Probability question

[zornwil]

Re: Probability question

 

Well, I can't give Cancer rep as apparently I gave it to him too recently - Cancer, I'll get you when I can, thanks for the extra above-and-beyond via email!

[superferret]

Re: Probability question

 

Using the following notation:

 

C(n,r) = n!/(r!*(n-r)!),

 

The I use N(m,r) for the number of times that 'r' is rolled on 'm' D6.

 

So N(m,r) is:

1, for r = m

m, for r = m+1

C(r-1,m-1) for r = m+2 to m+5

C(r-1,m-1) - m*C(r-7,m-1) for r = m+6 to (m*3.5, rounded down)

N(m,mn+m-r), for r > (m*3.5, rounded down)

 

The probability of rolling 'r' on 'm' D6 = N(m,r)/(6^m)

 

The cumulative probability, for calculating the probability of rolling 'r' or less on 'm' D6 is N(m+1,r+1)/(6^(m+1)).

[gojira]

Re: Probability question

 

I think I follow, but I wanted a more generalized solution (more than just d6).

 

So I wanted N (d, m, r), where d = number of sides on a given die.

 

You gave N(6, m, r) only.

 

N = 1 for m = r and r = 6^m also.

 

N = m for r = m+1 and r = 6^m-1, similar idea.

 

If C(r-1,m-1) is correct for r up to (6^m)/2 (I *think* that's what you're saying), then the probability curve is symmetric and the probably of rolls over (6^m)/2 can be found by using values already computed. No need for the other formula.

 

Brain hurting now. Must lie down and think more. *sigh* I wish I spent more CP on INT...

[gojira]

Re: Probability question

 

I think I'm getting this now. I'll try to work it out more fully, but some nice person in chat just pointed me here:

 

http://mathworld.wolfram.com/Dice.html

 

This appears to contain correct formulas. Superferet, I'm pretty sure your equations don't work.

 

For 4 dice, where the roll is 6, I get C(r-1, m-1) = C(6-1,4-1) = 5!/3!2! = (5*4)/2 = 10, which is incorrect. It should be 6 (easy to count).

 

I think the correct answer is an arithmetic series that isn't too hard to derive (still workin' tho). There's a formula for area though on that mathworld page that can be expressed as a simple polynomial. It's cool how mathematics actually does make things simple, once you get the details worked out.

[zornwil]

Re: Probability question

 

I really regret getting rid of my probability book from college - I used to know how to do this stuff.

 

Even if not correct yet, I really appreciate this, thanks so much!

[zornwil]

Re: Probability question

 

I think I'm getting this now. I'll try to work it out more fully, but some nice person in chat just pointed me here:

 

http://mathworld.wolfram.com/Dice.html

 

This appears to contain correct formulas. Superferet, I'm pretty sure your equations don't work.

 

For 4 dice, where the roll is 6, I get C(r-1, m-1) = C(6-1,4-1) = 5!/3!2! = (5*4)/2 = 10, which is incorrect. It should be 6 (easy to count).

 

I think the correct answer is an arithmetic series that isn't too hard to derive (still workin' tho). There's a formula for area though on that mathworld page that can be expressed as a simple polynomial. It's cool how mathematics actually does make things simple, once you get the details worked out.

No, it should be 10. That is in Cancer's results and also from brute-force investigation, as I checked:

 

1 1 1 3

1 1 3 1

1 3 1 1

3 1 1 1

1 1 2 2

1 2 1 2

2 1 1 2

1 2 2 1

2 1 2 1

2 2 1 1

[gojira]

Re: Probability question

 

Oops, you right. I missed the first four...