The odds of succesfully tiering an UL item up

[Johned]

You are at 53% on success and 98% on Salvage....3% increase to Success rate = 3% increase to salvage rate?

I have no idea. I was even wondering if the item itself might have an effect on the calculation of salvage rate. I also wonder if someone can get to 100% salvage rate. I hope we will know before too long.

 

[Oleg]

step 1
1) you succeed with prob a, that's your n=0
or
2) you don't succeed but save the item with prob (1-a) * b
or
3) you do not succeed and lose the item with prob (1-a) * (1-b)

This is the part where you are going wrong, as per IceCold's post above.

The probability of 1) is a
The probability of 2) is b-a
The probability of 3) is 1-b

Note that a + (b-a) + (1-b) = 1

This is of course assuming that it does work in the way that Marco stated.

 

[falkao]

This is the part where you are going wrong, as per IceCold's post above.

The probability of 1) is a
The probability of 2) is b-a
The probability of 3) is 1-b

Note that a + (b-a) + (1-b) = 1

This is of course assuming that it does work in the way that Marco stated.

ah I see now. There is a problem with conception. What probabilities does MA really state?

Terms

we have 4 distinct events
S+ event of success S- event of failure
L- item saved
L+ item lost

with respective probabilities: P(S+),P(S-),P(L-),P(L+)

To identify them I've used the following text from entropiagateway.com:

First Check:

When attempting to upgrade an unlimited item to the next major tier, a profession check is made against the Tier Upgrader profession. If the check is within the Success Rate percentage range the item is upgraded to the next major tier. The Secondary item (the copy of the upgraded item) is salvaged fully (meaning you get to keep the item).

So what does this mean?
We have a check that is performed on success rate. The success rate is the one given on the item. This rate seems to depend on the tier upgrade profession. Now what is a check?
I've interpreted it that way, that a uniform random number is drawn. If it's value
is below the success rate then you succeed otherwise you'll fail.
Hence P(S+) = success rate and P(S-) = 1- success rate


In the case of S-,

If the check is over the Success Rate percentage but within the Salvage Rate percentage range the item is not upgraded to the next major tier. The Secondary item is salvaged (you keep the item).

Now this sentence is very imprecise. "If the check is over but within the salavage rate" implies a second check. As the first check was a comparison with a random number I did assume the same for the second check. Therefore I did assume something like P(L+|S-). Now I see that it might also be P(L-,S-) + P(S+).

The advantage for MA is that only one random number needs to be drawn (less computing power).


Assuming your assumption is correct we would have

P(S+) = a (with a success rate, p in my notation)
P(S-) = 1- a
P(L-,S-) + P(S+) = b (with b salvage rate)
P(L+,S-) = 1-b ((1-s)*(1-p) in my notation)

from P(L-,S-) + P(S+) = b it follows that
P(L-,S-) = b - a

Now lets transform notations:

I had p = success rate = a
and

(1-s) = P(L+|S-) = P(L+,S-)/P(S-) = (1-b)/(1-a)

s = P(L-|S-) = P(L-|S-)/P(S-) = (b-a)/(1-a)


therefore my probability to lose an item while upgrading

P = (1-s)(1-p)/(ps + 1 -s) transforms to

(1-b)/(a - b + 1)


You had

P = 1-Sum(a(b-a)^n),

the limit of Sum(a(b-a)^n), n=0 to inf is a/(a-b+1)

P = 1 - a/(a-b+1) = (1-b)/(a-b+1)

which is the same as I had.

Hence we both had the same solution. The difference was, that you did interpret 1-b as P(L+,S-) and I have seen it as P(L+|S-).

Btw) the same result follows from

probability to lose an item while upgrading

step 1) 1-b
step 2) (b-a)(1-b)
step n) (b-a)^(n-1)*(1-b)

Hence the total probability is

sum((b-a)^(n-1)*(1-b)), n = 1 ..inf which is again

P = (1-b)/(a-b+1)

 

[falkao]

unfortunately when using Olegs interpretation of given probabilities we end up with quite higher item loss probabilities.

