[long] [images] More Dice Formulas (Help!)

[BryonD]

OK, calculate the average for 100d6, drop the lowest 65 and the highest 32.

About 13.4 and you can be confident that you will not get any results >15 or <12.

That is the answer if you are neither a programmer nor a mathematician, but an engineer.

 

[CRGreathouse]

EDIT: So, having done some testing on this one, it seems that this does work, but ONLY when dropping a single die. (k is, in fact, the number of dice kept, and this equation only works for k = n - 1)

For nds drop 1, the chance of rolling a given x (1<=x<=s) is (x^n-(x-1)^n)/(s^n). That should simplify that special case.

 

[CRGreathouse]

About 13.4 and you can be confident that you will not get any results >15 or <12.

That is the answer if you are neither a programmer nor a mathematician, but an engineer.

Let's see... 1:64 100 000 000 000 chance for an 18.

 

[BryonD]

Let's see... 1:64 100 000 000 000 chance for an 18.

We used this system and one of my players got three 18s.
I saw it, no cheating, I swear

 

[Zappo]

Dudes, you're making me ever happier and happier for my choice of using point buy.

 

[CRGreathouse]

We used this system and one of my players got three 18s.
I saw it, no cheating, I swear

(grin)

Well, that's only 1:17 540 000 000 000 000 000 000 000 000 000 000 000 000, after all... it could happen.

I bet if I used this method for my group, a certain player would still manage to get an 18... but that's more 'by hook or by crook' than a statement about the probability.

 

[Zappo]

Well, that's only 1:17 540 000 000 000 000 000 000 000 000 000 000 000 000, after all... it could happen.

It is somewhat scary to think that the chances of an unidentified comet wiping Europe from the face of the planet within the next 10 years are many orders of magnitude higher than that.

 

[Janx]

ByronD corrects my memory, and yes I believe the answer I got was 13.4 and not 13.5 (I had did all this a few years ago and my memory dulled).

For those wanting % chances of any roll and thinking my cheesy method (and it is cheesy) won't work, sadly, you're not a programmer. Once you get the loops set up, you can contain counts of the occurances of each number (an array indexed 3-18). Pretty simple stuff.

Janx

 

[Darkness]

Once you get the loops set up, you can contain counts of the occurances of each number (an array indexed 3-18). Pretty simple stuff.

That's right.

Me, I'd use an Excel spreadsheet to avoid the randomization - but then I am a multiclassed computer guy/math guy. (For larger problems I'd use something else, though; I wouldn't go so far to build a spreadsheet that lists and sums up all possible combinations of roll 41d6, drop the lowest 38. )

Still, a non-programmer would be hard-pressed to replicate these methods until taught in detail how to go about it - with programs they already have at their disposal or can download easily, for that matter. So unless you're going to do that, it's probably better to leave the math people to their calculations.

 

[BryonD]

Nevermind