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Zeltar
08-11-2008, 06:47 AM
Value of enchanted weapons guide:

This is a guide on the value of enchanted items.
Warning this contains heavy maths, you can look at the maths if you want, or ignore it and just see the conclusions.

I’m going to assume that the blacksmith is not rigged, and that there is a 50% chance of success. Yes some people are more lucky than others, I’m not getting into those discussions here.

So the question is, what is a high + enhanced item worth?

I’m going to take a probabilistic approach, the cost of an action is the probability of one outcome times the cost of the outcome. Yes there is statistical variation, you can be lucky one time and very unlucky the other time, but in average that would be the cost.

For those that like equations, it is: sum of n from 1 to N (chance_of_outcome_n *loss_of_outcome_n)

So chance of success of a +1 in the range of 1-4[B]

Cost of success=1 xen
Cost of failure and return to starting position =1 xen (xen for attempt, but it does not lose a lvl)

1st try gives success =1/2 chance of this outcome (50%)
Cost=1 xen
2nd try gives success =1/4 (25%)
Cost=2 xen
Nth try gives success =1/2^n chance
Cost=N xens

So it is sum(n/(2)^n) for n=1:inf
Now for those that have done series math’s, you will know that the sum of 1/2+2/4+3/8+4/16+5/32+…=2

So therefore the cost of going from 3 to 4 is 2 xens.
Therefore a +4/6ap is worth a +1/6ap and 6 xens.

[B]Enchanting to +5

Now when going to +5 a failure will take you back to +3. and as we previously showed it’ll take another 2 xens to get back to the starting point of +4

So we can say:
Cost of success=1 xen
Cost of failure and return to starting position=3 xen (xen for attempt, and 2 to get back to starting point of +4)

Now we have the sum of
1st try gives success =1/2 chance of this outcome (50%)
Cost=1 xen
2nd try gives success=1/4 (25%)
Cost=1+3 xen
Nth try gives success =1/2^n chance
Cost=1+3*N xens


So it is sum((1+3*n)/(2)^n) for n=1:inf
=4 xens (look at number theory)

If 1 is the cost of success, and b is the cost of failure then we have sum((1+b*n)/(2)^n) for n=1:inf
= 1+b (use similar to the first equation, but extract the multiple of b out of it)
Actually the probabilistic cost equals the cost of failure+cost of success.

So to get to +5 it takes a +4 and 4 xens.
I.e. +5/6ap is worth a +1/6ap and 10 xens

Enchanting to +6
Going to +1 raises the stakes, as failure will result in it droppping to +2 and having to re-enchant back to the starting point of +5.
Cost of success=1 xen
Cost of failure and return to starting position =9 xen (xen for attempt, and 8 to get back to starting point of +5 from +2)

On same reasoning as before it will take 10 xens to get to +6 from a +5.
Therefore a +6/6ap is worth a +1/6ap and 20 xens.

Enchanting to +7
Going to +7 raises the stakes a bit, as failure will result in needing to re-make the item from scratch.
Ok, I don’t think anybody is crazy enough to try enchanting to +7, but if you take the rish, I’ll calculate the probabilistic cost of making it.

Now for simplistic reasons, let’s say that a +1/6ap costs 2 xen (yeah market price fluctuates, but I need to account for it’s cost of a broken item in terms of xens)

Cost of success=1 xen
Cost of failure and return to starting position=1+cost +1/6ap+cost of enhancing it to +6/6ap=1+2+20=23 xens

Now you have to repeat this process for n to infinity (I’d hate to break a +6/6ap 3 times in a row)

So the probabilistic average cost of enchanting from +6 to +7 would be 24 xens.
Therefore a +7/6ap would be worth a +1/6ap and 44 xens.

Enchanting to +8
If anyone tries enchanting a +7 then they must be crazy, well good luck!
If you have an unlimited supply of xens and really want to try it, well this is what it’ll take on average.

Cost of success=1 xen
Cost of failure=1+cost of +1/6ap +44=47 xens
Cost of enchanting from +7 to +8=48 xens

Average cost of producing a +8/6ap would be worth a +1/6ap and 84 xens.

Enchanting to +9
These items only come from the old GM event, nobody would enhance to this. They are not worth this value in reality, but I’ll calculate it just for those that like it.
Cost of success=1 xen
Cost of failure=1+cost of +1/6ap +84=87 xens
Cost of enchanting from +8 to +9=88 xens

Average cost of producing a +9/6ap would be worth a +1/6ap and 172 xens.

Enchanting to +10
These items only come from the old GM event, nobody would enhance to this. They are not worth this value in reality, but I’ll calculate it just for those that like it.
Cost of success=1 xen
Cost of failure=1+cost of +1/6ap +172=175 xens
Cost of enchanting from +8 to +9=176 xens

Average cost of producing a +9/6ap would be worth a +1/6ap and 348 xens.

note, this accounts for the full cost of creating on average 16 items from scratch to have one survive through to +10.

Conclusion
The value of enchanted items are as follows:
+2/6ap=+1/6ap & 2 XENs
+3/6ap=+1/6ap & 4 XENs
+4/6ap=+1/6ap & 6 XENs
+5/6ap=+1/6ap & 10 XENs
+6/6ap=+1/6ap & 20 XENs
+7/6ap=+1/6ap & 44 XENs
+8/6ap=+1/6ap & 84 XENs
+9/6ap=+1/6ap & 172 XENs
+10/6ap=+1/6ap & 348 XENs

Yes it is a gamble that everyone takes when they enhance, if you try past +4 you have to be prepared to put in the xens incase you fail. Maybe you’ll get it first try and your items will be worth much more, maybe you’ll have bad luck.