SR	P(Upgrade)	P(Loss)
0.1	57.94%		42.06%
0.2	75.61%		24.39%
0.3	84.16%		15.84%
0.4	89.21%		10.79%
0.5	92.54%		7.46%
0.6	94.90%		5.10%
0.7	96.66%		3.34%
0.8	98.02%		1.98%
0.9	99.11%		0.89%
1	100.00%		0.00%

with SR = success rate
P(Upgrade) = probability to successfully upgrade the item without item loss
P(Loss) = probability of item loss during upgrade

 

[Coop_]

unfortunately when using Olegs interpretation of given probabilities we end up with quite higher item loss probabilities.

Just to further confirm Olegs interpretation and your new probabilities table:

:
Originally Posted by GameEducation
My understanding of Berthas+Marcos posts, and a humbe attempt to put it in different words:

One check, that we can think of as a 1-100 roll.

The combined success rate and salvage rates placed on a single 1-100 scale gives three ranges, the 1-50, 51-95 and 96-100 range (using current unskilled numbers).

Roll result gives three possibilities regarding the main issues:
1-50: Item upgraded, secondary item kept.
51-95: Item not upgraded, secondary item kept.
96-100: Item not upgraded, secondary item lost.
Is this way of putting it correct?

Yes, that is a good summary of the process.

 

[falkao]

Just to further confirm Olegs interpretation and your new probabilities table:

I guess that Olegs interpretation is the correct one as MA might have implemnted in that way to reduce computing cost (instead of drawing 2 numbers they use the same). The formula is the same but the parameter values are different.

When we assume that with (1 - Salvage rate) = P(L+,S-) the probabilty to fail and lose the item is given, then the probability to lose the item in the case of a failure is

P(L+|S-) = P(L+,S-) / P(S-)

Hence, with a salvage rate of 95% and success rate of 50%
P(L+|S-) = 10% in the first attempt. Therefore the new list has quite higher loss rates.

 

[Ace Flyster]

I am confused again surely if the following is correct.....

Roll result gives three possibilities regarding the main issues:
1-50: Item upgraded, secondary item kept.
51-95: Item not upgraded, secondary item kept.
96-100: Item not upgraded, secondary item lost.

Then the chance of losing an item is just 4/100

Created from a random number between 1 & 100

There for if someone has a 53% success and 98% chance of salvage

Then chance of success = 53%
Chance of Failure = 47%
Chance of losing second item = 2%

??

Rgds

Ace

 

[Oleg]

P = (1-b)/(a-b+1)

Agreed, good work

My rusty maths let me down when trying to simplify the infinite sum

I thought I used the sum formula on my figures in post 4 but I must have made an error because they don't match what I get using either formula now. They are close though.

So the probability of destroying an item before succeeding the upgrade with 54%/98% chances is 3.57%.

And if you do a whole armour set, the chance of destroying at least one piece is 22.48% [ = 1-(a/(a-b+1))^7 ]

 

[Ace Flyster]

Agreed, good work

My rusty maths let me down when trying to simplify the infinite sum

I thought I used the sum formula on my figures in post 4 but I must have made an error because they don't match what I get using either formula now. They are close though.

So the probability of destroying an item before succeeding the upgrade with 54%/98% chances is 3.57%.

And if you do a whole armour set, the chance of destroying at least one piece is 22.48% [ = 1-(a/(a-b+1)^7) ]

Oh crap you guys are talking cumulative

I think i will leave it to the maths masters

Rgds

Ace

PS thanks for the hard work

 

[Oleg]

I am confused again surely if the following is correct.....

Roll result gives three possibilities regarding the main issues:
1-50: Item upgraded, secondary item kept.
51-95: Item not upgraded, secondary item kept.
96-100: Item not upgraded, secondary item lost.

Then the chance of losing an item is just 4/100

Created from a random number between 1 & 100

There for if someone has a 53% success and 98% chance of salvage

Then chance of success = 53%
Chance of Failure = 47%
Chance of losing second item = 2%

??

Rgds

Ace

Not sure what you're asking here. The 53/47/2 is correct but remember that the 2% is included in the 47%. Also remember that this is the chance on a single click. What we are looking at here is the total chance of destroying the item before succeeding in the upgrade, which will require an unknown number of clicks.

 

[falkao]

Agreed, good work

My rusty maths let me down when trying to simplify the infinite sum

Wolfram Alpha does it for free

I thought I used the sum formula on my figures in post 4 ...

Must have overseen them.

