Side Bar: Let's work out the math for sorcerer empowered spell + flames of phelegothos
I want to do a fireball + firebolt sorcerer build and get the math right. I'm thinking empowered spell will be worth more damage than heightened but computing the reroll abilities is hard.
How does empowered and flames of phelogothos work together?
I think it depends whether you consider the rerolls you are doing to be "rolling damage", and whether "you must use the new rolls" overrides the text of the other ability that lets you reroll. If no to the first or yes to the second, FoP just lets you reroll 1s without counting those dice toward the CHA mod number of dice you get to reroll.
But I think it makes to allow the abilities to daisy chain: use empowered spell to reroll everything below average first, then any 1s that remain, reroll those using FoP.
Let's first ignore the limitations on the number of dice you can reroll, and work out the expected value of a single die using both abilities.
With Empowered alone, a d6 has a 1/2 chance of being rerolled, and a 1/2 chance of not being rerolled. If it's not rerolled, its expected value is 1/3 * (4 + 5 + 6) = 5. If it is rerolled, without FoP, its expected value is 3.5. So overall its expected value is 1/2 * 3.5 + 1/2 * 5 = 4.25.
With FoP alone, there's a 1/6 chance the die is a 1, and therefore gets rerolled, with an expected value of 3.5. There's a 5/6 chance it isn't, with an expected value of 1/5 * (2 + 3 + 4 + 5 + 6) = 4. So overall we get an expected value of 1/6 * 3.5 + 5/6 * 4 = 3.92.
Nesting FoP into Empowered, we get
1/2 * 3.92 + 1/2 * 5 = 4.46
But, if you're using empowered spell on a fireball, you might have more dice that are below average than you are allowed to reroll.
Suppose your CHA mod is 4.
Strictly speaking we need to consider each case separately, from 1 die below average to all 8, because conditioning on the number of dice you are rerolling changes the expected value of the remaining dice.
Dice that are rerolled have an expected value of 4.46; dice that aren't rerolled but could have been have an expected value of 5. Dice that aren't rerolled because we couldn't afford to reroll them have a really complicated expected value that depends on the number of other dice in this category, since we'll obviously reroll the lowest ones first.
The distribution of the number of dice, k, that we would want to reroll is just a binomial, with n = 8 and p = 1/2.
For k = 1 through 4, the expected value of the fireball is just:
k * 4.46 + (8 - k) * 5
For k = 5, the 5th die will usually be a 3 (probability nearly 87% there is at least one, from the Binomial(5,1/3)), but will sometimes be a 2 (probability nearly 13%). So ignoring the 0.4% chance that all 5 are 1s, the 5th die has an expected value of 0.87 * 3 + 0.13 * 2 = 2.87.
Then it gets hairier, since with k = 6, we need to consider the highest die and the second highest die separately...
To be honest, it gets complex enough at that point that it seems like just doing the simulation is the easiest way to go.