Also selling second hand never sells as much as the original price of making an item. I find second hand items generally sell for about 75% of the original making price (depending on demand). You also have to remember that the level requirement goes up by 1 at +2 +4, +6 etc, so highly enhanced second hand items have higher lvl requirements making them slightly less attractive than self-enhancing, though second hand is a good option if you’re short on money and don’t want to throw in all those xens.

Good luck with your enchanting!

CyberPhoenixSlayer
08-11-2008, 10:05 AM
Excellent guide, but using a crap of load of Xen Stone to get it at +10?
Kinda useless for me. I will stay at +5 max not more than that, too much risky at my advice to get it +10.

Character's names:
CyberClyde: LV 8X Templar
CyberSlayer: LV 6X Knight
Marshmalow: LV 4X Scout
????????: LV 4X ???????
CyberNeos: LV 1X Acolyte

Zeltar
08-11-2008, 01:44 PM
Excellent guide, but using a crap of load of Xen Stone to get it at +10?
Kinda useless for me. I will stay at +5 max not more than that, too much risky at my advice to get it +10.

Character's names:
CyberClyde: LV 8X Templar
CyberSlayer: LV 6X Knight
Marshmalow: LV 4X Scout
????????: LV 4X ???????
CyberNeos: LV 1X Acolyte

I agree, most people stop at +4, attempting +5 is feasable though the price goes up, only if people have bucket loadsof xens to blow do they ever bother taking a chance with +6. and anything past +6 is not worth attempting IMO as failure will cause your item to break.

sickness12
08-18-2008, 02:48 PM
I sawww a staff +10 it was wonderfull to seee that...

Krilm
08-18-2008, 10:47 PM
Great job Zeltar, very interesting and informative also. Keep it up!

BellaAnimorum
08-19-2008, 01:39 PM
I sawww a staff +10 it was wonderfull to seee that...

That item was a prize from the Amazing Adventure. Eventually though it will be useless, as it will expire.

wildswing
08-24-2008, 11:46 PM
There's an easier way to calculate said expectation,

T(n) + 1 xen -> p*T(s) + (1-p)*T(f)
where n is the current, s is the state when xenning successful, which is n+1, f is the state of failing, which depends on n

where n denotes the state of the equipment, or +n rather. Take the expectation of both sides

E[T(n)] + 1 = p*E[T(s)] + (1-p)*E[T(f)]

for n=[0,3]
E[T(n)] + 1 = 0.5*E[T(n+1)] + 0.5*E[T(n)]
after rearranging
E[T(n+1)] = E[(T(n)] + 2

for n=[4,5]
E[T(n)] + 1 = 0.5*E[T(n+1)] + 0.5*E[T(n-1)]
after rearranging
E[T(n+1)] = 2*E[T(n)] - E[T(n-1)] + 2

for n=[6,9]
E[T(n)] + 1 = 0.5*E[T(n+1)] + 0.5*E[T(0)]
E[T(n+1)] = 2*E[T(n)] - E[T(0)] + 2

A basic recursion in 3 different parts. Only assumption to make or to know about is the prices of worth of a +0/6ap, of course discounting the real chance of xen success rate and omittng the chance of breaking.

legendaryJ
09-13-2008, 03:25 PM
There's an easier way to calculate said expectation,

T(n) + 1 xen -> p*T(s) + (1-p)*T(f)
where n is the current, s is the state when xenning successful, which is n+1, f is the state of failing, which depends on n

where n denotes the state of the equipment, or +n rather. Take the expectation of both sides

E[T(n)] + 1 = p*E[T(s)] + (1-p)*E[T(f)]

for n=[0,3]
E[T(n)] + 1 = 0.5*E[T(n+1)] + 0.5*E[T(n)]
after rearranging
E[T(n+1)] = E[(T(n)] + 2

for n=[4,5]
E[T(n)] + 1 = 0.5*E[T(n+1)] + 0.5*E[T(n-1)]
after rearranging
E[T(n+1)] = 2*E[T(n)] - E[T(n-1)] + 2

for n=[6,9]
E[T(n)] + 1 = 0.5*E[T(n+1)] + 0.5*E[T(0)]
E[T(n+1)] = 2*E[T(n)] - E[T(0)] + 2

A basic recursion in 3 different parts. Only assumption to make or to know about is the prices of worth of a +0/6ap, of course discounting the real chance of xen success rate and omittng the chance of breaking.



lmao.:D:D:D crazy.

P17ch8l4cx
10-14-2008, 05:47 PM
there's an easier way to calculate said expectation,

t(n) + 1 xen -> p*t(s) + (1-p)*t(f)
where n is the current, s is the state when xenning successful, which is n+1, f is the state of failing, which depends on n

where n denotes the state of the equipment, or +n rather. Take the expectation of both sides

e[t(n)] + 1 = p*e[t(s)] + (1-p)*e[t(f)]

for n=[0,3]
e[t(n)] + 1 = 0.5*e[t(n+1)] + 0.5*e[t(n)]
after rearranging
e[t(n+1)] = e[(t(n)] + 2

for n=[4,5]
e[t(n)] + 1 = 0.5*e[t(n+1)] + 0.5*e[t(n-1)]
after rearranging
e[t(n+1)] = 2*e[t(n)] - e[t(n-1)] + 2

for n=[6,9]
e[t(n)] + 1 = 0.5*e[t(n+1)] + 0.5*e[t(0)]
e[t(n+1)] = 2*e[t(n)] - e[t(0)] + 2

a basic recursion in 3 different parts. Only assumption to make or to know about is the prices of worth of a +0/6ap, of course discounting the real chance of xen success rate and omittng the chance of breaking.

ahhhhhhhh too many numbers. My head hurts