5D6 drop the lowest two (math ?)

[Gwaihir]

This thread rocks. And to think I never believed my teachers that math would be important to know for real life.

 

[orsal]

Formula by Gauß:

1 + 2 + 3 + 4 + ... + N = N * (N + 1) / 2.

Proof for this? Look at: (1 + N) + (2 + (N - 1)) + ... + (N/2 + (N/2) + 1) if N is even. Similar case for N odd.

That formula predates Gauss by about 2000 years. The famous connection with Gauss is the anecdote of his rediscovering it when he was 8 years old.

 

[dylanarena]

I do hope you test your software against a brute-force method (please).

Absolutely: I checked it against the table of results posted earlier in this thread and against a brute-force run for a few combinations that weren't that table. It's got all of the latest revisions you guys found in your iterative process, and the addition of the probabilities of individual outcomes checks out empirically as well.

 

[dylanarena]

OK, the software package that calculates what has been discussed in this thread has been released:

http://cran.r-project.org/src/contrib/Descriptions/dice.html

It requires the use of the open-source R statistical programming environment, so there's a bit of an up-front investment of time (download/install R if you don't already have it, then install the package "dice"), but the package is there if anyone wants to play around with it.

(Note that I credit this thread with providing the context and foundation for the package's existence.)

Cheers,
Dylan

 

[Treebore]

I don't worry about the math. I let players play the character they want, including stats. If that means they give themselves all 18's, then so be it. Its not like its hard to account for high stats. Plus, I get to decide stats the same way for NPC's and monsters. So high stats are easy to "balance out".

So my players get to play the character they want, right down to their ability scores.

 

[1of3]

*raise_thread*

Hi everyone, I have a question about Nathan's formula. I take it, that's the formula as presented.

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I'm not sure about parentheses. I'd multiply binomials and (l-1)^r with the results of the Cs as well. Like this...

1/(n^d) SUM SUM SUM nCr (n-r)Cu (l-1)^r
( l(u-k+r) + (n-u-r)( (d+1)d-(l+1)l )/2 )

I may be wrong, of course. If so please tell me how.

 

[OnlineDM]

This is an interesting double-necro. A thread from 2003 was raised in 2007, and now again in 2012. I expect some sort of Halley's Comet effect, and we'll see it again around late 2017 or something.

But as for the topic itself, I was glad to see this thread; I'd been pondering the math behind 4d6 drop the lowest recently.

 

[3D]

Analytical solution

I haven't read all the posts, but here is my analytical solution, it is writen in python3, but it should also work with python2 if you change the print-function:


###############################################################


al=[]
s=0 # summ


# The Array "al" is filled with all possible combinations,
# it takes into account that rolling [1,1,1,1,2] got the same probability than rolling [2,1,1,1,1]

for a in range(1,7):
> for b in range(1,7):
>> for c in range(1,7):
>>> for d in range(1,7):
>>>> for e in range(1,7):
>>>>> al.append([a,b,c,d,e])

# sort the array and add the 3 highest throws

for i in range(6**5):
> al.sort()
> s=s+al[2]+al[3]+al[4]

## divide summ by the number of possibilitys

print(s/(6**5))

# solution: 13.43016975308642

#########################################

Even this is calculated with the pc its not brute-force, it takes each possibility (6^5) which can be rolled into account.

I hope this helps

 

[JiffyPopTart]

I rarely read, and even more rarely post on Enworld nowadays. I happened to hop on tonight since I am waiting for dinnertime.

I saw this thread title and thought to myself "I remember asking a question like that a long time ago."

This is truly a thread that doesn't seem to die.

DS

Good work everyone!

 

[GX.Sigma]

Ah, so this was before anydice.com?

http://anydice.com/program/1744