Would you want 'Advantage' on damage die if they exploded?

[Vogie]

The traditional exploding die is rolled again whenever its highest value is rolled.

If your character(s) class, subclass, feat or feature allowed your damage dice to explode, would you prefer to have advantage (roll twice, choose the highest) on the attack roll (as in 5e) or on the damage roll (a la Savage Attacker)?

[CheddarChampion]

For spellcasters, advantage on damage rolls, hands down. Unless you're talking only spells with attack rolls?

Can these dice explode multiple times? 6->6->3 or the like on 1d6?

For the fun factor, I'd probably go with advantage on damage rolls, the great weapon fighting style, and a greatsword. Rogues now have a competitor for most d6's rolled in one hit.

Speaking of rogues, advantage to hit all the time will proc sneak attack, so I'd take that as a rogue.

Finally, advantage to hit won't stack with other sources of advantage to hit. You'd be better off with advantage to damage in those cases.

For white room theorizing, the math is beyond my capacity during lunch break with only my phone.

[Composer99]

On the damage rolls. You already automatically hit on a roll of 20.

[sithlordnergal]

Damage rolls, no question about it. I tend to play Soradins, and when they smite, they smite hard. One in particular has a Rod or Lordly Might, which allows me to potentially paralyze targets for auto-crits. As such, when I do get those auto-crits, I tend to use Quicken spell to roll things like 2d8 slashing + 4d6 fire + 10d8 radiant + 6d8 thunder. That's a grand total of 18d8+4d6, I would LOVE to see those dice explode and let me roll even MORE as I blast things into oblivion.

[Lunali]

Speaking of rogues, advantage to hit all the time will proc sneak attack, so I'd take that as a rogue.

Really? I'd go the other way on rogue, if you're having trouble getting sneak attack on rogue, you aren't trying hard enough. Being able to explode the sneak attack dice would be massive.

[JNAProductions]

For reference, a d6 that explodes with advantage is worth about 4.6 damage, if my rough math is accurate.

[Reevh]

For reference, a d6 that explodes with advantage is worth about 4.6 damage, if my rough math is accurate.

I get 4.0833, assuming it can only explode once. AVERAGE(1,2,3,4,5,6+3.5), right?

[JNAProductions]

I get 4.0833, assuming it can only explode once. AVERAGE(1,2,3,4,5,6+3.5), right?

I assume it can explode infinitely, but that doesn't affect the math much.

And you forgot to factor in the advantage on the die roll.

Edit: Come to think of it, I kinda forgot to too! The result should be MORE than 4.6!

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Okay. So each d6 with advantage is 4.472, 2 is repeating.

So with an 11/36 chance of getting a 6, that means we add (11/36)*4.472, which is 1.367.

With an 11/36 chance of getting a 6 the second time, we add (11/1,296)*4.472, which is .038.

Further ones are so small as to not be worth adding, I think.

5.877 is the result, but we'll round up to the nearest 10th to represent the other explosions.

5.9

BAD MATH! IGNORE!

[Teaguethebean]

I assume it can explode infinitely, but that doesn't affect the math much.

Well technically if it can explode infinitely then the average is infinity damage as if there is one scenario that does infinity damage that is the average no matter what.

[JNAProductions]

Well technically if it can explode infinitely then the average is infinity damage as if there is one scenario that does infinity damage that is the average no matter what.

That's not how limits work.

As damage approaches infinity, the odds of achieving that damage grow infinitely smaller.

Edit: My math person says the result is 6.44.

And I did my math wrong. Redoing! (My error was that I used 11/36 and 11/1,296, NOT (11^2)/1,296.)

Edit II: Real math!

4.472+((11/36)*4.472)+((121/1296)*4.472)+((1331/46656)*4.472)+((14641/1679616)*4.472)=6.423

Which means 6.44 is probably close to the limit.

[Teaguethebean]

That's not how limits work.

As damage approaches infinity, the odds of achieving that damage grow infinitely smaller.

It is true the chance becomes infinitely small but if a single instance is infinity than the average also becomes infinity though logically we can disregard the chance.

[JNAProductions]

It is true the chance becomes infinitely small but if a single instance is infinity than the average also becomes infinity though logically we can disregard the chance.

