Analysis of the star force enhancement system for superior equips
We provide a way to compute the expected number of enhancements, hi,j ,
to go from i stars to j stars (0 ≤ i < j < 10) for superior equipment. We
also provide a way to compute the expected number of destroyed equips, di,j ,
while going from i stars to j stars.
Let Pi denote the probability of successfully enhancing at state i. Let Di
denote the probability of destroying an item while attempting to enhance at
state i. Let Fi := 1 − Di − Pi denote the probability of demotion or stability
while at state i.
Our model for the enhancements is described by the following two diagrams
F0
0
P0
0
1
n
P1
Pn
···
P8
9
P9
10
n+1
Dn
0
0Dn
Fn 0
Pn
(n = 1, . . . , 8).
(∗)
Fn+1
ñ
The role of the first diagram should be apparent, so we will only explain
the role of the latter diagram. The state ñ serves to act as state n while also
remembering that it arrived at state n from having backtracked once from
the last attempted enhancement. If one succeeds from state ñ, then one is
promoted to state n + 1, just as it would have been from state n. If one fails
at state ñ, then one is sent to state n. This is feature not quite like chance
time, because there is a missing step in reaching state n from a failure at
state ñ. This is remedied in the recurrence relations below.
For the remainder of this document, we fix the following ordering of states:
(nk ) := (n1 , . . . , n18 ) := (0, 1, 1̃, . . . , 8, 8̃, 9).
Let M be the Markov chain corresponding to the above diagram according
to this order.
1
Expected number of enhancements from i stars to 10 stars
We first calculate hi,10 for 0 ≤ i < 10. The recurrence relations for these
values are as follows


h0,10 = 21 (1 + h0,10 ) + 21 (1 + h1,10 )



1
1


h1,10 = 2 (1 + h0,10 ) + 2 (1 + h2,10 )
hi,10 = Fi (1 + h]
) + Si (1 + hi+1,10 ) + Di (1 + h0,10 ) (2 ≤ i ≤ 9) . (1)
i−1,10



hĩ,10 = Fi (2 + hi,10 ) + Si (1 + hi+1,10 ) + Di (1 + h0,10 )
(1 ≤ i ≤ 8)




h10,10 = 0
Let h := (hnk ,10 )t and let
u := ((1, . . . , 1) + (0, 0, P1̃ , 0, P2̃ , . . . , 0, P8̃ , 0))t .
Then the recurrence relations (1) are given by
(I − M )h = u ⇐⇒ h = (I − M )−1 u
and we find that
h0,10
h1,10
h2,10
h3,10
h4,10
= 285.846
= 283.846
= 279.946
= 273.345
= 263.158
h5,10
h6,10
h7,10
h8,10
h9,10
= 249.089
= 230.325
= 204.421
= 167.176
= 109.986
These values were rounded to the nearest third decimal place.
2
(1*)
Expected number of destroyed items from i stars to 10 stars
We now compute di,10 for 0 ≤ i < 10. The recurrence relations for the
expected number of destroyed items are as follows


d0,10 = 21 d0 + 21 d1



1
1


d1,10 = 2 d0 + 2 d2
(2)
di,10 = Fi d]
+ Si di+1 + Di (1 + d0,10 ) (2 ≤ i ≤ 9) .
i−1



dĩ,10 = Fi di + Si di+1 + Di (1 + d0,10 )
(1 ≤ i ≤ 8)



d
10,10 = 0
Define d := (dnk ,10 )t , let
v := (Dnk )t = (0, ..., 0, D5 , D5̃ , . . . , D8 , D8̃ , D9 )t ,
and lastly, let M0 denote the Markov chain M with its destruction entries
removed (that is, the entirety of the first column of M is removed with the
exception of the (1,1) and (2,1) entry). Then the recurrence relations (2) are
given by
(I − M0 )d = v ⇐⇒ d = (I − M0 )−1 v.
(2*)
We find now that
d0,10
d1,10
d2,10
d3,10
d4,10
= 2.09
= 2.09
= 2.09
= 2.09
= 2.09
d5,10
d6,10
d7,10
d8,10
d9,10
= 2.09
= 2.045
= 1.903
= 1.615
= 1.093
These values were rounded to the nearest third decimal place.
3
Going from hi,10 to hi,j and from di,10 to di,j
If you look at the definitions of hi,j and di,j , it’s plain that
hi,j = hi,10 − hj,10
and
di,j = di,10 − dj,10
for 0 ≤ i < j < 10.
An example
We may calculate the number of destructions d8,9 expected while enhancing
from 8 stars to 9 stars. As mentioned above, this is given by
d8,9 = d8,10 − d9,10 = .522.
Similarly, we find that the expected number of enhancements from 8 stars to
9 is
h8,9 = 57.19.
The value d8,9 is not at odds with the fact that the probability of destruction
going from 8 stars to 9 stars is .06. The value h8,9 indicates that the expected
number of attempts at enhancement from 8 to 9 is around 57. Consider a
fixed path of length 57 going from 8 stars to 9 stars for sake of concreteness.
This path taken going from 8 stars to 9 stars will rack up possibilities of
destruction at many places (possibly all at going from 8 stars to 9 stars, but
more likely other places as well).
4