Multistrike Logic

[FaFiend 0]

Hi,

 

Edit:

Ok, I found out what was wrong -> See my last post (post #7)

End Edit.

 

I'm sorry if this is the wrong subforum - I thought about putting it into general discussion, but then decided to put it here, because the multistrike stat was introduced with WoD.

 

I'm writing because I have some problems with multistrike.

Reading the tooltip, I worked out how it should work in theory, and I came up with the exact same things as Zagam in this thread:

https://www.icy-veins.com/forums/topic/6044-warlords-alpha-stat-summary-with-mathematical-analysis/

 

However, now that WoD is live, I actually wanted to test if the amount of multistrikes that I get is consistent with what is predicted by the model (which is two independent rolls with a probability equal to the multistrike chance).

 

Grabbing all mulstistrike gear I currently have, I made it to 8.21% multistrike on my feral druid.

Going to the training dummy in the garrison (which is a dungeon dummy, level 102 - if that makes any difference, don't know yet), I switched on autohit and accumulated a bit of statistics. Here are the results:

 

Hits.....:  2732

Crits....: 1121

Multis...: 441

 

The data was taken with the 'skada' AddOn. I checked that multistrike critical hits are actually counted as Crits and Multistrikes, so the numbers should be correct.

Thus, the relative amount of multistrikes is 441/(2732+1121) = 11.45%

Compared to what is expected from two independent chances of 8.21%, this is too low - from 2 times 8.21% I expect:

Chance for 0 Multistrikes: p(0) = 0,918*0,918 = 84.3%

Chance for 1 Multistrike..: p(1) = 2*(0,0821*0,918) = 15.06%

Chance for 2 Multistrikes: p(2) = 0,0821*0,0821 = 0.67%

So the amount of multistrikes should be 15.73%, shouldn't it?

From this I would expect 606 Multistrikes, and not 441...

 

Did anyone else do some tests on this?

I know I'm missing the statistical error, but it should not be very large, given the close to 4000 trials.

 

If anyone has a clue of what's going on (or what I am doing wrong here), please let me know.

 

Thanks & Cheers

Edited November 19, 2014 by FaFiend

[Ratanael 0]

I think you shouldn't consider each multistrike to be independent, but that if you fail the first one, you have no chance to trigger the second.

So instead of 2*(0,0821*0,918) for Multistrike = 1, you have only 0.0821*0.918,

 

So the calculation is: 7,53%+0,67%, which actually makes 8,2% of multistrike. (0.01% missing from not exact values)

 

And to reach 100%, just consider the event (non multistrike then multistrike) to be like no multistrike at all.

 

And 4000 is maybe not big enough ... who knows !

Edited November 19, 2014 by Ratanael

[Orthios 271]

I think you shouldn't consider each multistrike to be independent, but that if you fail the first one, you have no chance to trigger the second.

So instead of 2*(0,0821*0,918) for Multistrike = 1, you have only 0.0821*0.918,

 

So the calculation is: 7,53%+0,67%, which actually makes 8,2% of multistrike. (0.01% missing from not exact values)

 

And to reach 100%, just consider the event (non multistrike then multistrike) to be like no multistrike at all.

 

And 4000 is maybe not big enough ... who knows !

Not quite.  I've been meaning to do a write up on all of the RNG factors in WoW and how they effect you.  Every attack has a RNG roll for whether or not it crits, whether or not it has 0, 1, or 2 multistrikes, and whether or not those multistrikes crit (Yes this is a separate roll, based on my observations with MSBT), all based on their respective values.

 

With the OP's 8.21% MS chance, every attack has 2 8.21% MS rolls.  This means a RNG generates a number between 1 and 10000, and if that number is less than 821, then the MS rolls true, and you get a multistrike.  This happens twice.  A crit roll works the same way.  This means that the OP should expect to have 606 MS events out of the 3853 attacks (hits + crits), however, RNG is RNG, so I would say that this is within a suitable range to expect.  I'm not sure how large of a sample you would need for the law of large numbers to kick in, but I would imagine that with a smaller % chance for an event to be considered true (if you are looking for a true, that is), you would need a larger sample size.

[FaFiend 0]

Hi,

 

This means that the OP should expect to have 606 MS events out of the 3853 attacks (hits + crits), however, RNG is RNG, so I would say that this is within a suitable range to expect.

 

Ok, let's try to go through this

The sample size is N=2732+1121=3853.

The probability to multistrike (assuming two independent rolls with 8.21% each) is 15.7%.

The expected number of Multistrikes M is M=3853*15.73%=606.

The binomial error on the expected amount of multistrikes is therefore:

Delta(M)= SQRT(N * p * (1-p) ) = 22.6

 

Note that I actually don't know if this calculation is correct.

It is correct for one chance (binomial error), but I don't know if it works in the case of multistrike.

