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

[pawsplay]

Easier here meaning that you don't really have to think about how the problem works to do a Monte Carlo approach to solving it.

Correct, and the "approximation" can be easily done to a degree of precision as to be nearly indistinguishable from a formal answer.

 

[Torm]

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:

Is the average 2.7?

And if anyone knows right off, is the average of 2d6, drop lowest, 4.05?

And 2d8, drop lowest, should be 5.4?

They probably aren't, but if these are even close, I'm using a method that is a LOT simpler than the other math I've seen in this thread so far.....

EDIT: Nevermind, back to the drawing board.

 

[Ry]

Just to expand on the question: How did you obtain the (2k-1) part of the formula?

It has to do with the difference between squares. Take a look at the breakdown in Excel. See how the 20s fill the far right and bottom rows? Then the 19s are the box inside that, and the 18s are the box inside that, all the way down to 1?

Think about the size of those row/columns.

There are 20 20s across the bottom, and 19 more 20s on the far right. (i.e. 2 * 20, minus 1 because of the overlap in the bottom right corner).

Now we're done with the 20s, imagine they're gone and look at the 19s. There's 19 19s across the bottom, and 18 more of them along the far right. (i.e. 2 * 19, minus 1 because of the overlap in the bottom right corner).

The pattern repeats all the way to 1.

 

[orsal]

Thanks! This is indeed useful.

Just to expand on the question: How did you obtain the (2k-1) part of the formula?

How many ways can you get less than or equal to k? Since both dice must independently be less than or equal to k, there are k^2 ways that can happen.

How many ways can you get strictly less than k? That means less than or equal to k-1, so there are (k-1)^2 ways.

How many ways to get exactly k? That means less than or equal to k but not less than or equal to k-1. So subtract:
k^2 - (k-1)^2 = k^2 - (k^2-2k+1) = 2k-1.

Also, is can this be generalized for multiple (more than two) rolls keep the highest,

Sort of. You can apply similar reasoning, but the formulas become more complicated.

For three dice, the number of ways (out of n^3) to get exactly k will be

k^3 - (k-1)^3
= k^3 - (k^3 - 3k^2 + 3k -1)
= 3k^2 - 3k + 1

Likewise, at this stage, if you have m dice, and want to keep only the best single die roll, you'll have a polynomial of degree (m-1) to deal with.

Now, with two dice, notice that I needed to know the formulas for sums from k=1 to k=n
sum(k) = n(n+1)/2
sum(k^2)=n(n+1)(2n+1)/6

Well, with three dice, I'm also going to need to know
sum(k^3)=[n(n+1)/2]^2
and with four dice I'm going to to know the formula for sum(k^4), which I'd have to look up. One of the Bernoulli brothers figured out all of these sums back in the 17th century, but apart from sum(k) (which the ancient Greeks knew), they aren't easy. (The formula for sum(k) was famously rediscovered by an 8-year-old Carl Gauss; most 8-year-olds aren't up to rediscovering it themselves, but you can show Gauss' reasoning to any mathematically competent middle school student and they'll at least be able to follow it. Not so with any proof I know of sum(k^2) or higher.)

Actually, for a specific die, if you don't want the general formula, you can do it without the algebra. Just calculate the 20 (or however many sides on the die) probabilities independently, and then add. For a big die it's a lot of numbers, but you don't need to know the Bernoulli formulas. If you want to do as I did, give a formula for n-sided dice without specifying a particular value of n, you'll need the Bernoulli formulas.

multiple rolls drop the lowest or even multiple rolls drop X rolls?

That's a lot trickier. I can't think of an elegant way to compute all the probabilities you need systematically. When I wanted to figure out the distribution for best 3 out of 4 d6, I set up a spreadsheet to enumerate all 6^4=1296 possibilities and calculate the sum for each.

 

[hong]

Because it's easier.

No it's not.

 

[pawsplay]

No it's not.

One requires 400 permutations. The other requires five lines of code in the scripting language of your choice. *shrug*

 

[orsal]

multiple rolls drop the lowest or even multiple rolls drop X rolls?

That's a lot trickier. I can't think of an elegant way to compute all the probabilities you need systematically. When I wanted to figure out the distribution for best 3 out of 4 d6, I set up a spreadsheet to enumerate all 6^4=1296 possibilities and calculate the sum for each.

After I wrote that, it occurred to me there is a nifty way.

The sum of the best three out of four dice is the best die plus the sum of the middle two. The average result for the sum of the middle two dice is easy: it's 7. The reason is symmetry: since as many dice (one) are dropped from the top as from the bottom, the distribution will not be skewed toward either the high or low end of its range. So the mean is the same as the middle possible value, which is 7. The distribution is somewhat different than just rolling two dice and adding them -- more often close to 7, fewer extremely high or extremely low results -- but the average is the same.

So, by the method described above, you might be able to calculate that the average result for the top die out of 4 d6 is about 5.24. (I didn't actually calculate it, but I know the answer I'm going to get from the aforementioned spreadsheet, so I worked backwards.) The average of the sum is the sum of the averages, so 7+5.24=12.24. This is the average ability score generated by the standard method.

 

[hong]

One requires 400 permutations.

No, one requires 2 formulas and 1 application of copy-and-paste in the spreadsheet of your choice.

 

[orsal]

One requires 400 permutations. The other requires five lines of code in the scripting language of your choice. *shrug*

But five lines of code or fewer can generate all 400 permutations. Just nest one loop inside another.

Or, four columns of a 400-row spreadsheet can do the trick -- the first column is a series of 400 consecutive numbers (any decent spreadsheet can do that automatically), the next two extract the individual die results coded by the numbers in the first column, and the last adds them.

 

[pawsplay]

No, one requires 2 formulas and 1 application of copy-and-paste in the spreadsheet of your choice.

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