help

Hello Melody.

\(\begin{array}{|c|l|} \hline n \text{ dices} & \text{probability } \\ \hline 1 & \dfrac{2}{6^1}= \dfrac{2}{6^1} = \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{6}{3}}{6} = \dfrac{\dfrac{6}{ \dfrac{6}{2} }}{6} = \dfrac{2\cdot 6^{1-1}}{61} \\ \hline 2 & \dfrac{12}{6^2} = \dfrac{12}{36} = \mathbf{\dfrac{1}{3}}= \dfrac{\dfrac{36}{3}}{36}= \dfrac{\dfrac{36}{ \dfrac{6}{2} }}{36} = \dfrac{2\cdot 6^{2-1}}{6^2} \\ \hline 3 & \dfrac{72}{6^3} = \dfrac{72}{216}= \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{72}{3}}{216}= \dfrac{\dfrac{72}{ \dfrac{6}{2} }}{216} = \dfrac{2\cdot 6^{3-1}}{6^3} \\ \hline \color{red}4 & \dfrac{432}{6^4} = \dfrac{432}{1296}= \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{432}{3}}{1296}= \dfrac{\dfrac{432}{ \dfrac{6}{2} }}{1296} = \dfrac{2\cdot 6^{4-1}}{6^4} \\ \hline 5 & \dfrac{2592}{6^5} = \dfrac{2592}{7776}= \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{2592}{3}}{7776}= \dfrac{\dfrac{2592}{ \dfrac{6}{2} }}{7776} = \dfrac{2\cdot 6^{5-1}}{6^5} \\ \hline \ldots & \ldots \\ \hline n & \dfrac{2\cdot 6^{n-1}}{6^n} = \mathbf{\dfrac{1}{3}}= \dfrac{ \phi(6^n) } {6^n} \\ \hline \end{array} \)

\(2,12,72,432,2592, 15552,\ldots\) see: http://oeis.org/search?q=2%2C12%2C72%2C432%2C2592&sort=&language=german&go=Suche

Thanks Heureka,

Am I right in thinking you just used a formula.

I mean can you see the logic behind the forumula - I can't.

Hi guest.

How did you work all these totals out?

Hi Melody: There is this formula that is programmed into my computer that gives you instantaneous  answer when  you enter the total under "S" and the number of dice under "N" :

S=6; N=4; sumfor(k, 0, ((S - N)/6), ((-1)^k * (N nCr k) * (S - 6*k - 1) nCr (N - 1))

 /* This formula calculates the number of permutations for "N" dice */
 /* given a total in "S" */

P.S. If you don't use a programming language, I believe you could programm it as a "Macro" into a Spreadsheet.

Thanks

Yes maybe I could do it in spreadsheet if I wanted to but I was really trying to extract the logic.

When you answer using a program it is nice if you say that so everyone knows where your answers comes from. 

I like people doing this. I am not trying to discourage you at all. 

Hello Guest,

to generate your formula see the link: http://mathworld.wolfram.com/Dice.html

\(\text{The probability of obtaining p points (a roll of p) on n s-sided dice can be computed as follows. $\\$ The number of ways in which p can be obtained is the coefficient of $x^p$ in $\\$ $f(x)=\left(x+x^2+...+x^s \right)^n, \qquad (1)$ } \)

\(\text{since each possible arrangement contributes one term. $f(x)$ can be written as a multinomial series $\\$ $f(x) = x^n \left( \sum \limits_{i=0}^{s-1} x^i \right)^n \qquad (2) \\ = x^n\left(\dfrac{1-x^s}{1-x}\right)^n, \qquad (3)$}\)

\(\text{so the desired number c is the coefficient of $x^p$ in $\\$ $x^n(1-x^s)^n(1-x)^{-n}. \qquad (4)$}\)

\(\text{Expanding,$\\$ $x^n \sum \limits_{k=0}^n (-1)^k \dbinom{n}{k} x^{sk} \sum \limits_{l=0}^{\infty} \dbinom{n+l-1}{ l}x^l, \qquad (5)$ }\)

\(\text{so in order to get the coefficient of $x^p$, include all terms with $\\$ $p=n+sk+l. \qquad (6)$ }\)

\(\text{c is therefore $\\$ $c= \sum \limits_{k=0}^n (-1)^k \dbinom{n}{k} \dbinom{p-sk-1}{p-sk-n}. \qquad (7)$ } \)

\(\text{But $p-sk-n>0$ only when $k<(p-n)/s$, so the other terms do not contribute. Furthermore, $\\$ $\dbinom{p-sk-1}{ p-sk-n}= \dbinom{p-sk-1}{ n-1}, \qquad (8)$}\)

\(\text{so $\\$ $c= \sum \limits_{k=0}^{\Big\lfloor \dfrac{p-n}{s} \Big\rfloor} (-1)^k \dbinom{n}{ k} \dbinom{p-sk-1}{ n-1}, \qquad (9)$}\)

\(\text{where $\lfloor x \rfloor$ is the floor function, and $\\$ $\mathbf{P(p,n,s)= \dfrac{1}{s^n} \sum \limits_{k=0}^{\Big\lfloor \dfrac{p-n}{s} \Big\rfloor} (-1)^k \dbinom{n}{ k} \dbinom{p-sk-1}{ n-1}} \qquad (10)$}\)