# Average of exploding dice?

#### John Q. Mayhem

##### Villager
How do you work out the average roll on an exploding die? I'd really appreciate it if someone better with probability could give me a hand here For those who don't know, with exploding dice if you roll the maximum, you roll again and add it until you stop rolling the max.

#### R_kajdi

##### Villager
John Q. Mayhem said:
How do you work out the average roll on an exploding die? I'd really appreciate it if someone better with probability could give me a hand here For those who don't know, with exploding dice if you roll the maximum, you roll again and add it until you stop rolling the max.
Mathematically, you'd have to solve this problem by infinite series. Though I will say that you could just interate throught the first 4-5 terms and get a pretty close approximation.

The formula you'd use is the following:

D = # of sides on the die
A = average result ( (D+1)/2 for all platonic dice)

Summing from n=0 to infinity

A/(D^n)

Hopefully this should help. It should work as a formula even for relatively odd dice like doubling subes and the like, as long as you correctly determine the average result.

#### Fifth Element

##### Villager
Well, the average of an exploding d6 is about 4.2 based on some back-of-the-envelope calculations, though I can't provide a formula for it or guarantee that it's correct. It should involve limits though. My last math class was a long time ago...

#### Fifth Element

##### Villager
R_kajdi said:
Mathematically, you'd have to solve this problem by infinite series. Though I will say that you could just interate throught the first 4-5 terms and get a pretty close approximation.

The formula you'd use is the following:

D = # of sides on the die
A = average result ( (D+1)/2 for all platonic dice)

Summing from n=0 to infinity

A/(D^n)

Hopefully this should help. It should work as a formula even for relatively odd dice like doubling subes and the like, as long as you correctly determine the average result.
For n=0 to n=5, the sum for a d6 is 4.1995.

#### R_kajdi

##### Villager
Fifth Element said:
For n=0 to n=5, the sum for a d6 is 4.1995.
Yup. After a few interations, you can get pretty close to the convergence point. Certainly close enough for you to make a smart guess if you are sure it will converge instead of slowly wandering off into infinity like an angel's horn.

#### Numion

##### Villager
For a single dice it's

avg(exploding dice) = average(dX)/(1-P(max))

where average(dX) is the average result of the dice used, non-exploding, and P(max) is the probability of throwing the maximum.

So, for d6 it's 3,5/(5/6) = 4,2, d8 = 4,5/(7/8) = 5,1428 etc..

#### freyar

##### Explorer
The answers given so far end up being right, but I don't think the above posts have made it clear why. The average should be given by the value of a given result times the probability of that result, summed over all possible results. Let's break it down.

The probability that you roll max (on one die) N times but not max on the N+1st time is (1/D)^N*(D-1)/D for a D-sided die.

The result we should get for N max rolls is D*N + (D/2), where the D/2 is the average value of the last non-max roll (this is the same as the average roll of a D-1 sided die).

So we need the sum of this from N=0 to infinity. For simplicity, call x=1/D. Then we have
(1-x)*Sum[ x^N (N*D+D/2)].
Break the sum into two terms. The second is just a geometric series:
(1-x)*Sum[ x^N D/2] = (1-x)*(1-x)^-1 D/2 =D/2.
The first part is
(1-x)*Sum[ x^N N*D] = (1-x)*D*x*d(Sum[x^N])/dx = D*x*(1-x)/(1-x)^2 = D*x/(1-x).
Putting it all together, we have
D/2 + D*x/(1-x) = D/2 + D/(D-1) = D*(D+1)/2(D-1).
In the end, this is the same as results given by the other posters. For a d6, you get 4.2, only a little higher than average for a single roll, which is because it's hard to roll max more than once.

#### Driddle

##### Villager
freyar said:
The answers given so far end up being right, but I don't think the above posts have made it clear why. ...

((arcane equations to summon Cthulu))

... In the end, this is the same as results given by the other posters. For a d6, you get 4.2, only a little higher than average for a single roll, which is because it's hard to roll max more than once.
Why? ... People* don't want to know why when the solution deals with mathification. They just want an answer.

#### brehobit

##### Villager
There is a slightly easier way to solve this using a (simple) recurrence, but it works out to be the same as freyar's solution just skips a few steps...

(Oddly, I was just talking about this to a theory professor yesterday!)

=====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====

It is interesting that the expected value is the _same_ no matter what number you reroll on (highest, lowest, middle, whatever). This formula also makes it easy to do exploding dice if you do it for more than one number.

X=(D+1)/2+NX/D

Where N is the number of numbers that cause a re-roll.

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

Mark

#### John Q. Mayhem

##### Villager
Thanks, y'all! I appreciate it.

#### freyar

##### Explorer
Driddle said:
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?  Last edited:

#### megamania

##### Community Supporter
(exploding dice) x sides / average = accedrilin 12 headache #### Twowolves

##### Villager
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

##### Explorer
brehobit said:
=====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

##### Villager
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.

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#### Pyrex

##### Villager
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

##### Villager
Pyrex said:
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

##### Villager
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

##### Adventurer
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

##### Villager
Driddle said:
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 