Supposing I were TWFing a pair of keen kukris and had five attacks per round, the odds of any one attack being a confirmed crit are [hit-chance] x [threat-chance], or, if I hit on anything but a 1, (.95)(.3) =.285
That's relatively straight forward. --But is there for DPR formula for calculating multiple crits? I.e., the odds of getting two, or three? (The formula will also have account for declining chances over worse iteratives.)
If you want to calculate DPR, you calculate DPR for each attack separately (remembering about the declining hit chance) and add them up together.
If you're just interested of a chance of 2 or 3 critical hits (which isn't necessary to calculate DPR, unless you have some specific effect that occur only when you score two or more critical hits in a round), you can do as follows. Let's assume you have N attacks, and Pk is the probability of a critical hit on the k-th attack. Let Xk=Pk/(1-Pk).
It can be proven the probability of getting exactly 2 critical hits (no more, no less) is equal to
((X1+...+XN)^2-(X1^2+...+XN^2))/2 * (1-P1)(1-P2)*...*(1-PN)
For exactly 3 critical hits the probability is
((X1+...+XN)^3-3*(X1^2+...+XN^2)(X1+...+XN)+2*(X1^3+...+XN^3))/6 * (1-P1)(1-P2)*...*(1-PN)
If you're just interested of a chance of 2 or 3 critical hits (which isn't necessary to calculate DPR, unless you have some specific effect that occur only when you score two or more critical hits in a round....
What's going on is I have a mounted build sharing the teamwork feats Outflank, Pack Flanking, and Paired Opportunists, and a crit by one of the team will trigger an Outflank AoO from the other and then a Paired Opportunist AoO by the first. The math gets quite complex (at least it seems that way to me) due to a hard limit of AoOs per round, as well as the stricture that you can't earn more than one AoO for the same effect (so a crit generating an AoO due to Outflank which then results in another crit can't generate yet another AoO again due to Outflank, at least for that initially triggering event). So the proper DPR calc not only needs to account for available AoO slots, but also take into account sequential crits being ineligible for generating AoOs.
The math gets quite complex (at least it seems that way to me) due to a hard limit of AoOs per round, as well as the stricture that you can't earn more than one AoO for the same effect (so a crit generating an AoO due to Outflank which then results in another crit can't generate yet another AoO again due to Outflank, at least for that initially triggering event).
I'm very certain that this sentence:
This does not allow you to take more than one attack of opportunity against a creature for a given action.
is only there to prevent you from getting two AoO's from an action that would already give you an AoO normally.
So you would not get two AoO's from tripping someone and having Vicious Stomp and Paired Opportunists.Each critical hit in a Outflank sequence should provoke another AoO from both you and your ally.
Well, you're quoting the text of Paired Opportunists, not Outflank.
Vicious Stomp and the Fortuitous weapon property (et al) are cases where it's possible to get multiple AoOs off the same trigger effect. Ouflank, Paired Opportunists, and "Greater" maneuvers permit allies to take AoOs. Paired Opportunists explicitly proscribes reverberation between allies with that same shared teamwork feat.
So you would not get two AoO's from tripping someone and having Vicious Stomp and Paired Opportunists.
I don't have Vicious Stomp in the build, but I do have Greater Trip. (So a trip by me will trigger an AoO from the mount, and then I'll receive an AoO from shared Paired Opportunists because an adjacent ally took an AoO. Whether the mount then gets another AoO from their Paired Opportunists that was triggered by my Paired Opportunist AoO is another matter.)
Anyway, back to the matter at hand:
If you're just interested of a chance of 2 or 3 critical hits (which isn't necessary to calculate DPR, unless you have some specific effect that occur only when you score two or more critical hits in a round), you can do as follows. Let's assume you have N attacks, and Pk is the probability of a critical hit on the k-th attack. Let Xk=Pk/(1-Pk).
It can be proven the probability of getting exactly 2 critical hits (no more, no less) is equal to
((X1+...+XN)^2-(X1^2+...+XN^2))/2 * (1-P1)(1-P2)*...*(1-PN)
For exactly 3 critical hits the probability is
((X1+...+XN)^3-3*(X1^2+...+XN^2)(X1+...+XN)+2*(X1^3+...+XN^3))/6 * (1-P1)(1-P2)*...*(1-PN)
Just to be clear, Pk is the probability of a critical hit (not a threat), correct? (I.e., .285 if one is 95% to hit with a keen kukri, not .3)
...correct?
Just to be clear, Pk is the probability of a critical hit (not a threat), correct? (I.e., .285 if one is 95% to hit with a keen kukri, not .3)
...correct?
Correct, for the first hit. Subesequent attacks may have lower Pk.
It can be proven the probability of getting exactly 2 critical hits (no more, no less) is equal to
((X1+...+XN)^2-(X1^2+...+XN^2))/2 * (1-P1)(1-P2)*...*(1-PN)
For exactly 3 critical hits the probability is...
Here's my problem: it's the Outflank feat, in which crits trigger AoOs.
With a given situation of 5 attacks total and 2 unfulfilled AoOs available for any crits to trigger.... --I might get no crits, in which case nothing happens, or one or two crits, in which case one or two AoOs occur, or more than two crits, which will not result in additional damage, as all AoOs have already been taken.
The term "exactly" is what is throwing me off, because there could be up to five crits, but at most two of them matter.
I am supposing that I require formulas for....
1. {probability of exactly 1 crit}
...and....
2. {probability of 2 or more crits} (...which I assume is identical to {probability of any number of crits} minus {probability of exactly 1 crit} ...?)
--Is there an easier way to express that second formula without longhanding every single Nth constituent component separately?
1. {probability of exactly 1 crit}
...and....
2. {probability of 2 or more crits} (...which I assume is identical to {probability of any number of crits} minus {probability of exactly 1 crit} ...?)
Yes, the formula you give is correct, and it's equivalent to
{Probability of 2 or more crits}=1-{Probability of exactly 1 crit}-{Probability of no crits}
where
{Probability of no crits}=(1-P1)*(1-P2)*...*(1-PN)
and
{Probability of exactly 1 crit} = (X1+X2+...+XN)*(1-P1)*(1-P2)*...*(1-PN)
so
{Probability of 2 or more crits} = 1-(1+X1+X2+...+XN)*(1-P1)*(1-P2)*...*(1-PN)
Is there an applicable on-line calculator out there anywhere?
I'd feel like Ren trying to smash off the Happy Helmet trying to do all of that longhand, especially for a TWF winding up into blend mode with a half-down maneuvers and teamwork feats.