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Idun is rolling four ordinary six-faced dice, with the faces labelled 1 through 6.  What is the probability that her total score is divisible by 3?

 Nov 25, 2019
 #1
avatar+23575 
+3

Idun is rolling four ordinary six-faced dice, with the faces labelled 1 through 6. 

What is the probability that her total score is divisible by 3?

 

\(\begin{array}{|rcll|} \hline \text{ probability } &=& \dfrac{2\cdot 6^3}{6^4} \\\\ &=& \dfrac{432}{1296} \\\\ &=& \dfrac{1}{3} \\ \hline \end{array} \)

 

laugh

 Nov 25, 2019
 #2
avatar+105989 
+2

Warum ist das Heureka?  Ich verstehe nicht.   sad

 

Ich spreche Deutsch nur einen bisschen.  

Ich übe nur.

Antworten Sie daher bitte hauptsächlich auf Englisch.

Danke   laugh

 Nov 25, 2019
edited by Melody  Nov 25, 2019
edited by Melody  Nov 25, 2019
 #3
avatar+23575 
+3

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

 

 

laugh

heureka  Nov 25, 2019
edited by heureka  Nov 25, 2019
 #5
avatar+105989 
+1

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.

Melody  Nov 25, 2019
 #4
avatar
+1

Total of 6 = 10 permutations

Total of 9 = 56 permutations

Total of 12=125 permuations

Total of 15=140 permutations

Total of 18 = 80 permutations

Total of 21 = 20 permutations

Total of 24 = 1   permuation

GRAND TOTAL =432 / 6^4 = 1 / 3

 Nov 25, 2019
 #6
avatar+105989 
+1

Hi guest.

How did you work all these totals out?

Melody  Nov 25, 2019
 #7
avatar
+2

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.

Guest Nov 26, 2019
 #8
avatar+105989 
+1

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. 

Melody  Nov 26, 2019
 #9
avatar+23575 
+2

Hello Guest,

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

laugh

heureka  Nov 26, 2019

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