Rolling rare charms with 6 skull recipie

[gorey]

If I take a rare jewel, found from hell cows, and reroll it with hopes of getting a usable one, what parts reroll?

For instance, do the # of mods also reroll or if there are 4 mods on the jewel before then there will also be 4 mods after?

If I remember correctly the item level does not change with this recipie, is that correct?


*Can't correct the title from 'charm' to jewel.

[inanefedaykin]

The ilevel goes down with the 6xpskull recipe.

[Starcrunch]

When using this formula the new ilvl is as follows:

new_ilvl = 0.4*ilvl + 0.4*clvl

This means that if you repeatedly reroll an item the new ilvl is asymptotic to (2/3)*clvl. If the current ilvl < (2/3)*clvl the new ilvl will in fact be higher than the current ilvl (with Hell Cows ilvl is certain to decrease, even with a clvl 99 character, as the Hell cow level is area lvl 81, IIRC).

-Starcrunch

[gorey]

So getting a decent lld jewel (lvl 30 or less) will continue to be possible if i roll it with my with my lvl 91 char. Thanks for the help on that.

What about the # of mods?

[Dodgydave]

It will be possible to get a jewel of req lvl 30 or less with a lvl 91 charater but not most likely.

The jewel ilvl will settle at 60 with a lvl91 character which will mean the jewel will be able to have up to req lvl 45 mods, in theory.

If you want lvl 30 or less jewels the ideal jewel ilvl would be 40 I do believe. Therefore you need a level 60 character for the rerolling.

(Of course I could be horribly wrong as I often am! :wink: )

[mephiztophelez]

What about the # of mods?
having 4 mods before the re-roll does NOT guarentee 4 mods after the re-roll.

you've "re-rolled" it, see? everything is up for grabs again.

[DudSpud]

When using this formula the new ilvl is as follows:

new_ilvl = 0.4*ilvl + 0.4*clvl

This means that if you repeatedly reroll an item the new ilvl is asymptotic to (2/3)*clvl.Help me out with the math, please (it's been a while...). Isn't it asymptotic to 2/5 clvl?

Thanks in advance,
DudSpud

[mephiztophelez]

Help me out with the math, please (it's been a while...). Isn't it asymptotic to 2/5 clvl?

Thanks in advance,
DudSpud

no. its (0.4 x ilvl) + (0.4 x clvl)

ergo, assuming an initial ilvl of 30 and a clvl of 90:

(0.4 x 30) + (0.4 x 90) = (12) + (36) = 48. so the new jool can have affixes up to lvl 48.

[DudSpud]

no. its (0.4 x ilvl) + (0.4 x clvl)

ergo, assuming an initial ilvl of 30 and a clvl of 90:

(0.4 x 30) + (0.4 x 90) = (12) + (36) = 48. so the new jool can have affixes up to lvl 48.The key to my question is "asymptotic." I'm guessing that you have taken Latin 101 ("ergo") but not yet trig (9th grade here in the States). My question is, what is the shape of the curve if you repeatedly re-roll the "jool." The gist is that the ilvl will continue to diminish, and so too the "new ilvl" with each re-roll. Eventually, it will reach a point where the re-roll will always be the same, so that new ilvl = old ilvl. This point is called the asymptote. My (aged) mind tells me the asymptote is at 2/5clvl, but my gut says it should be higher, and I cannot figure out where my math is wrong. Hence, my question.

Thanks,
DudSpud

[mephiztophelez]

err, remember, it's a two part formula. you can increase the clvl of the re-rolling char to "offset" the lowering ilvl from the 0.4 x bizzo.

btw: it's been ~20 years since i was in 9th grade. and i sucked at latin.

[DudSpud]

err, remember, it's a two part formula. you can increase the clvl of the re-rolling char to "offset" the lowering ilvl from the 0.4 x bizzo.clvl max = 99

btw: it's been ~20 years since i was in 9th grade. and i sucked at latin.25 years here...

But getting back to Starcrunch's assertion that the asymptote is at ilvl= 2/3clvl, he's(she's) right. I just can't figure out how that value was arrived at.