Thus, (assuming this is not too far off), the expectation in this sample is to have

M_Expected = 606 +/- 22.6

 

The deviation from the observation is (606-441)/22.6 = 7.3 times the expected statistical error, which is extremely unlikely.

 

 

I think you shouldn't consider each multistrike to be independent, but that if you fail the first one, you have no chance to trigger the second.

So instead of 2*(0,0821*0,918) for Multistrike = 1, you have only 0.0821*0.918,

 

So the calculation is: 7,53%+0,67%, which actually makes 8,2% of multistrike. (0.01% missing from not exact values)

 

If this is true (the second one only has a chance to occur if the first occured), then the probabilities for multistrikes will be:

p(0) = 0,918 = 91.8% (fixed thanks to Ratanael)

p(1) = 0,0821 = 8,21% (only one roll to check whether the first one occurs)

p(2) = 0,0821*0,0821 = 0,67%

 

i.e. the chance to multistrike is p = 8,88%, leading to an expected 342 Multistrikes, which is less than observed.

If the probabilities are calculated like you say, then the expectation is 316.

 

The uncertainties here are (taking again the total chance as the probability):

Delta(M, p=8,88%) = 17.7

Delta(M, p=8,21%) = 17.0

 

Here, the observations are also off by more than 7 times the statistical uncertainty (assuming the uncertainty was calculated correctly).

 

__________

 

Can someone please check whether the error is reasonable?

Also, does anyone observe way more mulstistrikes than expected?

If it is really a statistical fluctuation, it should be equally likely to observe more than expected as it is to observe less than expected.

 

I think I will run another test, and yet another one on a level 60 dummy... just to make sure

 

Edit 1:

In another test (again with the 8.21% on a level 102 dummy) I get the following:

Hits: 4001

Crits: 1559

Observed Multistrikes: 656

-> Sum Trials: 5560

-> Multistrike chance: 11,8%

 

Expected Multistrike Chance: 15.73%

Expected Multistrikes: 875 +/- 23

Error obtained in the same way as described above (i.e. not sure if it is correct).

 

However, the total sample (of a 8.21% chance on a level 102 dummy) now has 9413 Trials, in which 1097 multistrikes were counted,

The multistrike fraction is thus 11.65%.

Expected were 15.73%, i.e. 1482 multistrikes, if the model of two independent checks was correct.

 

Edit 2:

I did a smaller test on a level 60 dummy in Stormwind.

I get the following:

 

Hits: 707

Crits: 294

Observed Multistrikes: 113

-> Sum Trials: 1001

-> Multistrike chance: 11,3%

 

Expected Multistrike Chance: 15.73%

Expected Multistrikes: 158 +/- 9.7

 

Looks like the target level has no effect.

 

End Edit.

 

 

Cheers

Edited November 19, 2014 by FaFiend

[Ratanael 0]

If this is true, then the probabilities for multistrikes will be:

p(0) = 0,918*0,918 = 84.3%

p(1) = 0,0821 = 8,21% (only one roll to check whether the first one occurs)

p(2) = 0,0821*0,0821 = 0,67%

 

i.e. the chance to multistrike is p = 8,88%

 

 

 

you should consider for p(1) the second roll too, no ? I don't have that much memories from school. P(1) is the chance to get only one multistrike, and not to land a multistrike, not caring about what happend behind, neither to get at least one multistrike, then the formula would be 1-p(0)

 

Still, i don't know how it works, i never really get interested in such things.

Edited November 19, 2014 by Ratanael

[FaFiend 0]

you should consider for p(1) the second roll too, no ?

 

Ah, sorry. That's a copy/paste error.

If the second one cannot proc if the first one didn't proc, then p(0) should be simply 1-p, i.e. 0,918 in this case.

But bear in mind that these are the numbers for the model that says that the second strike can only go off if a first multistrike fired.

The calculations for the two independent rolls are in the very first post of this thread.

Thanks for spotting this though, I've fixed it :)

 

 

I just did a smaller test on a level 60 dummy and edited it into the post above.

[FaFiend 0]

Ok, I found the problem...

 

Turns out that Skada actually counts multistrikes as multistrike and as hit or crit.

Thus, the number of hits and crits contain the multistrikes, and building the ratio of those numbers leads to wrong results... silly me

 

After yet another test, for the 8.21% multistrike, i.e. 15.7% expected amount of multistrikes I get:

 

Number of Hits: 1300

Number of Crits: 409

Sum of Hits and Crits: 1709

Number of Multistrikes in there: 229

 

Attacks without multistrike: 1480

Multistrike percentage: 229/1480 = 15.47%

 

Expected from model: 233, i.e. 15.74%

Error on expectation: 14, i.e. 0.9%

 

Thus, I'd say that multistrike indeed works as figured out by Zagam (and others)

 

Sorry for the spam

But now I can go to bed happily, knowing that this is solved

 

Thanks & cheers!