What happens if a charakter is affected by a misfortune hex and uses Perfect Strike?
How many attack rolls would a 10th-level zen archer or weapon adept make and which one applies?
Perfect Strike lets you re-roll and take the highest once per round.
Misfortune makes you re-roll and take the lowest any time you roll.
So here's how I see it going down.
Guy rolls attack.
Guy re-rolls attack from Perfect Strike. He takes the higher number.
Guy re-rolls because of Misfortune. He takes the lowest number of the three.
Seems the simplest way to go about it without houseruling. Misfortune's wording is very hard as opposed to Perfect Strike. He MUST take the lowest roll, rather than he MAY take the highest roll.
My personal houserule would be that they cancel each other out. Guy rolls once, normally.
As I see it, Perfect Strike allow you to roll two d20s and choose the best.
Each of these d20s are rolled twice, due to misfortune.
So he rolls two pairs of d20, choosing that lowest of the pair. Afterwards he get that highest result between the two pairs.
This way, the effect of both abilities apply. It might be more functional to have them cancel eachother out, but it doesn't take Zen Archer Perfect Strike into account.
Guy rolls attack.
Guy re-rolls attack from Perfect Strike. He takes the higher number.
Guy re-rolls because of Misfortune. He takes the lowest number of the three.
I thouhgt of that, but you could also argue it's the other way around. First re-roll because of Perfect Strike and then because of Misfortune.
My personal houserule would be that they cancel each other out. Guy rolls once, normally.
This sounds like a good idea.
So he rolls two pairs of d20, choosing that lowest of the pair. Afterwards he get that highest result between the two pairs.
This works too, but a 10th-level zen archer ends up with rolling two sets of 3d20, so chances are high for getting a decent result. Plus, you have to roll 6 dice for a single attack. Might be closer to the rules, however.