statistics by dropping dice rolls..

okay, so, we all play D&D right? if not, quick explanation of terms:
a "d6" is a six-sided (cube) dice, which (unsurprisingly) contains the numbers 1, 2, 3, 4, 5 and 6 arranged with one value per face.

the term "1d6" means "roll 1 six-sided die" (1d6 has a sample space of {1...6}. "2d6" means "roll 2 six-sided dice and add their values together" sample space: {2...12}

the average of 1d6 is equal to (1+6)/2 = 7/2 = 3.5
the average of 3d6 is equal to 3*3.5 = 10.5

historically, character stats in D&D are generated via rolling 3d6. when 3rd edition was released oh-so-many years ago, they replaced the standard "3d6" generation method with "4d6, drop lowest roll"

Before we tackle the main event, let's take a simpler problem: 2d6 drop lowest.

for 2d6 drop lowest, the average is approximately 4.47. We can find this out by summing all 36 possible outcomes and then dividing by the number of possible outcomes. it's dirty and ugly, but trivial for 2 dice. 3 dice has a lot more combinations (6*6*6=216) :(

so, for 3d6, 4d6 and above, is there a more elegant solution?

wtf is d&d

the more elegant solution would be to write a program.

I don't know if there's actually a nice equation for conditional stuff like this, so your best bet is probably to try to find a pattern in the answers.

nonlinear dynamics, ai

>>2 Dungeons and Dragons

One solution might be to find the probability of a certain number being "dropped" from the equation. I'd break it down like this:

Roll 4 dice, subtract lowest
=(average result of 4d6) - (average number subtracted from this test)

P(any number rolled being dropped) = 1 - P(all other numbers rolled are equal to or higher than this number)

I'm sure you can figure something out from this.

Let's say the lowest number is dropped from 4 rolls. If this lowest number is one, then we have:

1***,

where the * can be 1 through 6.
If there are 0 1s amongst the ***, then there are (4,3)*5^3 choices ((4,1) denotes the binomial coefficient of 4 choose 1).
Likewise for j=1,2,3 1s amongst the ***, there are (4,3-j)*5^(3-j). So, there are sum((4,j)*5^j,j=0..3) ways to roll 4 6-sided dice where the 1 is dropped.

Generalizing, there are sum((n,j)*(s-k)^j,j=0..n-1) ways to roll n s-sided dice where k is dropped.

So, the probability of dropping k from n rolls of s-sided dice is found by P(k)=1/s^n*sum((n,j)*(s-k)^j,j=0..n-1).

Averaging these gives sum(k*P(k),k=1..s).

Thus the average value rolled after rolling n s-sided dice and dropping the lowest value should be

n*(s+1)/2-1/s^n*sum(sum(k*(n,j)*(s-k)^j,j=0..n-1),k=1..s),

which gives values for 6-sided dice:

2 dice -> avg. = 4.64
3 dice -> avg. = 8.49
4 dice -> avg. = 12.25,
etc.

Ack, just noticed in error. In the final expression, I put the k inside the wrong summation.

So, the the average value rolled after rolling n s-sided dice and dropping the lowest value should instead be

n*(s+1)/2-1/s^n*sum(k*sum((n,j)*(s-k)^j,j=0..n-1),k=1..s),

which gives values for 6-sided dice:

2 dice -> avg. = 4.47
3 dice -> avg. = 8.46
4 dice -> avg. = 12.24,
etc.

Sorry for the mistake. I think this is correct now. It at least agrees with the previous computation of the 2d6 case.

Actually, taking another look it wasn't just "misplacing" the k the gave the incorrect answers as that would leave the expression the same. The first time, I was summing with the order of summation reversed as well and likely made an error regarding this instead.

Anyway...

The expression above simplifies into one involving a single sum without binomial coefficients.

n*(s+1)/2-1/s^n*sum(k*((s-k+1)^n-(s-k)^n),k=1..s),

where n is the number of dice and s is the number of sides.

The sum can even be further simplified due to somewhat of a telescoping phenomenon. Using 4d6 as an example, the sum is (6^4-5^4)+2*(5^4-4^4)+3*(4^4-3^4)+4*(3^4-2^4)+5*(2^4-1^4)+6*1^4. Simplifying and regrouping yields 1^4+2^4+3^4+4^4+5^4+6^4, a power sum.

So, the average roll using n s-sided dice while dropping the lowest value becomes

n*(s+1)/2 - 1/s^n*sum(k^n,k=1..s).

It can also be written as

n*(s+1)/2 + ζ(-n,s+1)/s^n,

where ζ(s,a) is the Hurwitz zeta function.

>>6 & >>9
You guys win at life.

>>6 & >>9
You guys win at life.

>>10 & >>11
You fail at posting.

>>10
Never did anything beyond basic math huh?

Calculus > Discrete math

calculus is pretty nuts if you think about it

calculus is for fuckers

discrete math owns you
espcially because there are usually like 2 or 3 different classes that are all basically discrete math that you can use to easily pass... discrete math, formal logic, etc

discrete math owns you
espcially because there are usually like 2 or 3 different classes that are all basically discrete math that you can use to easily pass... discrete math, formal logic, etc

>>9 -san,

thank you. all my character generation are belong to you. :)

What's great about mathematics is that there is no room for opinion.

>>20
then you arent doing the right mathematics

whats the point in simply reusing other peoples discoveries

>>21
A solid proof does not allow for any opinion, whether it is newly discovered or ages old.

>>22
opinion leads to theory leads to proof leads to discovery

>>23
No.

>>23 No, proof leads to proof leads to proof leads to proof, etc. Then we can evaluate the results of these proofs for discovery.

bump