Which yields higher average damage, an Empowered or Twinned Chaos Bolt?

Question

Chaos bolt from UA had the nifty feature of bouncing targets in the event of your 2d8 damage rolling up doubles.

While some people complained about the relatively low damage output, no one highlighted the interesting interplay between chaos bolt and the empowered/twinned spell metamagics, that I saw.

Basically, this would seem to give you a much higher chance of a ricochet, since you can reroll either one of the two d8s on each hit, effectively a 25% chance of ricochet in addition to the higher average damage.

Conversely, Twinned Spell makes two separate attacks, and while each individual attack has a 1/8th chance to ricochet, you're still at a 15/64th's chance of at least one ricochet and already dealing more damage on average with the first attack.

Assuming you can only choose one of these two metamagics, which would yield the most average damage after accounting for ricocheting?

Assuming you could add both metamagics to the spell, what would be your average damage and number of ricochets, assuming you always select the lowest damage die to reroll?

Answer 1

Computing the ricochet rate is actually simpler than it looks:

• The chance of 2d8 NOT rolling doubles is (1*7/8)=7/8=87.5%
• If we get to choose whether to reroll one of the two dice we multiply that AGAIN by the chance of rolling doubles, giving us 7/8*7/8=49/64=~76.6%
• On a critical hit (5% of the time), you roll a total of 4d8, remembering that ANY double results in a ricochet.
• The chance of 4d8 NOT rolling doubles is (1*7/8*6/8*5/8)=41%
• If we get to choose whether to reroll one of the four dice we multiply that AGAIN by the chance of rolling doubles on the fourth die, giving us 41*62.5%= 25.6%
• The upshot is that a standard chaos bolt will ricochet 12.5% of the time, but an empowered chaos bolt will ricochet 23.4% of the time.
• On a critical hit, a standard chaos bolt will ricochet 59% of the time, while an empowered chaos bolt will ricochet at a whopping 74.4% rate!

In order to figure out HOW MANY times we'd ricochet on average, we use a version of the multiplier effect. Simply put, our average number of targets hit is 1/(chance of no ricochet).

• For a regular bolt, that's 1/(7/8)=8/7=~1.14, a twinned spell bolt multiplies this by 2 for 2.28
• For an empowered bolt with one reroll, that's 1/(49/64)=64/49=~1.31, 2.62 as a twinned spell
• For a crit from a regular bolt, that's 1/(.41)=2.44 targets, 4.88 as a twinned spell
• For an empowered crit, our average is 1/(.266)=3.9 targets, 7.8 as a twinned spell.

Assuming every attempted attack hits, your expected number of targets including the 5% chance of a crit is:

• (1.14*95%+2.44*5%)=1.21 for a regular chaos bolt
• (1.21*2)=2.42 for a twinned chaos bolt
• (1.31*95%+3.9*5%)=1.44 for an empowered chaos bolt
• (1.44*2)=2.88 for a twinned, empowered chaos bolt.

For our first two results, the expected damage is the number of targets multiplied by the average damage roll of 2d8 (which is 9). So a standard bolt deals 10.89 damage while a twinned bolt deals 21.78 on average.

Figuring out the average damage on the empowered bolt is... more complex, since you wouldn't reroll when your initial rolls were doubles, you have to compute the probability of a rerolled die exceeding the original damage die, and recompute the average value for the rerolled die.

For the case of an empowered hit from a chaos bolt, we can divide our calculation into three parts:

1. The expected value of rolling doubles
2. The expected value of either the die dropped or the die kept in the event doubles are not rolled
3. The expected value of the replacement die roll

Note that we don't reroll doubles, nor is there never a point where we don't reroll without them - we are optimizing for ricochet and thus will ALWAYS go for doubles.

1. This is fairly straightforward, since the probability of rolling doubles is 1/8 and the probability of rolling any particular set of doubles is 1/8th of that (1/64). The sum of (1+1)/64 + ... (8+8)/64 is identical to 2 times the average roll ((1+2+3+4+5+6+7+8)/8) times the probability of rolling doubles (1/8). The result is an EV of 9/8 for the doubles.
2. This gets a little bit more complicated. We need to compute the probability for each possible number that it will remain when the other die is dropped.
   • There is no chance for us to have a 1 remaining when dropping a die, since that would mean the other die was also a 1 and we would not choose to reroll.
   • For a 2, we could have rolled (2,1) or (1,2) out of 64 options, so we have a 1/32 chance that our highest die before reroll is a 2.
   • The probability formula for having a highest side k on two n sided dice is 2*(k-1)/n^2. Note that summing this formula across all sides and the probability of rolling doubles adds to 1, since you will either have doubles OR the remaining die.
   • The expected value of the die remaining after drop is 5.25
3. For the reroll, our expected value is the average roll (9/2) times the probability of not rolling doubles (7/8), which is 63/16, about 3.94

Adding all of these together, we get an EV of 10.31 for an empowered Chaos Bolt, a 15% increase in damage. But what about a critical hit? Computing the "lowest die" probability is a significantly more complicated problem that I'm still working on...

Placeholder until I can craft a computational model to explain the last part of the question.

Answer 2

Since a Twinned chaos bolt can't exist,^[1],[2] we can set its expected damage output to zero in any calculation.

That means that an Empowered chaos bolt (which can exist) is necessarily more optimal.