Solution: Apply the stars and bars principle with 17 stars and 4 bars. The 4 bars must be placed among the 16 spaces between the bars to simulate dice outcomes. This can be done in C(16,4) ways, but this includes outcomes where at least some of the 5 bins contain more than 6 stars. Suppose (WLOG) the first bin contains more than 6 stars. To count the number of ways this can happen suppose the first bin contains 6 stars at the outset, then apply stars and bars to the remaining 11 stars. This yields C(10, 4) ways to distribute the 11 stars into the 5 bins. Since this scenario can occur for each of the 5 bins, there are 5∙C(10,4) of these outcomes in all. Note that all outcomes in which two bins containing more than 6 stars were counted twice here, so we can apply the principle of inclusion-exclusion to restore these outcomes to the count. There are C(5, 2) ways to choose the two bins containing more than 6 stars and for each of these cases we can apply stars and bars to count the number of ways of distributing the remaining 5 stars into the 5 bins. This yields C(5, 2))∙C(4, 4)) outcomes to be added back. Therefore, the required probability is (C(16, 4)) - 5∙C(10, 4)) + C(5, 2))∙C(4, 4)))/6^5 =780/7776=65/648≈.1003 which is consistent with published probability tables.

$NOTE \quad :-\quad \\ { X }^{ m }\quad |\quad f\left( x \right) \quad \quad Denotes\quad the\quad coefficient\quad of\quad { X }^{ m }\quad in\quad f\left( x \right)$

$\quad \quad \quad \quad Using\quad Multinomial\quad theorem\quad \\ calculation\quad of\quad total\quad number\quad of\quad possible\quad cases.\\ Let\quad { X }_{ i }\quad denotes\quad the\quad number\quad appearing\quad on\quad { i }^{ th }\quad Dice\\ { X }_{ i }\in \quad \{ 1,2,3,4,5,6\} \\ \sum _{ i=1 }^{ 5 }{ { X }_{ i } } =17\\ { X }^{ 17 }\quad |\quad { (x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 }+{ x }^{ 6 }) }^{ 5 }\\ { X }^{ 12 }\quad |\quad { (1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 }) }^{ 5 }\\ { X }^{ 12 }\quad |\quad { \left( \frac { { x }^{ 6 }-1 }{ x-1 } \right) }^{ 5 }=780\\ Total\quad number\quad of\quad cases\quad are\quad =\quad { 6 }^{ 5 }\\ Therefore\quad Probability=\frac { 780 }{ { 6 }^{ 5 } } \\$

matlab

D&D-Notation: $m{d}N$ is a shorthand for "Throw $m$ dice with $N$ equal surfaces labeled $1;\ldots;N$ and calculate their sum."

In this problem, we want to calculate the probability distribution function $p_5(n)$ for a $5d6$ roll and evaluate at $n=17$ . The dice are independent from each other. Let's begin with one die:

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One die

The pdf for a $1dN$ roll is easy to write down. Using the discrete unit step function $r_0(n)$ , we get rid of the two cases: $\begin{aligned} p_1(n)&=\begin{cases} \frac{1}{N}:&1\leq n\leq N\\ 0:&\text{else} \end{cases}=\frac{1}{N}\left(r_0(n-1)-r_0(n-(N+1))\right)=:q_1(n-1)&&&&\left|r_0(n):=\begin{cases} 1:&n\geq 0\\ 0:&\text{else} \end{cases}\right. \end{aligned}$

The function $q_1(n)$ gets rid of the translation by 1 to the right.

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The general case - $m$ dice

What about an $mdN$ roll? The pdf for each die is $p_1(n)$ , of course. and all dice are independent from each other - but we are interested in their sum! Recall that the pdf of the sum of $m$ independent random variables is the convolution of their pdf's: $\begin{aligned} p_m(n)&=\underset{m\times p_1(k)}{ \underbrace{ \left(p_1(k)*\ldots *p_1(k)\right)(n) } }=\left(q_1(k-1)*\ldots *q_1(k-1)\right)(n)=\left(q_1(k)*\ldots*q_1(k)\right)(n-m)=:q_m(n-m)\\ q_m(n)&:=\underset{m\times q_1(k)}{ \underbrace{ \left(q_1(k)*\ldots *q_1(k)\right)(n) } }=\frac{1}{N^m}\sum_{l=0}^m (-1)^l\binom{m}{l} r_{m-1}(n-lN)\qquad(*) \end{aligned}$

The last step can be shown by induction over $m$ : $\begin{aligned} m=1:&&q_1(n)&=\frac{1}{N}(r_0(n)-r_0(n-N))\\ m\rightsquigarrow m+1:&&q_{m+1}(n)&=(q_m(k)*q_1(k))(n)\underset{(*)}{=}\frac{1}{N^m}\sum_{l=0}^m (-1)^l\binom{m}{l} (r_{m-1}(k-lN)*q_1(k))(n)\\ &&&\underset{(**)}{=}\frac{1}{N^{m+1}}\sum_{l=0}^m (-1)^l\binom{m}{l} (r_{m}(n-lN)-r_{m}(n-(l+1)N))\\ &&&=\frac{1}{N^{m+1}}\left( r_m(n) + \sum_{l=1}^{m} (-1)^l\left[ \binom{m}{l}+\binom{m}{l-1} \right]r_m(n-lN) + (-1)^{m+1}r_m(n-(m+1)N) \right)\\ &&&=\frac{1}{N^{m+1}}\sum_{l=0}^{m+1} (-1)^l\binom{m+1}{l}r_m(n-lN) \end{aligned}$

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The special case - five dice

In the special case of a $5d6$ roll, the pdf evaluated at $n=17$ is given by $\begin{aligned} p_5(17)&=q_5(17-5)=q_5(12)=\frac{1}{6^5}\sum_{l=0}^5 (-1)^l\binom{5}{l} r_{4}(12-6l)=\frac{1}{6^5}\left( \binom{5}{0}\binom{12+4}{4}-\binom{5}{1}\binom{6+4}{4}+\binom{5}{2}\binom{4}{4} \right)\\ &=\frac{780}{6^5}\approx\boxed{0.1003} \end{aligned}$

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Theorem: Let $m\in\mathbb{N}_0,\:n\in\mathbb{Z}$ and $r_0(n)$ be the discrete unit step function. Then $\begin{aligned} r_m(n)&:=r_0(n)\cdot\binom{n+m}{m}&\Rightarrow &&r_{m+1}(n)&=(r_m(k)*r_0(k))(n)&&&&\left|r_0(n):=\begin{cases} 1:&n\geq 0\\ 0:&\text{else} \end{cases}\right.&&&&(**) \end{aligned}$

Prove it directly for $n<0$ and by induction over $n$ for $n\geq 0$ !

No# Dice = n (i.e5) Range of Sum n >= k <= 6n (5 to 30). Written Program to Print Number of Possible condition for each SUM Num[5]=1 Num[6]=5 Num[7]=15 Num[8]=35 Num[9]=70 Num[10]=126 Num[11]=205 Num[12]=305 Num[13]=420 Num[14]=540 Num[15]=651 Num[16]=735 Num[17]=780 Num[18]=780 Num[19]=735 Num[20]=651 Num[21]=540 Num[22]=420 Num[23]=305 Num[24]=205 Num[25]=126 Num[26]=70 Num[27]=35 Num[28]=15 Num[29]=5 Num[30]=1

780/(6^5) = 0.1003