DudSpud

Edit - nevermind, I see now. Solve the equation for clvl where ilvl=0.4ilvl+0.4clvl and you get ilvl=2/3clvl. Sorry to bother you.

[mephiztophelez]

Edit - nevermind, I see now. Solve the equation for clvl where ilvl=0.4ilvl+0.4clvl and you get ilvl=2/3clvl. Sorry to bother you.

errrrrrr, i dunno that this will work. theres two different values being expressed by "ilvl" in this formula: ilvl(out) and ilvl(in). i dont think you can simply solve for "ilvl" in this way. (mind you, been a long time since i did math either)

but the short answer (i would imagine) is the minimum ilvl you can get too is ilvl1.

[Ce Olba]

no. its (0.4 x ilvl) + (0.4 x clvl)

ergo, assuming an initial ilvl of 30 and a clvl of 90:

(0.4 x 30) + (0.4 x 90) = (12) + (36) = 48. so the new jool can have affixes up to lvl 48.

Tell me, why separate the 0.4s? You can just take it as the common divisor and write it as 0.4x(ilvl+clvl)? It's still correct according to the axioms of real numbers. The one that says that a(b+c) = ab + ac. So why write it like that?

Is it just me or is it really interesting that you can actually draw a formula for the level of your character based on the ilvl of the item before and after re-rolling? Not to mention that you can do other funny stuff with it, such as draw a formula for ilvl based on the new ilvl and your clvl. Yes I know, useless in practice, but fun.

[mephiztophelez]

Tell me, why separate the 0.4s? You can just take it as the common divisor and write it as 0.4x(ilvl+clvl)? It's still correct according to the axioms of real numbers. The one that says that a(b+c) = ab + ac. So why write it like that?

your point being?

dammit, i really do have to update that ignore list.

[krischan]

Help me out with the math, please (it's been a while...). Isn't it asymptotic to 2/5 clvl?

Thanks in advance,
DudSpud

If we assume that the ilvl converges to some value (I will skip proving that :wink3:), the level won't change anymore from a certain point on, i.e. old ilvl = new ilvl

ilvl=0.4*ilvl+0.4*charlvl ==> (1-0.4)*ilvl=0.4*charlvl ==> ilvl=0.4/0.6*charlvl=2/3*charlvl

That's not an exact mathematical proof, of course. BTW, the game does the math with integers, so rounding might result to lower ilvls than the formula suggests.

[sangfagel]

Help me out with the math, please (it's been a while...). Isn't it asymptotic to 2/5 clvl?

Thanks in advance,
DudSpud

Almost :azn: - it IS asymptotic to 2/5 clvl +1 (since ilvl cant be <1 afaik).

After a number of rolls (0.4 x ilvl) part will be closer and closer to zero. 0.4 clvl is constant assuming that it´s about the same toon.

[krischan]

Almost :azn: - it IS asymptotic to 2/5 clvl +1 (since ilvl cant be <1 afaik).

After a number of rolls (0.4 x ilvl) part will be closer and closer to zero. 0.4 clvl is constant assuming that it´s about the same toon.

So you assume that 0.4*clvl is approaching zero and your conclusion is that it's 2/5*charlvl+1 ? That's not close to zero which means your considerations are contradicting, i.e. you are wrong :tongue:

As I explained above, it settles at 2/3*charlevel with the mentioned rounding effect.

[Ava]

OMG. What a math dispute. Why not just give an example?

Assume a lvl91 character.

An illvl1 jewel would give you:
roll -- ilvl (rounded down)
1 ---- 36
2 ---- 50
3 ---- 56
4 ---- 58
5 ---- 59
6 ---- 60
7 ---- 60 (will converge to this)

An illvl99 jewel would give you:
roll -- ilvl (rounded down)
1 ---- 76
2 ---- 66
3 ---- 62
4 ---- 61
5 ---- 60
6 ---- 60 (will converge to this)

So for a lvl91 character, the ilvl of the jewel will converge to 60, no matter what you start with (that's called asymptotic and it's a calculus term) which is roughly 2/3rd of the current clvl. It's not that hard to believe to that this applies to all levels (you could do the numbers for other levels if you don't believe it).

