Average of exploding dice?

[freyar]

Why? ... People* don't want to know why when the solution deals with mathification. They just want an answer.

EDITOR'S NOTE: Be careful that when you post generalizations about "people," someone will assume you mean "ALL people" instead of "SOME people." So use modifiers as often as possible to avoid that confusion. Or better yet, stop generalizing. ... People do that far too often.

Well, SOME people are ridiculous math nerds.

Edit: And, besides, who says I'm not trying to summon Cthulu?

 

[megamania]

(exploding dice) x sides / average = accedrilin 12 headache

 

[Twowolves]

I'm pretty sure www.savagepedia.com has a link that has all this worked out for you.

I'd include the link, but I already sealed the envelope.

 

[freyar]

=====Start argument====
Let X be the expected value using a D-sided exploding die.

X=(D+1)/2+X/D -- First term is the expected value of a single roll, second addresses the "exploding" part.

Solve for X.
=====End argument====

Huh, how does that second term work? I must think in too straightfoward a manner.

By the way, I've tried to solve it the way freyar did, and never managed it. Very nice!

Thanks, though I'm much more interested in slick ways to solve things! I just used brute force.

 

[InVinoVeritas]

The average roll of an exploding die N is equal to the average roll of a regular N-1 die, N-1/N of the time, plus (N plus the average roll of an exploding die N), 1/N of the time.

The average roll of a regular N-1 die is (N-1+1)/2, or N/2.

Let X(N) be the average roll of an exploding die N.

Then,
X(N) = N/2 * (N-1)/N + (N + X(N)) * 1/N
X(N) = (N-1)/2 + (N + X(N)) * 1/N
X(N) = (N-1)/2 + 1 + X(N)/N

(N-1)/N * X(N) = (N-1)/2 + 1
(N-1)/N * X(N) = (N-1+2)/2
(N-1)/N * X(N) = (N+1)/2

X(N) = (N+1)/2 * N/(N-1)

For a d6, that's

X(6) = 7/2 * 6/5 = 42/10 = 4.2

X(20) = 21/2 * 20/19 = 420/38 = 11.05263

It's freyar's result, without the geometric series.

 

[Pyrex]

Unless I'm misunderstanding your math IVV, you're only accounting for the first "explosion" of a die (i.e., values 7-12 on a d6) without accounting for the value of multiple "explosions" (i.e., roll 6,6,6,4)

Granted, only the first (mabye the second) are really all that significant...

 

[Numion]

Unless I'm misunderstanding your math IVV, you're only accounting for the first "explosion" of a die (i.e., values 7-12 on a d6) without accounting for the value of multiple "explosions" (i.e., roll 6,6,6,4)

Nah, he's right. The math corresponds to his sentence:

"The average roll of an exploding die N is equal to the average roll of a regular N-1 die, N-1/N of the time, plus (N plus the average roll of an exploding die N), 1/N of the time."

The underlined, expanded, is the whole sentence again. And the same part of that expanded sentence can be expanded again. And again, ad infinitum. The math simplifies the answer since geometric series have a simple analytic answer.

Granted, only the first (mabye the second) are really all that significant...

True dat.

 

[Driddle]

It's the averaging aspect that would seem to make this pretty simple to figure out, as I understand it. Of course I may be wrong, because I'm no longer a mathematician and I haven't put a LOT of thought into it...

But we're looking at an average die value of 3.5, right? And then 1/6th of that -- representing an "exploding" result. And then 1/6th of 1/6th of 3.5 on top of that. And 1/6th of 1/6th of 1/6th on top of that, etc. It doesn't take long for the iterations to whittle down to an insignificant fractional difference.

Eh?

 

[Agamon]

Okay, smarty pantses, what are the odds your die will explode when you roll it?

For those that don't know, I mean blow up good.

 

[Numion]

But we're looking at an average die value of 3.5, right? And then 1/6th of that -- representing an "exploding" result. And then 1/6th of 1/6th of 3.5 on top of that. And 1/6th of 1/6th of 1/6th on top of that, etc. It doesn't take long for the iterations to whittle down to an insignificant fractional difference.

Eh?

When you're calculating your WFRP (which uses exploding dice) Free Lance's damage output, you want to get your bragging rights correct, no matter how infinite the calculations