What is the Average Result of 2d20, Drop the Lowest.

[hong]

True enough, but the monte carlo method is easily adaptable to different die types.

So is the enumeration method.

Now if you were talking about different NUMBERS of dice, that would be different.

 

[pawsplay]

In any case, the tools of modern technology offer two relatively simple methods of determining the number.

 

[Rabelais]

Just a question... is it a Troll to ask a Statistics question on a D&D board?

Yes... yes it is.

QED

 

[Darkness]

Just trying a quick little experiment in my head, sans calculator, and I have a question for the people who said they did 2d4, drop lowest:

Possible combinations for 2d4, drop lowest:
Die 1/Die 2 = Result
1/1 = 1
1/2 = 2
1/3 = 3
1/4 = 4
2/1-2 = 2 (note that this is 2 combinations)
2/3 = 3
2/4 = 4
3/1-3 = 3 (note that this is 3 combinations)
3/4 =4
4/any =4 (note that this is 4 combinations)

(1+2+3+4+2*2+3+4+3*3+4+4*4)/16 = (10+4+7+9+4+16)/16 = 50/16 = 3.125

 

[orsal]

So is the enumeration method.

Now if you were talking about different NUMBERS of dice, that would be different.

Ah, I see. You set up your spreadsheet by using rows for one die and columns for the other. Works very well, and more efficient than my way for two dice, but can't be applied to more dice.

My method, however, works for any number of dice, as long as you have enough rows on your spreadsheet. For m dice of n sides each, you need m^n rows. In the first column put the numbers from 0 up to m^n-1; use the next m columns to give the base-n representation of the number in the first column. You then get every possible dice combination with dice labelled from 0 to n-1. Add one to each die, and you have standard dice.

 

[kromelizard]

It's an algorithm.

Is it also an equation?

 

[pawsplay]

Is it also an equation?

You could equate it to something, yes. It is not, however, algebraic, unless you first melt down the average results with probabilities (the other method mentioned above, not the one I suggested).