[sangfagel]

So you assume that 0.4*clvl is approaching zero and your conclusion is that it's 2/5*charlvl+1 ? That's not close to zero which means your considerations are contradicting, i.e. you are wrong :tongue:

As I explained above, it settles at 2/3*charlevel with the mentioned rounding effect.

I never wrote that "0.4*clvl is approaching zero". Please read my post again - it´s short.

Let´s look at this very simple formel again:

new_ilvl = 0.4*ilvl + 0.4*clvl

I consider a case when an item with base quality 1 is rolled multiple times by the same character.

Since the character is the same all the time, clvl is the same all the time and the value [0.4*clvl] = constans.

Now, if we use the symbol "ilvl" for "starting item level" - before the first roll, and symbol "x_ilvl" for "new item level after x rolls" we can re-write the formel:

x_ilvl= [(0.4)^x]*ilvl + 0.4*clvl

The part of equation "[(0.4)^x]*ilvl" is approching 0 (since 0.4<1)

The second part, "0.4*clvl" - remains the same

So, we have to do with a function asymptotic to 2/5 clvl.

---------------

I was wrong anyway :laugh: - in the part I wrote "2/5 clvl +1". *DudSpud was right - it´s just "2/5 clvl".

[DudSpud]

*DudSpud was right - it´s just "2/5 clvl".I really don't hear this often enough, but...

In this case, I think Krischan is correct and Teh Spud is wrong - if we assume that the ilvl plateaus, which makes intuitive sense but I cannot prove why. I can also not see what is wrong with sangfagel's reasoning either.

This is why I became a biologist...


Thanks,
DudSpud

[NASE]

it comes down to 0.6666, and only to that number and no other.

lets see:

ilvl_n = 0.4*ilvl_(n-1) + 0.4*C
ilvl_n = 0.4*(0.4*ilvl_(n-2)+0.4*C)+0.4*C
This is no more then a substitution of ilvl_(n-1) = 0.4*ilvl_(n-2) + 0.4*C
Now we get ride of some of those '()'.
ilvl_n = 0.4*0.4*ilvl_(n-2) + 0.4*0.4*C+0.4*C.
Again the same routine.
ilvl_n = 0.4*0.4*(0.4*ilvl_(n-3)+0.4*C) + 0.4*0.4*C+0.4*C.
ilvl_n = 0.4³*ilvl_(n-3)+0.4³*C + 0.4²*C+0.4*C.
If we this thing n times, it not that hard to see what will happen.
ilvl_n = 0.4^n*ilvl_1 + sum(0.4^i,i=1..n)*C.
That last thing is just C*(0.4+0.4²+0.4³...+0.4^(n))

As already said, a number smaller then 1 will eventually go to 0. 0.4^infinity = 0
The sum is a bit harder, though still not difficult, I'm not going to this now, though if I must, I'll do it anyway. The sum comes do to.

sum(0.4^i,i=1..infinity) = -1+1/(1-0.4) = 2/3.



You may remember this sum in the following maner.
sum(0.4^i,i=0..infinity) = 1+sum(0.4^i,i=1..infinity) = 1/(1-0.4) = 5/3.

[krischan]

I never wrote that "0.4*clvl is approaching zero". Please read my post again - it´s short.
OK, lets do that. Here it is:

Almost :azn: - it IS asymptotic to 2/5 clvl +1 (since ilvl cant be <1 afaik).

After a number of rolls (0.4 x ilvl) part will be closer and closer to zero. 0.4 clvl is constant assuming that it´s about the same toon.
I underlined the relevant part.

Anyway, you meant to say that 0.4^N*ilvl approaches zero.
Let´s look at this very simple formel again:

new_ilvl = 0.4*ilvl + 0.4*clvl

I consider a case when an item with base quality 1 is rolled multiple times by the same character.

Since the character is the same all the time, clvl is the same all the time and the value [0.4*clvl] = constans.

Now, if we use the symbol "ilvl" for "starting item level" - before the first roll, and symbol "x_ilvl" for "new item level after x rolls" we can re-write the formel:

x_ilvl= [(0.4)^x]*ilvl + 0.4*clvl

That's wrong. You cannot rewrite the formula like that.

