\(% 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 this condition be \(C_1\). Since the game ends in the first round, the score must remain to be zero, and therefore \[ \Prob (N = 0 \mid C_1) = 1, \] and for all other \(n \in \NN \) where \(n \neq 0\), \[ \Prob (N = n \mid C_1) = 0. \]
This means the p.g.f. for \(N\) conditional under \(C_1\) is just simply \(G(t \mid C_1) = \Prob (N = 0 \mid C_1) \cdot t^0 = 1\).
Denote the condition where the score is increased by \(1\) as \(C_3\). Since in the first round the game score increased by one, and after the first round it is just as if this was a new game, so for all \(n \in \NN \), we must have \[ \Prob (N = n \mid C_3) = \Prob (N = n - 1), \] and \[ \Prob (N = 0 \mid C_3) = 0. \]
Hence, \[ G(t \mid C_3) = \sum _{n = 0}^{\infty } \Prob (N = n \mid C_3) \cdot x^n = \sum _{n = 1}^{\infty } \Prob (N = n - 1) \cdot t^n = t \cdot \sum _{n = 0}^{\infty } \Prob (N = n) \cdot t^n= tG(t). \]
Since in the first round, one of \(C_1, C_2\) and \(C_3\) must happen, we must have that \[ G(t) = \Prob (C_1) \cdot G(t \mid C_1) + \Prob (C_2) \cdot G(t \mid C_2) + \Prob (C_3) \cdot G(t \mid C_3) = a + b G(t) + ctG(t). \]
Hence, rearranging gives \[ (1 - b - ct) G(t) = a, \] and hence \[ G(t) = \frac {a}{(1 - b) - ct} = \frac {a / (1 - b)}{1 - ct / (1 - b)} \]
Hence, using the infinite expansion, we have \begin {align*} G(t) & = \frac {a}{1-b} \cdot \sum _{k = 0}^{\infty } \left (\frac {ct}{1 - b}\right )^k \\ & = \sum _{k = 0}^{\infty } \frac {a}{1 - b} \cdot \frac {c^k}{(1 - b)^k} \cdot t^k \\ & = \sum _{k = 0}^{\infty } \frac {ac^k}{(1 - b)^{k + 1}} \cdot t^k. \end {align*}
But the coefficient before \(t^n\) is precisely the probability \(\Prob (N = n)\). This means \[ \Prob (N = n) = \frac {ac^k}{(1 - b)^{k + 1}}, \] as desired.
We know that \(\mu = G'(1)\). We can find that \[ G'(t) = \frac {ac}{[(1 - b) - ct]^2}, \] and evaluating this at \(t = 1\) gives \[ \mu = G'(1) = \frac {ac}{(1 - b - c)^2} = \frac {ac}{a^2} = \frac {c}{a}. \]
Therefore, we have \(c = \mu a\) \begin {align*} \Prob (N = n) & = \frac {a c^k}{(a + c)^{k + 1}} \\ & = \frac {a (\mu a)^k}{(a + \mu a)^{k + 1}} \\ & = \frac {a \mu ^k a^k}{a^{k + 1} (1 + \mu )^{k + 1}} \\ & = \frac {\mu ^k}{\mu ^{k + 1}}, \end {align*}
as desired.