Direct Hit; how much is "enough"?

[nand]

That being said, the real question in practice is more complicated to answer (and depends strongly on the game), because the constant factor on each stat plays an important role in determining the stat weights within a bounded region (say, from 0 to 3000). The previous image illustrates that perfectly - the only reason direct hit rating outscales critical hit rating for low stat amounts is because the constant factor on it is better. So in some games it's still best to ignore all stats except the best one, if that stat is so good that it's better than the rest even after the "opportunity cost"; and it seems like this might be the case for FFXIV.

I wanted to think some more about this problem. To use a simple model, say we have two linear multipliers with relative weights A and B, depending on the amount of stat (a) or (b) invested. Also, assume that we have constant offsets +1, i.e. the multiplier for stat A looks like 1 + A·a where a represents the amount of stat we have. Finally, to use the same general assumption as in the previous post, let's assume we have a bounded amount T of total stat available to us, and the only thing we control is the distribution x, i.e. a = x·T and b = (1-x)·T.

We now arrive at a result that looks like (1 + A·x·T)(1 + B·(1-x)·T). This expands (via WolframAlpha) into this second-order polynomial: -ABT²x² + ABT²x + (AT - BT)x + BT + 1

In order to simplify this formula somewhat so we can squint at it from a distance, let's combine T and x by rewriting it in terms of a = Tx again. We can also kill off the constant term because we're only interested in the local maximums of this polynomial, resulting in -ABa² + ABTa + (A-B)a. If we furthermore observe that T = a+b we can simplify this into AB·ab + (A-B)·a. Some observations:

1. For small T, the (A-B)·a term is what rewards us for having more of the better stat.
2. For large T, the AB·ab term is what punishes us for having either stat close to zero.

To figure out the best long-term ratio, we need only look at the limiting growth behavior, which depends only on the quadratic AB·ab term. Since AB is constant, we need only maximize a·b = T² · x · (1-x), which (by symmetry) is obviously maximal exactly when a = b, i.e. x = 0.5 - same as if the stats had been equally good.

To figure out the best ratio near T = 0, we need only look at the linear term, which is (A-B)·a = (A-B)·x·T (as a function of T), therefore the best slope is achieved by maximizing (A-B)·x, i.e. putting everything into the best stat.

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