average result from an open ended dice roll

[silentounce]

Yeah, I read where you said, "The guy who said it adds .5 is wrong."

But I notice you edited that out.

Maybe after you actually read the thread.

Yeah, after I read it for the fourth or fifth time and understood what they were doing with that. You notice that the last time I edited that post was before I commented on yours and before you replied to me? So, if you saw that then you must have known that I reread things and discovered the error, no? Yet you still accused me of not reading the thread when you had pretty good proof right there that not only did I discover myself in error, but I actually corrected myself. How many people do that on here? See that they are wrong and correct themselves? Yet you still decided to insult me by saying I didn't read the thread?

 

[Nifft]

I was hoping for a formula approaching the simplicity of (N+1)/2 (though, clearly, not so simple).

Nah, once you throw in nonlinear stuff like min/max within a single roll (rather than a cascade of independent rolls), a lot of simplifications go away.

But d6 isn't all that terrible to enumerate -- 4d6 only has 1,296 combinations!

Cheers, -- N

 

[Wulf Ratbane]

Nah, once you throw in nonlinear stuff like min/max within a single roll (rather than a cascade of independent rolls), a lot of simplifications go away.

But d6 isn't all that terrible to enumerate -- 4d6 only has 1,296 combinations!

Cheers, -- N

I put away my calculations ages ago, but IIR how I did it before, it was to iterate the odds that any subsequent roll would be higher than a given result.

Adding in open-endedness of any kind would change those results.

I can say for certain that the open-ended results skewed the averages much higher than I was comfortable with, which is why I put the work aside.

 

[Nifft]

I put away my calculations ages ago, but IIR how I did it before, it was to iterate the odds that any subsequent roll would be higher than a given result.

Adding in open-endedness of any kind would change those results.

Ooo, yeah, then you're compounding the problem: max-drop-low means you roll more dice and skew towards the high end, and reroll means high rolls are disproportionately rewarded.

Why would you want to combine them, though?

Cheers, -- N

 

[Wulf Ratbane]

Ooo, yeah, then you're compounding the problem: max-drop-low means you roll more dice and skew towards the high end, and reroll means high rolls are disproportionately rewarded.

Why would you want to combine them, though?

Cheers, -- N

As you well know, I'm willing to try out all kinds of crazy .

 

[drothgery]

Yes... Without having a strong foundation in probability, my solutions all tend toward brute force calculations. (Seriously-- some really ugly Excel sheets full of matrices of results.)

I had the math once, but since I never used it after taking the classwork, I don't remember how to do it. So my tendency is to code out a probabilistic solution (i.e. run something 10,000 times or so and look at the averages).