\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Let \(X\) be the random variable for the outcome of one die roll. It has probability distribution \(\Prob (X = x) = \frac {1}{6}\) for \(x = 1, 2, \ldots , 6\).
Therefore, \(R_1\) follows the probability distribution \(\Prob (R_1 = x) = \frac {1}{6}\) fir \(x = 0, 1, \ldots , 5\), since \(R_1 = X \modulo 6\).
This means that \[ G(x) = \frac {1}{6} \left (1 + x + x^2 + x^3 + x^4 + x^5\right ). \]
\(R_2 = (X_1 + X_2) \modulo 6 = ((R_1)_a + (R_1)_b) \modulo 6\), and notice that, \[ G(x)^2 = \frac {1}{36} \left (1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + 5x^6 + 4x^7 + 3x^8 + 2x^9 + x^{10}\right ). \]
Therefore, combining the terms with the same powers modulo 6, we get \[ G_{R_2}(x) = \frac {1}{36} \left ((1 + 5) + (2 + 4)x + (3 + 3)x^2 + (4 + 2)x^3 + (5 + 1)x^4 + 6 x^5\right ) \] which simplifies gives \(G(x)\), as desired.
Therefore, since \(R_{n} = (X_1 + X_2 + \ldots + X_n) \modulo 6 = (R_{n - 1} + R_1) \modulo 6\), by mathematical induction, we can conclude that the probability generating function for \(R_n\) is always \(G(x)\).
This means that the probability of \(R_n\) being a multiple of 6, is \[ \Prob \left (6 \divides R_n\right ) = \frac {1}{6}. \]
Notice that \(G_1(x)\), the probability generating function for \(T_1\) must be \[ G_1(x) = \frac {1}{6} \left (1 + 2x + x^2 + x^3 + x^4\right ). \]
Therefore, notice that \[ G_1(x)^2 = \frac {1}{36} \left (1 + 4x + 6x^2 + 6x^3 + 7x^4 + 6x^5 + 3x^6 + 2x^7 + x^8 \right ), \] and combining the powers with the same remainder modulo 5, we have \[ G_2(x) = \frac {1}{36} \left (7 + 7x + 8x^2 + 7x^3 + 7x^4\right ) = \frac {1}{36} \left (x^2 + 7y\right ) \] where \(y = 1 + x + x^2 + x^3 + x^4\), as desired.
Expressing \(G_1\) in terms of \(y\), we have \[ G_1(x) = \frac {1}{6} (x + y). \]
Experimenting with \(G_3\), we notice \begin {align*} G_1(x) \cdot G_2(x) & = \frac {1}{6^3} (x + y)(x^2 + 7y) \\ & = \frac {1}{6^3} (x^3 + 7xy + x^2y + 7y^2). \end {align*}
But notice that up to the congruence of the powers modulo \(5\), we have \(x^n y\) will simplify to simply \(y\), and \[ (x + y)^2 = x^2 + y^2 + 2xy = x^2 + 7y \] from \(G_1(x)^2 = G_2(x)\) implies that \(y^2\) simplifies to \(5y\).
Therefore, \[ G_3(x) = \frac {1}{6^3} (x^3 + 7y + y + 7 \cdot 5y) = \frac {1}{6^3} (x^3 + 43y). \]
Now, we assert that \[ G_n(x) = \frac {1}{6^n} (x^{n \modulo 5} + \frac {6^n - 1}{5}y). \]
The base case is shown in \(G_1\), and now we do the inductive step. Assume that \[ G_k(x) = \frac {1}{6^k} (x^{k \modulo 5} + \frac {6^k - 1}{5}y) \] for some \(k \in \NN \).
\begin {align*} G_{k + 1}(x) & = G_k(x) \cdot G_1(x) \\ & = \frac {1}{6^k} \cdot \left (x^{k \modulo 5} + \frac {6^k - 1}{5}y\right ) \cdot \frac {1}{6} \cdot (x + y) \\ & = \frac {1}{6^{k + 1}} \cdot \left (x^{k \modulo 5} \cdot x^{1} + x^{k \modulo 5} \cdot y + x \cdot \frac {6^k - 1}{5}y + \frac {6^k - 1}{5} y^2\right ) \\ & = \frac {1}{6^{k + 1}} \cdot \left (x^{(k + 1) \modulo 5} + y + \frac {6^k - 1}{5}y + \frac {6^k - 1}{5} \cdot 5 y\right ) \\ & = \frac {1}{6^{k + 1}} \cdot \left (x^{(k + 1) \modulo 5} + \left (\frac {6^k - 1}{5} + 6^k\right )y\right ). \end {align*}
What remains to prove is that \[ \frac {6^k - 1}{5} + 6^k = \frac {6^{k + 1} - 1}{5}, \] but this is straightforward since this is just trivial algebra.
So our assertion is true, and \[ G_n(x) = \frac {1}{6^n} (x^{n \modulo 5} + \frac {6^n - 1}{5}y). \]
Now, the probability of \(5 \divides S_n\) is the coefficient of \(x^0\) (the constant term) in \(G_n(x)\).
If \(5 \notdivides n\), \(x^{n \modulo 5}\) is not \(x^0\), and therefore the only term that contributes to the constant term comes from \(y\), therefore \[ \Prob \left (5 \divides S_n\right ) = \frac {1}{6^n} \cdot \frac {6^n - 1}{5} = \frac {1}{5} \left (1 - \frac {1}{6^n}\right ), \] as required.
If \(5 \divides n\), then \(x^{n \modulo 5}\) will be \(x^0 = 1\) contributing to the probability, hence \[ \Prob \left (5 \divides S_n\right ) = \frac {1}{6^n} \cdot \left (1 + \frac {6^n - 1}{5}\right ) = \frac {1}{5} \left (1 + \frac {4}{6^n}\right ). \]