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This question is mostly for hypothetical fun because setting all of these up isn't super practical (unless it's a 1-on-1 campaign perhaps?)

But with all of these stacking, how likely are you to be hit by an enemy? (let's assume they have Hexblade's Curse on them for Armor to work)

This probably depends on AC and to-hit from the enemy too, so let's just take a look at a decent 13 AC, with the enemy having a +7 to-hit. It's up to you if you add to hit or AC too to really look at how hard it is to hit something with absurd AC + all these defensive buffs, or how much high to-hit really helps.

Someone_Evil
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Cyanic
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1 Answers1

13

This is fairly straight forward to work through, because each effect is independent, meaning we can just multiply the individual probabilities together.

  • Blink gives a 50% chance for our caster not to be on the plane where they can be attacked. (Though of note, if we're not there, our assailant is free to attack someone else or Ready their action for our return. We're ignoring that.)

  • Mirror Image gives a \$\frac{1}{4}\$, \$\frac{7}{20}\$, or \$\frac{1}{2}\$ chance to be hit through depending on the number of duplicates. I'll be assuming we have 3, but it's fairly easy to lower that if desired.

  • Armor of Hexes gives another 50% to miss (that's what 4 or higher on a d6 means).†

  • Blur gives the resulting attack (should it go through everything else) disadvantage.

The remaining thing is to work out the chance to hit with +7 against 13 with disadvantage, which we can rephrase as the chance of two 2d20 both being 6 or higher (i.e. needs to be one of the 15 outcomes from 6 to 20). The probability of one being so is \$\frac{15}{20}=\frac{3}{4}\$ and the probability of both is \$\left(\frac{3}{4}\right)^2\$. Bringing everything together:

$$ \left(\frac{3}{4}\right)^2\times\frac{1}{4}\times\frac{1}{2}\times\frac{1}{2}=\frac{9}{256}\approx0.035=3.5\% $$

†: Armor of Hexes technically applies after you are hit, however this only affects the profitability you'll have to use your reaction, not the final probability for being hit.

Exempt-Medic
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Someone_Evil
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  • Might be fun to work out the integrals for the general case of +n to hit against AC k. – wzbillings Jun 11 '20 at 18:57
  • @wz-billings That sounds like it can be achieved by substituting (3/4)^2 with the answers to https://rpg.stackexchange.com/q/14690/48827. Although the general formula wouldn't be that complicated either. ((21-k+n)/20)^2, with some stipulations to constrain the value to between (1/20)^2 and (19/20)^2. – BBeast Jun 12 '20 at 00:59