Is there a universal "occasionally attacks twice" rate for Magian weapons?
Appealing to Occam's razor, one could assert that the lack of duplicate entries for "occasionally attacks twice" (OAT) in the .DATs means that all the Magian weapons with that trait have the same OAT rate. While I do not know whether that assertion is actually true, at least I can look at various pieces of published evidence to see if this notion of a universal rate has any traction.The track record of English-language FFXI sources is dismal: there is exactly one forum post (that I am aware of) that presents any data that can be used to estimate the OAT rate. The "general consensus" is that it's 40%. (By comparison, the Joyeuse rate is 45%.) As far as I know, this is the only evidence that English-language users cite or allude to when making claims about the Magian OAT rate, which is pathetic, but in line with the natural incuriosity of the FFXI sheep.
Another data set that someone shared with me, concerning the OAT rate of a Magian great axe (Luchtaine) with a 19% base double attack rate, showed 344 double attacks out of 689 total attack rounds. Based on these counts, an interval estimate of the Magian OAT rate (given 95% confidence) is (.3357284, .4279119), which is also consistent with the idea of a 40% OAT rate. (Reasoning leading to the OAT rate estimation is similar to that for virtue weapons I discussed previously.)
Among Japanese sources, there is more data but an annoying lack of statistical consistency, if this one blog post is to be taken as a summary of all pieces of evidence regarding the Magian OAT rate. They can be grouped into two categories: evidence consistent with a 40% rate and evidence consistent with a rate higher than 40% but lower than 50% ("statistically significantly," what a gauche phrase). The foolish conclusion that the OAT rate is 43.75%, based on an idiotic pooling of the data, has no traction.
The attack distribution of "occasionally attacks 2-3 times" Magian weapons
Given the above discussion of the lack of reliable information on the Magian OAT rate, the prospect of getting reliable data concerning the attack distribution of Magian weapons that "occasionally attack 2-3 times" (OA2-3T) appears poor. In fact, there is one set of count data (source) for Magian OA2-3T hand-to-hand with MNK (whatever its actual name is for the weapon, I don't give a fuck) that can shed light on the matter, but it can only do so provided that the attack distribution associated with OA2-3T is the same for all Magian weapons and the data are actually credible. The counts are as follows (302 total):2 attacks: 57
3 attacks: 98
4 attacks: 81
5 attacks: 48
6 attacks: 16
7 attacks: 2
In order to obtain estimates of the attack distribution probabilities for Magian OA2-3T, a probability model needs to be specified and estimation based on this model.
Let H denote the number of attacks in a given attack round. Let πn denote the probability of n = 1, 2, 3 attacks of a single hand in an attack round, and that the sum of the probabilities equals 1, and also let k denote the probability of a kick attack in an attack round. Provided that the number of attacks of one hand, the number of attacks of the other hand, and the number of kick attacks (all in a given attack round) are mutually independent, the probability mass function of H is
and 0 otherwise.
Aside: "Why do you care about kicks?" is a valid question. The answer is that the data were collected with a parser. Just as WAR cannot have a 0% double attack rate, MNK cannot have a 0% kick attack rate, and kparser cannot make the distinction between a kick and a punch (nor should we expect that kind of distinction to be made). Surely, a person can tell the difference, but why would you expect anyone to count manually when a parser is available? The occurrence of kicks does not provide any useful information about the attack distribution of an OA2-3T weapon (but can help validate the probability model), so all kicks do is introduce undesirable variability to the proceedings, but you can't do anything about it (other than get the data using PUP).
With the above data and probability model, maximum likelihood estimation can proceed. Of immediate concern is whether to assume that the kick attack rate, given 5/5 Kick Attack merits, is actually 17.5%. (Of course, I could let the kick attack rate be yet another parameter to estimate, but estimating four parameters with a sample size of 302 is not really that helpful.) People who play monk are generally fucking retarded, but I'll just use that rate. Using numerical methods, a set of point estimates and 95% simultaneous confidence intervals (Bonferroni, too lazy to care about other methods) is generated:
p.hat ci.lower ci.upper
[1,] 0.4795746 0.4160971 0.5430521
[2,] 0.3377920 0.2492500 0.4263339
[3,] 0.1826334 0.1283014 0.2369655
Assuming the 17.5% kick attack rate is valid (the weakest assumption by far in my view, to go along with all the other assumptions upon which the analysis is based), the probability distribution of attacks for Magian OA2-3T is obviously not the same as that for the likes of Ridill, Mercurial Kris, and Soboro Sukehiro. The alleged 30:50:20 ratio for 1-3 attacks obviously does not agree with the data (and the corresponding estimates). Given the data, the multi-attack probability (including 2 and 3 attacks) could be 1/2, partitioning to 3/10 for two attacks and 1/5 for three attacks.
To put it another way, the ratio of 1-3 attacks could be 50:30:20 for Magian "occasionally attacks 2-3 times" weapons (generalizing from hand-to-hand to all weapons), and that's what I'll stand by until other data persuasively rejects that working hypothesis.
Addendum: numerical maximum likelihood estimation
Analytical MLE for the above case is a complete waste of time if it is even possible, so I tapped out an R script for the purposes of numerical estimation.ll <- function(p,X,k) {
X2 = X[1]; X3 = X[2]; X4 = X[3]; X5 = X[4]; X6 = X[5]; X7 = X[6]
p1 = p[1]; p2 = p[2]
ll = -(X2*log(p1*p1*(1-k)) +
X3*log(2*p1*p2*(1-k) + p1*p1*k) +
X4*log((2*p1*(1-p1-p2)+p2*p2)*(1-k)+2*p1*p2*k) +
X5*log(2*p2*(1-p1-p2)*(1-k) + (2*p1*(1-p1-p2)+p2*p2)*k) +
X6*log((1-p1-p2)*(1-p1-p2)*(1-k) + 2*p2*(1-p1-p2)*k) +
X7*log((1-p1-p2)*(1-p1-p2)*k))
return(ll)
}
counts = c(57,98,81,48,16,2)
est = optim(c(.05,.05),ll,X=counts,k=.175,hessian=T,control=list(reltol=1E-40))
fim = solve(est$hessian);
p.hat = c(est$par,1-sum(est$par))
se = c(sqrt(diag(fim)),sqrt(sum(diag(fim))+2*fim[1,2]))
ci.lower = p.hat - qnorm(1-.05/(2*3))*se
ci.upper = p.hat + qnorm(1-.05/(2*3))*se
cbind(p.hat,ci.lower,ci.upper)
k = .175
fitted = c(p.hat[1]*p.hat[1],2*p.hat[1]*p.hat[2],2*p.hat[1]*p.hat[3] + p.hat[2]*p.hat[2],2*p.hat[2]*p.hat[3],p.hat[3]*p.hat[3],0)*(1-k) +
c(0,p.hat[1]*p.hat[1],2*p.hat[1]*p.hat[2],2*p.hat[1]*p.hat[3] + p.hat[2]*p.hat[2],2*p.hat[2]*p.hat[3],p.hat[3]*p.hat[3])*k
chisq.test(counts,p=fitted)