So the probability of destroying an item before succeeding the upgrade with 54%/98% chances is 3.57%.

yep

And if you do a whole armour set, the chance of destroying at least one piece is 22.48% [ = 1-(a/(a-b+1)^7) ]

the 22.48% is correct but the formula is 1-[1-(a/(a-b+1)]^7

 

[falkao]

I am confused again surely if the following is correct.....

Roll result gives three possibilities regarding the main issues:
1-50: Item upgraded, secondary item kept.
51-95: Item not upgraded, secondary item kept.
96-100: Item not upgraded, secondary item lost.

Then the chance of losing an item is just 4/100

no, the chance per click in your example is 5/100 as the 96 in 96-100 is included.

 

[Oleg]

the 22.48% is correct but the formula is 1-[1-(a/(a-b+1)]^7

That one's not right either is it?

I misplaced a bracket in the previous post, it should be 1-(a/(a-b+1))^7

 

[falkao]

What is the best form to show the formula as we know now that with

a = success rate
b = salvage rate

P = probability to destroy secondary item till upgrade

is

P = 1 - a/(a-b+1) = (1-b)/(a-b+1)

Should we use
P = 1 - a/(a-b+1)

or

P = (1-b)/(a-b+1)

Both do need 5 numbers to be inserted and both do use the same number of a's or b's. There are less parenthesis in the first and there is the advantage to get the probability of success as well.

 

[Oleg]

P = 1 - a/(a-b+1) looks good to me

 

[falkao]

That one's not right either is it?

I misplaced a bracket in the previous post, it should be 1-(a/(a-b+1))^7

that's correct, I forgot a 1 as well Holly shit!!

1-[1 - 1 - (a/(a-b+1))]^7 = 1-[a/(a-b+1)]^7

 

[falkao]

I have no idea. I was even wondering if the item itself might have an effect on the calculation of salvage rate. I also wonder if someone can get to 100% salvage rate. I hope we will know before too long.

It was mentioned that the success rate is related to the Untier Profession. Higher standing = higher success rate. Since success rate is going up, also salvage rate has to go up sooner or later or success rate is capped at salvage rate.

 

[Immortal]

I guess that Olegs interpretation is the correct one as MA might have implemnted in that way to reduce computing cost (instead of drawing 2 numbers they use the same). The formula is the same but the parameter values are different.

When we assume that with (1 - Salvage rate) = P(L+,S-) the probabilty to fail and lose the item is given, then the probability to lose the item in the case of a failure is

P(L+|S-) = P(L+,S-) / P(S-)

Hence, with a salvage rate of 95% and success rate of 50%
P(L+|S-) = 10% in the first attempt. Therefore the new list has quite higher loss rates.

congratulations on catching up hence why I used a+b+c=1 a=success, c=(1-salvage) b=salvage-success

before the 'bug' was removed where we only knew that 50/95 existed, I had done a previous calculation based upon getting to tier V https://www.planetcalypsoforum.com/...1-how-tier-unlimited-item-19.html#post2228722

and least now we agree on both the maths, and woflramalpha

 

[falkao]

A formal summary.

We have to solve a state model where there are two absorbing states
S1) Item upgraded or
S2) secondary item lost

and one intermediate state

S3) item not upgraded and secondary item saved.

With the given success rate a and salvage rate b the following probabilities are defined:

P(Item Upgraded|S3) = a
P(Item Lost|S3) = 1-b

P(Item saved|S3) = b-a

What we are looking for are the terminal probabilities P(Item Upgraded) and P(Item Lost).

The solution so far was to calculate the limiting probabilities of every path from 1 to inf. This took quite some time. We could have been much faster as follows.

As there are only two absorbing states (i.e. P(Item saved) = 0) the total terminal probability will be the sum of P(Item Upgraded|S3) and P(Item lost|S3)) =

a + (1-b)

The terminal probabilities must sum up to 1 and hence we will have

P(Item Upgraded) = a/(a + 1 -b)
P(Item Lost) = (1-b)/(a + 1 -b)


Please observe that when b = 1
P(Item Upgraded) = 1 and P(Item Lost) = 0.

This is what a terminal model is.

The expected mean number of attempts to reach a final state is

1/(a +1 -b),

and to succeed (indifferent from any possible item loss) is 1/a.