First roll: ilvl(1) = 1*0.4+charlvl*0.4 = clvl*0.4
Second roll: ilvl(2) = 0.4*ilvl(1) + 0.4*clvl = 0.4*0.4*clvl + 0.4*clvl = 0.4^2*clvl+0.4*clvl
Third roll: ilvl(3) = 0.4*ilvl(2) + 0.4*clvl = 0.4*(0.4*0.4*clvl + 0.4*clvl) + 0.4*clvl = 0.4^3*clvl+0.4^2*clvl+0.4*clvl

and so on. The formula for N tries is clvl multiplied by the sum of all powers of 0.4 from 1 to N. You can prove that via "full induction" (*). If you wish, I can do that in an even more elaborate posting.

For x>1, 1/(x-1) = sum((1/x)^N) with N=1..infinity. Everybody who studied mathematics will agree to that, I guess.

That means, with x=2.5, 1/x=0.4 and sum(0.4^N)= 1/(2.5-1)=1/1.5 = 2/3, q.e.d.

(*) I'm not sure is it's the right English word. It's an axiom which says that if you can prove that (1) a statement is true for N=1 and (2) you can prove the conclusion "if it's true for N-1, then it's true for N as well", then the statement is true for any natural number.

Edit: Damn, NASE stole my show :laugh:

[sangfagel]

It´s of course right... I promise to not touch any formulas more today ... May be I should call a neurologist :rolleyes:

[Ax2Grind]

I had a comparative list written down in a file, probably an old text that I didn't save after a drive failure and/or re-install. So I rewrote it just for you, now stop arguing.

A CLvl99 converges at ILvl65, not 66 (i.e. 2/3), strictly due to the rounding, which - as asked about earlier - makes the result {2/3} +/-1 (minimum of 1). Each CLvl below 100 reduces the character-based calculation by .4, and when that drops below an integer post-rounding, the old item level drops by .4 and an additional .4 if it drops below an integer in the process. You can also see the pattern with regard to ILvl every ten CLvls. This, of course, has nothing to do with the supposed 75% of ALvl deciding RLvl, so make sure the ALvls are accurate and do a separate calculation as to what CLvl you need for the best chance at a LLD item.

