average result from an open ended dice roll

[Nifft]

Nothing! In fact, as a professional mathematician, I couldn't agree more. It's just that after working out a solution to Feldspar's question, I was a bit surprised at how much it simplified after such a small change to the problem.

I'm sure I'm not the first to have thought of it, but a colleague & I were trying to come up with novel statistical questions to ask when we interviewed people, and that's one of mine.

Does it have an official name?

Cheers, -- N

 

[GSHamster]

I took a slightly different approach to the original question and got the same n/2 + n/(n-1) formula.

For Nifft's question, when I replace the term that represents the accumulation of 6s with one for 5s, things simplify quite a bit and I get: (n+2)/2, or 4 when n=6.

I thought this must be wrong (since it's so simple), so I double-checked my work and ran simulations for various die, each with 10^7 trials, and the results agree with the formula.

I can "show my work" if anyone's interested....

Bah, you're absolutely right. I made a silly mistake when summing. That -1 in my formula should be -2, which simplifies down to (n+2)/2.

 

[Feldspar]

when you roll a 6, you record a 5 and roll again

Walter: Mark it five, Dude. Next roll.
Smokey: What? I rolled a six ... gimme the white erase pen, I'm marking six.
Walter: This is not Nam, this is gaming. There are rules.

 

[Feldspar]

In fact, as a professional mathematician ...

So, in all of this, the thing that interested me the most was sort of stumbling onto how to find the average length of the chains, basically that:
sigma(from i =1 to infinity) 1/(n^i) converges to 1/(n-1).

I thought that was way interesting. Does that have a name? I barely remember anything from the class I took on infinite series besides the simple summation of a set from 1 to n.

So, I'll also mention that what started me down this path was the new save mechanism in 4th Ed D&D and wondering, on average, how many 'ticks' of the ongoing effect they'll take. Now, if they had written it as "10 or below fails" then I'd have the answer. Is there a quick easy way to solve sigma(from i=1 to infinity) (m/n)^i? If not, does anyone know the value for 9/20 specifically?

 

[Obryn]

You'd already know this if you'd played a lot of Earthdawn!

-O

 

[Flatus Maximus]

So, in all of this, the thing that interested me the most was sort of stumbling onto how to find the average length of the chains, basically that:
sigma(from i =1 to infinity) 1/(n^i) converges to 1/(n-1).

I thought that was way interesting. Does that have a name? I barely remember anything from the class I took on infinite series besides the simple summation of a set from 1 to n.

As malraux indicated above, sigma(i=0 to infinity){1/n^i}=1/(1-1/n) is the special case of the geometric series where a=1 and r=1/n. If you want to start the series at i=1, just subtract off the i=0 term's value from both sides of the given equation.

So, I'll also mention that what started me down this path was the new save mechanism in 4th Ed D&D and wondering, on average, how many 'ticks' of the ongoing effect they'll take. Now, if they had written it as "10 or below fails" then I'd have the answer. Is there a quick easy way to solve sigma(from i=1 to infinity) (m/n)^i? If not, does anyone know the value for 9/20 specifically?

As above, you have a geometric series (starting at i=1), this time with a=1 and r=m/n, so you get (m/n)/(1-m/n), which simplifies to m/(n-m). It should be noted that we must have |r|<1, otherwise the formula gives nonsense; assuming m and n are positive, you'd need m<n. Finally, if m=9 and n=20, then you'd get 9/11=.818181....

Regarding the average number of 'ticks': A random variable that gives the number of failed "Bernoulli trials" before the first success has a geometric distribution. Here, the Bernoulli trials are "roll a d6," success (which stops the process) is not rolling a 6, and failure (which keeps the process going) is rolling a 6. If r is the probability of success (here 5/6), then the average number of rolls is 1/r=6/5.

Hope that helps.

 

[Flatus Maximus]

I'm sure I'm not the first to have thought of it, but a colleague & I were trying to come up with novel statistical questions to ask when we interviewed people, and that's one of mine.

Does it have an official name?

Cheers, -- N

Asking interviewees novel statistical questions? I think that's called EVIL.

No, seriously, does what have an official name? if you're referring to the d5+1 open-ended rolling, then I don't think so, but I could be wrong....

 

[Jhaelen]

You'd already know this if you'd played a lot of Earthdawn!

Hehe, yup, I've been waiting for someone to mention Earthdawn.

I really like Nift's variant, though. I wish I would have thought of that when starting to play Earthdawn.
I also made a modification to Earthdawn's system: I eliminated the d20 from the progession. It was too 'swingy' for my taste, so the highest die you'd ever roll is a d12. The most complicated part was adjusting the table with target numbers for the degrees of success.

Otherwise I rather liked the mechanism from a mathematical point of view. Lots better than, say, Shadowrun...

 

[Wulf Ratbane]

Using Nifft's open-ended version, does this bring the average result of 2d6 up to 8? 3d6 to 12?

Or is the result more complicated?

 

[meomwt]

Here's another problem for the maths-heads.

Tunnels & Trolls used an open-ended rolling system for saving throws. Roll 2d6, if you got a double (any double, even snake-eyes), you got to roll again and add the two rolls together.

And as long as you rolled doubles, you could carry on rolling and adding.

What would be the average roll for that situation?