average result from an open ended dice roll

[Feldspar]

Some game systems make use of open ended dice rolling - that is, everytime you roll the max you get to roll again adding to the total. If the system called for you to roll 2d6 open ended you could conceivably roll:



for a grand total of 23.

This got me wondering about how to calculate the average value that an open ended roll produces. For example, we all know that the average value produced by rolling a d6 is 3.5 ... but what is the average value produced by rolling a d6 open ended? And what is the value for other dice types?

So ... after fiddling around with numbers on some scratch paper for a little bit I think I've found a general formula. I'm sure that this site has some grognards who *know* the answer for sure and I'm hopeing some will see this post and can tell me if I'm right or correct me if not.

Anyway, my belief is that the general function for the average open ended roll of an n-sided dice is: n/2 + n/(n-1)

For a six sided dice, that formula produces a result of 4.2 - and funny enough that's the value I get from a computer program I wrote a long time ago to sim open ended d6 roles.

The first part of the formula (n/2) is the average for the terminator roll, the one that failed to carry the series forward.

The second part represents the contribution of the max value chaining along. So we have that max value * the average length of the chain. I aver (and I have only some sketchy scribbled calculations and faith to back it up) that the average length of chain of events with a probability of 1/n approaches 1/(n-1) as you push towards infinity.

So ... am I right?

 

[Nifft]

I don't see anything wrong there. Someone more fresh in the math can tell you exactly which Poisson thingy reduces to that formula.

The more interesting case is what I call d5+1, which means: when you roll a 6, you record a 5 and roll again. This closes some annoying holes in the PDF you'd normally get with open ended dice (for example: it's impossible to get a result of exactly 6, or any multiple of 6), and brings the expected result of each rolled d6 to exactly 4.

Cheers, -- N

 

[malraux]

Ooooo math.

So the formula is gonna look something like this.

P = Sum(probability of die being rolled * Average value of the die)

For the first few terms, it looks like this:

1*N/2 + (1/N)*N/2 + (1/N)^2 * N/2.....

or N/2 * ( (1/N)^0 + (1/N)^1 + (1/N)^2....)

All that is the Geometric series.

When the n in the sum goes to infinity,

So, using those values, a = N/2 and r = 1/N. So the final answer is the average value will be (N/2)/(1 - 1/N).

Crap, I realized, that the N/2 from above should really be (N+1)/2 for the average die value. It doesn't change anything other than the final formula in any important way. So the average value should be ((N+1)/2)/(1-1/N). For a 6 sided die, it is indeed 4.2

 

[GSHamster]

Feldspar, your equation is correct. Though most people will approach it from a different angle and get Malraux's version.

Nifft, your case gives: (N^2 + N - 1) / (2N - 2) by my math.

 

[Flatus Maximus]

Feldspar, your equation is correct. Though most people will approach it from a different angle and get Malraux's version.

Nifft, your case gives: (N^2 + N - 1) / (2N - 2) by my math.

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....

 

[Lanefan]

The more interesting case is what I call d5+1, which means: when you roll a 6, you record a 5 and roll again. This closes some annoying holes in the PDF you'd normally get with open ended dice (for example: it's impossible to get a result of exactly 6, or any multiple of 6)

That's exactly how we've always treated any sort of open-ended roll, regardless of die size; and for just that reason.

and brings the expected result of each rolled d6 to exactly 4.

Which is 0.5 more than the usual average of 3.5. A friend of mine who is much better at this mathmology stuff than I am worked out that for any die size, open-ended rolling increases the overall average by 0.5. This was relevant in one game I was in, where instead of criticals we used open-ended rolls for all damage, and we were wondering how much difference it made.

Lanefan

 

[Flatus Maximus]

That's exactly how we've always treated any sort of open-ended roll, regardless of die size; and for just that reason. Which is 0.5 more than the usual average of 3.5. A friend of mine who is much better at this mathmology stuff than I am worked out that for any die size, open-ended rolling increases the overall average by 0.5. This was relevant in one game I was in, where instead of criticals we used open-ended rolls for all damage, and we were wondering how much difference it made.

Lanefan

My formula can be rewritten: (n+2)/2 = (n+1)/2 + 1/2, which is what your friend got. More evidence that I didn't screw up....

 

[Nifft]

(n+2)/2, or 4 when n=6.

I thought this must be wrong (since it's so simple)

What's wrong with neat & simple?

It's my favorite little dice discovery for exactly that reason!

Cheers, -- N

 

[Nifft]

for any die size, open-ended rolling increases the overall average by 0.5.

Yup. I like the d6 -> d5+1 version because rerolls are common, and when you consider multiple dice, you get a normal bell curve on the downside of the mode, but a nice gradual slope on the upside. Essentially, every +5 DC gets a "step" on a per-die basis, so each re-roll has the potential to be as meaningful as it feels.

Cheers, -- N

 

[Flatus Maximus]

What's wrong with neat & simple?

It's my favorite little dice discovery for exactly that reason!

Cheers, -- N

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.