CLvl99 = ILvl65 ({39.6}+{26.0})
CLvl98 = ILvl65 ({39.2}+{26.0})
CLvl97 = ILvl63 ({38.8}+{25.2})
CLvl96 = ILvl63 ({38.4}+{25.2})
CLvl95 = ILvl63 ({38.0}+{25.2})
CLvl94 = ILvl61 ({37.6}+{24.4})
CLvl93 = ILvl61 ({37.2}+{24.4})
CLvl92 = ILvl60 ({36.8}+{24.0})
CLvl91 = ILvl60 ({36.4}+{24.0})
CLvl90 = ILvl60 ({36.0}+{24.0})
CLvl89 = ILvl58 ({35.6}+{23.2})
CLvl88 = ILvl58 ({35.2}+{23.2})
CLvl87 = ILvl56 ({34.8}+{22.4})
CLvl86 = ILvl56 ({34.4}+{22.4})
CLvl85 = ILvl56 ({34.0}+{22.4})
CLvl84 = ILvl55 ({33.6}+{22.0})
CLvl83 = ILvl55 ({33.2}+{22.0})
CLvl82 = ILvl53 ({32.8}+{21.2})
CLvl81 = ILvl53 ({32.4}+{21.2})
CLvl80 = ILvl53 ({32.0}+{21.2})
CLvl79 = ILvl51 ({31.6}+{20.4})
CLvl78 = ILvl51 ({31.2}+{20.4})
CLvl77 = ILvl50 ({30.8}+{20.0})
CLvl76 = ILvl50 ({30.4}+{20.0})
CLvl75 = ILvl50 ({30.0}+{20.0})
CLvl74 = ILvl48 ({29.6}+{19.2})
CLvl73 = ILvl48 ({29.2}+{19.2})
CLvl72 = ILvl46 ({28.8}+{18.4})
CLvl71 = ILvl46 ({28.4}+{18.4})
CLvl70 = ILvl46 ({28.0}+{18.4})
CLvl69 = ILvl45 ({27.6}+{18.0})
CLvl68 = ILvl45 ({27.2}+{18.0})
CLvl67 = ILvl43 ({26.8}+{17.2})
CLvl66 = ILvl43 ({26.4}+{17.2})
CLvl65 = ILvl43 ({26.0}+{17.2})
CLvl64 = ILvl41 ({25.6}+{16.4})
CLvl63 = ILvl41 ({25.2}+{16.4})
CLvl62 = ILvl40 ({24.8}+{16.0})
CLvl61 = ILvl40 ({24.4}+{16.0})
CLvl60 = ILvl40 ({24.0}+{16.0})
CLvl59 = ILvl38 ({23.6}+{15.2})
CLvl58 = ILvl38 ({23.2}+{15.2})
CLvl57 = ILvl36 ({22.8}+{14.4})
CLvl56 = ILvl36 ({22.4}+{14.4})
CLvl55 = ILvl36 ({22.0}+{14.4})
CLvl54 = ILvl35 ({21.6}+{14.0})
CLvl53 = ILvl35 ({21.2}+{14.0})
CLvl52 = ILvl33 ({20.8}+{13.2})
CLvl51 = ILvl33 ({20.4}+{13.2})
CLvl50 = ILvl33 ({20.0}+{13.2})
CLvl49 = ILvl31 ({19.6}+{12.4})
CLvl48 = ILvl31 ({19.2}+{12.4})
CLvl47 = ILvl30 ({18.8}+{12.0})
CLvl46 = ILvl30 ({18.4}+{12.0})
CLvl45 = ILvl30 ({18.0}+{12.0})
CLvl44 = ILvl28 ({17.6}+{11.2})
CLvl43 = ILvl28 ({17.2}+{11.2})
CLvl42 = ILvl26 ({16.8}+{10.4})
CLvl41 = ILvl26 ({16.4}+{10.4})
CLvl40 = ILvl26 ({16.0}+{10.4})
CLvl39 = ILvl25 ({15.6}+{10.0})
CLvl38 = ILvl25 ({15.2}+{10.0})
CLvl37 = ILvl23 ({14.8}+{9.2})
CLvl36 = ILvl23 ({14.4}+{9.2})
CLvl35 = ILvl23 ({14.0}+{9.2})
CLvl34 = ILvl21 ({13.6}+{8.4})
CLvl33 = ILvl21 ({13.2}+{8.4})
CLvl32 = ILvl20 ({12.8}+{8.0})
CLvl31 = ILvl20 ({12.4}+{8.0})
CLvl30 = ILvl20 ({12.0}+{8.0})
CLvl29 = ILvl18 ({11.6}+{7.2})
CLvl28 = ILvl18 ({11.2}+{7.2})
CLvl27 = ILvl16 ({10.8}+{6.4})
CLvl26 = ILvl16 ({10.4}+{6.4})
CLvl25 = ILvl16 ({10.0}+{6.4})
CLvl24 = ILvl15 ({9.6}+{6.0})
CLvl23 = ILvl15 ({9.2}+{6.0})
CLvl22 = ILvl13 ({8.8}+{5.2})
CLvl21 = ILvl13 ({8.4}+{5.2})
CLvl20 = ILvl13 ({8.0}+{5.2})
CLvl19 = ILvl11 ({7.6}+{4.4})
CLvl18 = ILvl11 ({7.2}+{4.4})
CLvl17 = ILvl10 ({6.8}+{4.0})
CLvl16 = ILvl10 ({6.4}+{4.0})
CLvl15 = ILvl10 ({6.0}+{4.0})
CLvl14 = ILvl8 ({5.6}+{3.2})
CLvl13 = ILvl8 ({5.2}+{3.2})
CLvl12 = ILvl6 ({4.8}+{2.4})
CLvl11 = ILvl6 ({4.4}+{2.4})
CLvl10 = ILvl6 ({4.0}+{2.4})
CLvl9 = ILvl5 ({3.6}+{2.0})
CLvl8 = ILvl5 ({3.2}+{2.0})
CLvl7 = ILvl3 ({2.8}+{1.2})
CLvl6 = ILvl3 ({2.4}+{1.2})
CLvl5 = ILvl3 ({2.0}+{1.2})
CLvl4 = ILvl1 ({1.6}+{0.4})
CLvl3 = ILvl1 ({1.2}+{0.4})
CLvl2 = ILvl1 ({0.8}+{0.0})
CLvl1 = ILvl1 ({0.4}+{0.0})
Edit: I forgot to answer the mod's question about what levels for LLD jewels directly, so anyone correct me if I'm wrong.

