The Loot Model
In this section I will describe the loot model I’ve got so far:
As explained in the previous chapters, global loot shows to be a mixture of several exponential distributions. Estimation can be done numerically we MLE (maximum likelihood) or EM (expectation maximization). Unfortunately there are some pitfalls. With 3 exp distributions one has to estimate 5 parameters and one needs a good initial guess to not end up in a local minimum. Moreover, the mixture probabilities will be correlated with the other parameter estimates, and hence we have a suboptimal situation in which estimation becomes a nightmare.
To model a mixture of exponentials a generalized pareto distribution (GPD) can be used. It has 3 parameters and looks like the following:
S(x) =(1 + s*(x-t)/k)^(-1/k)
with location parameter t, scale s and shape k. Furthermore x>=t and k, s > 0.
(There is also a form of the GPD where k can be less than 0, this is however a situations that is not of interest here).
The mean of the GPD is
m = t + s/(1-k)
The advantage of the GPD over the mixture is that I have only 2 parameters to estimate instead of 5, since t can and should be chosen a priori. So I have a more stable situation.
Now let’s try it out:
Fig. 1: Ambu Y global loot in PED. Source starfinders loot tracker.
Click to enlarge
As you see in Fig. 1 the model seems to do a great job. I had to use 3 exp. distributions to get a similar fit before. However, I had to limit loot to 1000 PED to get a good fit and I explain later why.
We have found now a model that can handle global loot rather easily, but what exactly did we calculate? Let’s go one step back. Global loot is truncated at 50 PED. So we have got parameters of a truncated distribution. For those readers that are not interested in the hard stuff, it would be better to jump directly to the next figures.
A left truncated distribution is nothing else as
ST(x) = S(x)/S(T) where T is the left truncation point.
If we resolve this term, we get another GPD (so a similar property as the exp dist has). It can be shown that that the mean of ST(x) is
mT = T + (s + k(T-t))/(1-k)
Having estimates for the parameters of ST(x), we have already an estimate of k and from mT we get estimates for s. So far so good.
Now let’s apply this to real data:
Fig. 2: Parameter estimates for s and t according to HP
Click to enlarge
(please note: Fig. 2 is the result of several fits using first, suited mobs (HP > 2000) to derive k. Having k we can calc s as mentioned above.)
As one can see, the relation is linear. So the parameters of the GPD’s for the different mobs do depend on HP. Moreover, we get a formula how to calc those parameters from HP. Parameter k itself does not depend on HP and is estimated to be .25. S can be estimated as 2/100 * HP (PED). The shift parameter t seems to be identically to s but with some constant added. This was what I used in my first model. However, it turns out that the constant term might only be sample related and I therefore removed it.
We have now a model for global loot below 1000 PED. Loot over 1000 PED is very rare. For Ambu we observe values from 1100 till 1600 (probability in global loot p = .002), , and 15000 till 20000 (p = .0005). So it seems we have two further loot distributions that can’t be estimated atm, due to lack of data. However we can use their means and probabilities. This is also the reason why I did the first fit for loot below 1000 PED.
We know now that global loot does depend on HP. From Jimmy’s post we know that this holds also for normal loot. The best approximation is an exponential distribution with mean HP/10 PEC, so HP/1000 in PED and that there is a min loot corresponding to HP/2000 PED.
One of the most important things to know about loot is how much is returned. This is expressed mathematically as the expectation of the loot distribution.
The loot distribution can be modeled as
L(x) = c +p1*S1(x) + p2 *S2(x) + … + pn*Sn(x)
This is a mixture distribution with n terms. The expectation of L(x) is nothing else as
E(x) = c + p1 * E1(x) + … + pn*En(x)
so the sum of the individual weighted expectations.
To make things short. From global loot I was further able to derive a 4th distribution that lies between normal and global loot. All in all global loot for a mob with HP > 1000 is a mixture of 6 distributions. This is depicted in the following table:
Table 1: Loot model for HP >1000 in PEC
Code:
Class p lower upper mean untruncated Truncation
C0 1.0000 HP/20
C1 .97500 HP/20 3*HP/10 HP/10 .05
C3 .00375 HP 12*HP (HP+HP/(1-k) .005
C4 .02119 2*HP 28*HP 2*HP+2*HP/(1-k) .003
C5 .00005 HP*1.5*100
C6 .00001 HP*15*100
p is the probability of the given class. Lower/upper are the lower/upper loot limits. The given mean is the untruncated mean and truncation is the probability at which S(x) is truncated. P has not the final values. Also the truncation must be verified when I have more data.
Next I will show what to do with it.
The Means
To get an estimate for the expectation of L(x) we need the means per class. It was no surprise for me to detect upper limits. I would have done it the same way, because otherwise an extreme large loot would be possible. Since C1-C4 do have an upper limit, we need again some math.
An upper truncated S(x) = SU(x) will have the following form
SU(x) = (S(x) – S(U))/(1-S(U)), where U is the upper limit.
The expectation
EU(x) is the integral from 0 to U of
EU(x) = Int[x * fU(x)]0..U,
where fU(x) is the density of the truncated distribution and Int[..]a..b is the notational form for an integral.
Solution for C1:
EC1(x) = HP/10 * [1-pu*(1-ln(pu))], where pu is the truncation probability
Solution for C3:
Was rather tricky but I found it to be
EC3(x) = 1/(1-pu)*[HP+HP/(1-k)-pu*[HP+HP(pu^-k+k-1)/(k*(1-k))]]
Solution for C4:
Analog to C3
EC4(x) = 1/(1-pu)*[2*HP+2*HP/(1-k)-pu*[2*HP+2*HP(pu^-k+k-1)/(k*(1-k))]]
Example:
Ambu y assuming 1.6 * HP effective dmg done
E(Ambu) =
HP/20 +
.975 * HP/10 * .8 +
.00375 * HP * .2.26 +
.02119 *2*HP * 2.28 +
.00005 * HP*1.5*100 +
.00001 * HP*15*100
= 419.1 PEC
So mean loot for ambu is about 420 PEC. 1010 * 1.6 HP will cost 539 PEC, so your expected loss per ambu is 119.5 PEC (22.2%) or expected return rate will be 77.8%.
In the coming chapter I will show the validation of the model.
tbc ...