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)