What's ∞/∞? Because that's what you're looking at, basically.

I'm not very good at calculus, but I know enough to know that you're incorrect in this instance.

[Teaguethebean]

What's ∞/∞? Because that's what you're looking at, basically.

I'm not very good at calculus, but I know enough to know that you're incorrect in this instance.

Fair point. Thank you for informing me. I hadn't thought of that all this time.

[JNAProductions]

Fair point. Thank you for informing me. I hadn't thought of that all this time.

No problem! Math is hard, and probably more relevant to this, unintuitive! It's easy to goof up because something will SEEM correct, when in fact it is not.

[Aelyn]

Further ones are so small as to not be worth adding, I think.

5.877 is the result, but we'll round up to the nearest 10th to represent the other explosions.

5.9

The best way to manage these iterations is to say that the damage (x) of one exploding d6 is:

x = 1*(1/36) + 2*(3/36) + 3*(5/36) + 4*(7/36) + 5*(9/36) + 6*(11/36) + x*(11/36)

Or

25x/36 = 1/36+6/36+15/36+28/36+45/36+66/36
25x = 161
x=6.44

Yup, an infinitely-exploding d6 with advantage does, on average, more than 6 damage.

(If you're wondering why it's that much higher than your approximation, the error in your maths is that the second explosion is 121/1296, not 11/1296)

EDIT: and I see you've now done all that before I could.

[Reevh]

That's not how limits work.

As damage approaches infinity, the odds of achieving that damage grow infinitely smaller.

Edit: My math person says the result is 6.44.

And I did my math wrong. Redoing! (My error was that I used 11/36 and 11/1,296, NOT (11^2)/1,296.)

Edit II: Real math!

4.472+((11/36)*4.472)+((121/1296)*4.472)+((1331/46656)*4.472)+((14641/1679616)*4.472)=6.423

Which means 6.44 is probably close to the limit.

If it can only explode once, I come to 5.83866...

[NNescio]

That's not how limits work.

As damage approaches infinity, the odds of achieving that damage grow infinitely smaller.

Edit: My math person says the result is 6.44.

And I did my math wrong. Redoing! (My error was that I used 11/36 and 11/1,296, NOT (11^2)/1,296.)

Edit II: Real math!

4.472+((11/36)*4.472)+((121/1296)*4.472)+((1331/46656)*4.472)+((14641/1679616)*4.472)=6.423

Which means 6.44 is probably close to the limit.

6.44 is exactly the limit (of convergence for this particular geometric series). Well, at least for the way you defined the problem.


The best way to manage these iterations is to say that the damage (x) of one exploding d6 is:

x = 1*(1/36) + 2*(3/36) + 3*(5/36) + 4*(7/36) + 5*(9/36) + 6*(11/36) + x*(11/36)

Or

25x/36 = 1/36+6/36+15/36+28/36+45/36+66/36
25x = 161
x=6.44

Yup, an infinitely-exploding d6 with advantage does, on average, more than 6 damage.

(If you're wondering why it's that much higher than your approximation, the error in your maths is that the second explosion is 121/1296, not 11/1296)

EDIT: and I see you've now done all that before I could.

The other, but related way is to evaluate it as a infinite geometric series (E = 4.2 * 36. Which might be more intuitive for some people (especially those who remember a/(1-r) ). It's effectively the same thing anyway, because the sum of an infinite geometric sequence uses the same recursive trick to evaluate a self-similar expression.

As for the general case for any arbitrary dX:

E(explode(max(dX,dX))) = E(max(dX, dX)) / (1 - [2X - 1]/X^2)
= (4X - 1)(X + 1)/6X * X^2 / (X-1)^2
= X(4X - 1)(X + 1)/6(X - 1)^2

As for the case of a simple exploding die, without 'advantage' or taking max/mins:

E(explode(dX)) = X(X + 1)/2(X - 1)

Though, from the way thread OP specified 'advantage', which is to say it would work like Savage Attacker, the "take the highest part" would come after exploding die, not before. In which case I think the average (expected value) is around ~5.8 instead of 6.44 exact. There's probably an elegant way of calculating this analytically but right now I'm too exhausted to rearrange infinite nested sequences.