I've been doing some back of the envelope calculation regarding crafting with the forge, and it seems to suggest that choosing the Mid value may actually be the best long term choice in conservation of crystals. Typically, the trend for crystals required is 1X for Low, 2X for Mid, and 3X for High.
Assuming Mid has a 70% chance of succeeding (the most costly ones at least), the expected value (number of crystals needed) looks like this:
0.7*2X + 0.3*0.7*4X + 0.3^2*0.7*6X + 0.3^3*0.7*8X + ...
where each term in this expression represents the probability of success on try number 1, 2, 3 and so forth. Mathematically, it would be:
summation[n, from 0 to infinity, (0.3^n)*0.7*(n+1)*(2X)]
Wolfram Alpha states this summation converges to 2.857X, which is less than the 3X required of the High forge cost.
It can also be expressed more generally as (1-P)^n * P * (n+1) * (2X) where P is the probability of a successful forging; for an 80% success rate, this converges to 2.5X.
These calculations don't take into account the fact that you could win a very valuable Prize Box for using the High amount though. Anyone want to back up my math and assumptions on expected value?
yes you have the maths correct, but there was a much quicker way of working it out
2 / 0.7 = 2.85714... which is just working out the expected value
and assuming 30% is the low option you would have 1 / 0.3 = 3.33333
but at 40% it's 1 / 0.4 = 2.5
so yes the mid option does seem like the best one if it is 70%
however I have noticed that it's not always 70% for the mid value
but i just want no hassles and a chance at the forge box.