Average of exploding dice?

[John Q. Mayhem]

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]

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]

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]

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]

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]

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]

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]

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]

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]

Thanks, y'all! I appreciate it.