A coin is tossed repeatedly. If a head turns up, the player gets $1$ point; if a tail turns up, the player gets $2$ points. What is the probability that the player gets a total of exactly $n$ points?
My attempt: I tried to partition $n$ by $1$′s & $2$′s . Thus if $n=2k$ for some $k\in\mathbb{N}$, there are either $k$ number of $2$′s & $0$ number of $1$′s or $(k−2)$ number of $2$′s & $2$ number of $1$′s or etc. So the number of possible cases is $\binom{2k}{k}\cdot\binom{2k}{0}+\binom{2k}{k-1}\cdot\binom{2k}{2}+⋯+\binom{2k}{0}\cdot\binom{2k}{2k}$. Thus the probability of obtaining one of the possible cases is $\dfrac{1}{\binom{2k}{k}+\binom{2k}{k-1}⋅\binom{2k}{2}+⋯+\binom{2k}{0}⋅\binom{2k}{2k}}$. Similarly, if $n=2k−1$ for some $k\in\mathbb{N}$ then the number of possible cases can be evaluated.
The answer in the book is $\dfrac12+\dfrac16\left(1−\left(−\dfrac12\right)^{n−1}\right)$.
Does the expression I've evaluated simplifies to this answer after some algebraic computations or am I in a completely wrong direction?
Let's consider the last throw in a successful series. Player had either $n-1$ points and threw head, or had $n-2$ points and threw tails. So, the probability: $$ p_n = \frac12p_{n-1}+\frac12 p_{n-2} $$
We solve this recursive equation by assuming $p_n = z^n$ and fidning eigenvalues $z_1=1$, $z_2=-1/2$. The probability is the sum of those: $$ p_n = C_1z_1^n+C_2z_2^n = C_1 + C_2\left(-\frac12\right)^n. $$ Finally, we find constants from the initial conditions: $p_0=1$, $p_1=1/2$: $$ p_n = \frac23+\frac13\left(-\frac12\right)^n $$