Exploding Dice Averages

[Yunru]

So I've been slowly recreating DnD from the ground up, and I've been looking into exploding dice.
You'll have to do the summing yourself, but here it is, the formula for working out the average of a dX that explodes on Y numbers:
http://www.HostMath.com/Show.aspx?Code=\sum_{n=0}^\infty \frac{y^n}{2x^{n+1}}(2nx+1+x-y)(x-y)

EDIT: I've made a mistake early on before simplifying, gimme a sec to correct.

EDIT 2: Fixed. Doesn't simplify as much any more though :/

EDIT 3: Gods damn it, no it's not.

EDIT 4: Done. Did the entire thing again from scratch.

EDIT 5: it, I give up :/

EDIT 6: Got it?

 

[FrogReaver]

Isn’t it just a geometric series with the mean of the dice as a and r being the chance you have to explode?

 

[Yunru]

Isn’t it just a geometric series with the mean of the dice as a and r being the chance you have to explode?

Don't know what half of those words mean.

For simplicity, a die that explodes is treated as it's highest value by the way.

So we've got:

(1/X)(1+...+(X-Y)+ (Y/X)(1+...+(X-Y)+ (Y/X)(1+...+(X-Y)+ (Y/X)(1+...+(X-Y)+ (Y/X)(...))))

But I keep failing at turning that into a Sum to Infinity... that's actually accurate.

 

[cbwjm]

That looks crazy complicated. I'm not really a math guy but looking at that, no wonder it was difficult to get right. I spent something like 20 minutes the other day trying to understand and simplify the equation to calculate the area of a hex. Trying to work out the average of an exploding die would be well beyond me.

 

[FrogReaver]

Look up geometric series

 

[Yunru]

Gods be damned that took too long. I kept misplacing nx.

 

[FrogReaver]

Now that I'm on my computer:

The Mean of an exploding d6 when the dice explodes on 6:

=1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 +6*1/6 + 1/6(1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 +6*1/6 + 1/6(…))
= 3.5 + 3.5*1/6 + 3.5*(1/6)^2 + … (This is a geometric series and the sum is defined below)
= 3.5 * 1/(1-1/6)
= 3.5*6/5
= 4.2

If you were to explode on a 5 or 6 then:
= 3.5 * 1/(1-2/6)
=3.5*6/4
=5.25

So then in general the mean of an n sided exploding dice that explodes on x numbers with mean m of normal non-exploding n sided dice is:
m * n/(n-x)

 

[MechaPilot]

Now that I'm on my computer:

The Mean of an exploding d6 when the dice explodes on 6:

=1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 +6*1/6 + 1/6(1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 +6*1/6 + 1/6(…))
= 3.5 + 3.5*1/6 + 3.5*(1/6)^2 + … (This is a geometric series and the sum is defined below)
= 3.5 * 1/(1-1/6)
= 3.5*6/5
= 4.2

If you were to explode on a 5 or 6 then:
= 3.5 * 1/(1-2/6)
=3.5*6/4
=5.25

So then in general the mean of an n sided exploding dice that explodes on x numbers with mean m of normal non-exploding n sided dice is:
m * n/(n-x)

Thank you for presenting the math in a more easily functional format.

I'm surprised the increase in average damage isn't that large.

I had been toying with the notion of replacing the natural 20 crit with exploding weapon damage, and people have warned me that the difference would be quite significant because the maximum value of a damage die will pop up more often than a natural 20.

Armed with this knowledge I might just attempt to swap natural 20 crits with exploding damage dice.

 

[Yunru]

Now that I'm on my computer:

The Mean of an exploding d6 when the dice explodes on 6:

=1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 +6*1/6 + 1/6(1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 +6*1/6 + 1/6(…))
= 3.5 + 3.5*1/6 + 3.5*(1/6)^2 + … (This is a geometric series and the sum is defined below)
= 3.5 * 1/(1-1/6)
= 3.5*6/5
= 4.2

If you were to explode on a 5 or 6 then:
= 3.5 * 1/(1-2/6)
=3.5*6/4
=5.25

So then in general the mean of an n sided exploding dice that explodes on x numbers with mean m of normal non-exploding n sided dice is:
m * n/(n-x)

How would rerolling all 1s and 2s be factored into it?
It effective turns 1dx into 2+1d(x-2), but how to account for that?

EDIT: Ah! Just use a higher average.

 

[FrogReaver]

How would rerolling all 1s and 2s be factored into it?
It effective turns 1dx into 2+1d(x-2), but how to account for that?

EDIT: Ah! Just use a higher average.

You would also need to adjust the n/(n-x) part. Rerolling 1's and 2's like with GWM also increases your chance of landing on one of the exploding numbers. So you would just need to do 1/(1-x) where x is your chance of landing on an exploding number with rerorolling 1's and 2's taken into account.