CLvl58-59 converges on ILvl38 jewels for RLvl30 affixes. ILvl39 will could get you "Brown", but not "of Fervor," since that's magic-only.
CLvl38-39 converges on ILvl25 jewels for RLvl18 affixes. One higher and you'll get several capable on rares, but remember "of Carnage" isn't possible.
CLvl20-22 converges on ILvl13 jewels for RLvl9 affixes. There are no ILvl14 affixes, but then there's no way to converge on that.

[krischan]

It´s of course right... I promise to not touch any formulas more today ... May be I should call a neurologist :rolleyes:
Thanks to you for not calling me a nerd :smiley:

Sometimes I like to brag with mathematics, to prove that I'm right while not noticing that 90% of the others will stop with reading at line 3 or so :rolleyes:

[Starcrunch]

The key to my question is "asymptotic." I'm guessing that you have taken Latin 101 ("ergo") but not yet trig (9th grade here in the States). My question is, what is the shape of the curve if you repeatedly re-roll the "jool." The gist is that the ilvl will continue to diminish, and so too the "new ilvl" with each re-roll. Eventually, it will reach a point where the re-roll will always be the same, so that new ilvl = old ilvl. This point is called the asymptote. My (aged) mind tells me the asymptote is at 2/5clvl, but my gut says it should be higher, and I cannot figure out where my math is wrong. Hence, my question.

Thanks,
DudSpud

No my degree is mathematics and it tends to 2/3*clvl. It's a reasonably trivial result of linear algebra (the transformation matrix is {{1,0},{0.4, 0.4}}, take this to the nth power and take the limit as n->infinity and apply it to the vector {{clvl},{ilvl}}.)

The math:
{{1,0},{0.4,0.4}}^n = {{1,0},{0.4+0.4^2+0.4^3+...0.4^n, 0.4^n}}; the limit as n->infinity of the above matrix is {{1,0},{2/3,0}} (this is why it is asymptotic). Multiplying this times the vector {{clvl},{ilvl}} yields the resultant vector {{clvl},{2/3*clvl}}; it converges to 2/3*clvl.

EDIT: for not reading the whole thread and being overly nerdy I see that krischan has provided a lot clearer explanation, devoid of linear algebra.

-Starcrunch

[crawlingdeadman]

wow lotsa math flying around in here. i keep seeing this thread and really wanted to point out that there's no rare charms... i know it's about jewels and rolling rares in general, but maybe the title could be changed to keep from confusing idiots like me... :hide:

[thepandafactor]

The easiest math for this is just an ordinary differential equation. Then you can see the convergence for all possible initial conditions.

[thepandafactor]

Someone PMed me so I'll just calculate it out here:
C = clvl,
I = ilvl.

So:

dI/dt = -0.6I + 0.4c

solving, that means:

I = 2/3 C + A e^-0.6 t,

t = number of times you reroll.
A depends on the initial Ilvl, A = initial Ilvl - 2/3 clvl.
So as t -> inf, ilvl = 2/3 clvl as previously stated.

I tihnk this way is clearer for everyone.

[Dodgydave]

Someone PMed me so I'll just calculate it out here:
C = clvl,
I = ilvl.

So:

dI/dt = -0.6I + 0.4c

solving, that means:

I = 2/3 C + A e^-0.6 t,

t = number of times you reroll.
A depends on the initial Ilvl, A = initial Ilvl - 2/3 clvl.
So as t -> inf, ilvl = 2/3 clvl as previously stated.

I tihnk this way is clearer for everyone.

That's how I would have doen it! :wink:

[krischan]

I tihnk this way is clearer for everyone.
:laughing:

It needs knowledge about differential equations to understand the proof and why the presented differential equation describes the problem at all. My proof probably isn't much better at that, however.