Editing Guides/Probability (section)

=== Amount of items from a fixed set of ingredients ===
Here, we'll look at how many items you can get from a fixed set of ingredients, say for 100 or ''n'' items if you try until you run out of ingredients.<br />
We're going to calculate this in steps:
* make ''n'' attempts to mix the item
** get the expected number of items (see expected value above)
** lose the expected number of items
* repeat with the left-over ingredients (from non-critical fails) until there are no more ingredients left.
The expected number of produced items in the first ''n'' attempts is ''n'' * p where p is your chance of success,
 p = (your level)/(2 * recommended level),
use 99% or some other value from your experience if your level is equal to or higher than 2 times the recommended level. The expected number of critical fails is 1/3 of the remaining ingredients leaving you with 2/3 of (n - n*p) items to attempt in the second run. In this second run we expect to get
 2/3 * (n - n*p)*p items,
lose 1/3 of the unsuccessful attempts and have
 2/3 of 2/3 * (n - n*p) - 2/3 * (n - n*p)*p = (2/3)^2 * (1 - p)^2 * n 
ingredients left, now we see that by induction that after the ''k''th repetition of our process we will have (2/3)^k * n * (1-p)^k ingredients left, we will lose 1/3 * (2/3)^(k-1) * n * (1 - p)^k, the total of items made is thus the sum from j=0 to k-1 of (2/3)^j * n * p * (1 - p)^j. The theory of [http://en.wikipedia.org/wiki/Series_(mathematics) infinite series] tells us, that the total of items made after infinite attempts (don't worry, this won't be necessary as there are no fractions of items) you will have
 n * p / (1 - (2/3 * (1-p))) items.
See below for [[#Examples|examples]].

==== Estimate of the necessary attempts ====
As there are no fractions of items, say a half [[Steel Bar|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made
 n * p * (1 - (2/3)^k(1 - p)^(k+1)) / (1 - 2/3 * (1 - p)) items.
To determine the step ''k'' at which to stop, we look for the ''k'' at which the left over ingredients are less than 1. Above we found that after ''k'' steps we have (2/3 * (1 - p))^k * n items left, which is less than one if 
 k > - ln(n)/ln(2/3 * (1 - p)).
The difference between number of attempts up to and including step ''k'' and ''n'' tells us how much extra food we'd have to buy if we want to use ''n'' ingredients. This number is, as it turns out, what we're really after, as we've been working with fractions of items and expected values the whole time and expecting some sort of guaranteed number of items would be neglecting the random nature of the underlying processes.

So, to be on the safe side, always work with conservative numbers here.

To determine the sum of steps taken, we figure that we made ''n'' attempts in the first round, and we had (2/3 * (1 - p))^k * n items left (k=1) after the ''k'' step, we attempt to make them in the second round, i.e. after the 2nd round we'll have made n + 2/3 * (1 - p) * n attempts, the same formula that we used above then gives us, that after ''k'' steps we will have made
 n * ((1 - (2/3 * (1 - p))^k ) / (1 - 2/3 * (1 - p)) ) attempts.
Since (1-p)^k tends to zero for large ''k'' we can simplify the last formula to
 n / (1 - 2/3 * (1 - p)) = number of produced items / p.

See below for [[#Examples|examples]].