Solution: Note that a player can score $n$ in the following ways:

• by throwing a tail when score is $n-2$, and
• by throwing a head when score is $n-1$ .

Also note that these two are mutually exclusive. Since they are mutually exclusive, the probability will be \begin{align*} P(a) + P(b) & = \frac{1}{2} p_{n-2} + \frac{1}{2} p_{n-1} \\ & = \frac{1}{2}\left( p_{n-2} + p_{n-1} \right) . \end{align*} Thus we proved that \[ p_n = \frac{1}{2}\left( p_{n-1} + p_{n-2} \right). \]

We now find an expression for $p_n$. Note that the above expression can be re-written as (by adding $\frac{1}{2}p_{n-1} $ on the both sides) \begin{align*} p_n + \frac{1}{2}p_{n-1} & = p_{n-1} + \frac{1}{2}p_{n-2} \\ & = p_{n-2} + \frac{1}{2}p_{n-3} \\ & = p_{n-3} + \frac{1}{2}p_{n-4} \\ & = \cdots \quad \cdots \\ & = p_2 + \frac{1}{2}p_1. \end{align*}

Let us find $p_1$ and $p_2$. As $p_1$ is the probability of getting exactly $1$ score, which can be obtained by a head in the first throw and hence $p_1 = \frac{1}{2}$. The score $2$ can be obtained in two ways.

• Head in the first and second throw.
• Tail in the first throw.

Thus, \[ p_2 = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} = \frac{3}{4}. \] Therefore, we have \begin{align*} p_n + \frac{1}{2} p_{n-1} & = \frac{3}{4} + \frac{1}{2} \times \frac{1}{2} \\ & = \frac{3}{4} + \frac{1}{4} = 1 \\ \implies p_n & = 1 - \frac{1}{2}p_{n-1}. \end{align*} Therefore, \[ \textcolor{teal}{\boxed{ p_n = 1 - \frac{1}{2}p_{n-1},\quad n\geq 2, \text{ and } p_1 = \frac{1}{2}. }} \]