stats nerds will arrive and help.

[Ashera]

WTB real nerd who can show us how statistics works.

[murkly gurgles]

Well one conclusion is that given the crazy rarity of 3 physical statuses landing at once (I spent an additional 20min firing away and again, still haven't seen 3 statuses at once), any weapon with over 3 status % chances can be classified as having a redundancy - ESPECIALLY if one of those statuses is resisted by the opposite type of damage you are doing (magic or physical).

The problem is that those extra status effects are taking the place of other potential bonuses. The power on the weapon would be far better distributed if instead of a 3rd % status effect, you choose a weapon one with 1 or 2(max) % statuses and higher power in other areas. Things like additional attributes, an extra rune spot, +% to crit etc. would be much more value than a 3rd status effect.

I would say for example if you're running a heavy physical dps oriented group, 2 physical based effects on a weapon are still good and not hogging too much space in the item's bonus power distribution, because the chances of 2 landing at the same time are actually reasonable.


**** TL DR ----- >>
While 2 status effect %'s can be really good, 1 single % status effect (in line with your damage type) and higher bonuses in other important areas is probably the most optimal imo. (having 3 seems kinda wasteful)

[Ptolemy]

Nerd here.
You have a 38% chance to land one effect, a 10% chance to land two, and a 1% chance to land all 3 for a total of 49% chance to land at least one.

Assuming the effects are statistically independent

Edit:
Math: odds of only 1 effect = odds that one specific effect succeeds and the other 2 fail * number of ways that can happen (3 since we have 3 combinations: success fail fail, fsf, and ffs) = .2*.8*.8*3 = .384

Odds of 2 = .2*.2*.8*3 = .096
The 3 here is (ssf,sfs,fss)

Odds of 3 = .2*.2*.2= .008
Only one combination here (sss)

[murkly gurgles]

Originally posted by Ptolemy:

Nerd here.
You have a 38% chance to land one effect, a 10% chance to land two, and a 1% chance to land all 3 for a total of 49% chance to land at least one.

Assuming the effects are statistically independent

Edit:
Math: odds of only 1 effect = odds that one specific effect succeeds and the other 2 fail * number of ways that can happen (3 since we have 3 combinations: success fail fail, fsf, and ffs) = .2*.8*.8*3 = .384

Odds of 2 = .2*.2*.8*3 = .096
The 3 here is (ssf,sfs,fss)

Odds of 3 = .2*.2*.2= .008
Only one combination here (sss)

Ptolemy, you're the man! Thx!

[Full_Regalia]

Originally posted by Ptolemy:

Nerd here.
You have a 38% chance to land one effect, a 10% chance to land two, and a 1% chance to land all 3 for a total of 49% chance to land at least one.

Assuming the effects are statistically independent

Edit:
Math: odds of only 1 effect = odds that one specific effect succeeds and the other 2 fail * number of ways that can happen (3 since we have 3 combinations: success fail fail, fsf, and ffs) = .2*.8*.8*3 = .384

Odds of 2 = .2*.2*.8*3 = .096
The 3 here is (ssf,sfs,fss)

Odds of 3 = .2*.2*.2= .008
Only one combination here (sss)

You’re a godwoken! Did you use source to figure this one out?

[Avstah]

So I was right eh?

.008 = 1/125

[Muted Kobold]

Mathmatically speaking. Your chances for all three effects to land at the same time are incredibly low. You have a decent chance of having a permutation of two where the order does not matter. And an even better chance for just one.

This seems like normal behavior to me.

[murkly gurgles]

So when considering power bonuses distributed on an item, we could all agree that having 3 total status %chance effects is rather useless, when that effect could otherwise be distributed as a useful "+"(to relevant attribute/skill/rune-slot/etc).

Question now, how do the actual chances work for 2 different possible status effects (keeping 20% each on the tooltip) on a weapon? Should be a pretty simple formula, but I'm terrible at math lol

[Ptolemy]

No worries.
All these are statistically independent meaning the success or failure of one doesn’t affect the odds of the other succeeding.
Which means we can use the formula:
For events A and B
Probability (A and B) = proability(A) * probability(B)

We also need to know that if we have Prob(A), then Prob(not A) = 1-Prob(A)

So we just need to 1) find the different scenarios we classify as success, 2) find the odds of those scenarios 3) add those together.
1)
So for any 1 status applying, the two scenarios are stat 1 succeeds and stat 2 fails, or stat 2 succeeds and stat 1 fails.
2)
The odds that stat 1 succeeds is .2
But we also need after that for the other to fail, so we multiply it by (1-.2) (the odds that the other stat fails)
So stat 1 succeeding alone happens 16% of the time,
And the same can be said for stat 2.
3)
16% + 16% = 32%

Odds of both:
.2*.2= .04

Odds of neither:
.8*.8= .64

And we can see that we’ve covered all of the possible scenarios because:
.64 + .04 + .32 = 1

Everything checks out.

[wendigo211]

It's a binomial distribution. The probability, p(k,n) of getting k of n effects to apply with a probability of success, q, is:
p(k,n)=n!/k!/(n-k)!*q^k*(1-q)^(n-k).
If we have different probabilities for each effect, then we have to consider individual cases. However, we can use the bionomial distribution to model the outcomes as long as:

1. There are only two outcomes for each effect (success/failure)
   
2. The result of current effect is not contingent on the result of the previous effect (statistical indepdence)
   
3. The probability of the outcomes are the same for each effect (identical distributions)

See, high school probability is useful for something after all.

[jdash417]

Originally posted by wendigo211:

It's a binomial distribution. The probability, p(k,n) of getting k of n effects to apply with a probability of success, q, is:
p(k,n)=n!/k!/(n-k)!*q^k*(1-q)^(n-k).
If we have different probabilities for each effect, then we have to consider individual cases. However, we can use the bionomial distribution to model the outcomes as long as:

1. There are only two outcomes for each effect (success/failure)
   
2. The result of current effect is not contingent on the result of the previous effect (statistical indepdence)
   
3. The probability of the outcomes are the same for each effect (identical distributions)

See, high school probability is useful for something after all.

Dude that is one heluva high school, I am thinking of going back to csu and getting my money back .

[Ptolemy]

Originally posted by jdash417:

Originally posted by wendigo211:

It's a binomial distribution. The probability, p(k,n) of getting k of n effects to apply with a probability of success, q, is:
p(k,n)=n!/k!/(n-k)!*q^k*(1-q)^(n-k).
If we have different probabilities for each effect, then we have to consider individual cases. However, we can use the bionomial distribution to model the outcomes as long as:

1. There are only two outcomes for each effect (success/failure)
   
2. The result of current effect is not contingent on the result of the previous effect (statistical indepdence)
   
3. The probability of the outcomes are the same for each effect (identical distributions)

See, high school probability is useful for something after all.

Dude that is one heluva high school, I am thinking of going back to csu and getting my money back .

Sometimes math has a way of making stuff sound a lot scarier than it is.

n!/k!/(n-k!)
is just a nice way to find the number of combinations of events that result in n successes out of k events.

Then it’s just
Number of combinations * probability of the combination.

The probability of the combination is (probably of success ^ # of successes) * (probability of failure ^ number of failures)