1. Define the problem and initial conditions:
- Let $p_n$ be the probability of scoring exactly $n$ points.
- Initial conditions:
$$p_0 = 1 \quad \text{(no toss - certainty to get 0 points)}$$
$$p_1 = \frac{1}{2} \quad \text{(one fair toss of heads)}$$

2. Establish the recurrence relation:
- The probability of scoring $n+2$ points can be achieved by:
- Scoring $n+1$ points and then getting a head (probability $\frac{1}{2} p_{n+1}$)
- Scoring $n$ points and then getting a tail (probability $\frac{1}{2} p_n$)
- Therefore, the recurrence relation is:
$$p_{n+2} = \frac{1}{2} p_{n+1} + \frac{1}{2} p_n$$

3. Solve the characteristic equation:
- The characteristic polynomial for the recurrence relation $p_{n+2} = \frac{1}{2} p_{n+1} + \frac{1}{2} p_n$ is:
$$2\lambda^2 - \lambda - 1 = 0$$
- Solving this quadratic equation:
$$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$$
$$\lambda_1 = 1, \quad \lambda_2 = -\frac{1}{2}$$

4. Form the general solution:
- The general solution to the recurrence relation is:
$$p_n = \alpha \cdot 1^n + \beta \left( -\frac{1}{2} \right)^n = \alpha + \beta \left( -\frac{1}{2} \right)^n$$

5. **Determine the coefficients $\alpha$ and $\beta$:**
- Using the initial conditions:
$$p_0 = 1 \implies \alpha + \beta = 1$$
$$p_1 = \frac{1}{2} \implies \alpha - \frac{1}{2} \beta = \frac{1}{2}$$
- Solving these equations:
$$\alpha + \beta = 1 \quad \text{(1)}$$
$$\alpha - \frac{1}{2} \beta = \frac{1}{2} \quad \text{(2)}$$
- Multiply equation (2) by 2:
$$2\alpha - \beta = 1 \quad \text{(3)}$$
- Add equations (1) and (3):
$$\alpha + \beta + 2\alpha - \beta = 1 + 1$$
$$3\alpha = 2 \implies \alpha = \frac{2}{3}$$
- Substitute $\alpha = \frac{2}{3}$ into equation (1):
$$\frac{2}{3} + \beta = 1 \implies \beta = 1 - \frac{2}{3} = \frac{1}{3}$$

6. **Write the final expression for $p_n$:**
- Substituting $\alpha$ and $\beta$ back into the general solution:
$$p_n = \frac{2}{3} + \frac{1}{3} \left( -\frac{1}{2} \right)^n$$
- Simplifying:
$$p_n = \frac{1}{3} \left( 2 + \left( -\frac{1}{2} \right)^n \right)$$

The final answer is $\boxed{ \frac{1}{3} \left( 2 + \left( -\frac{1}{2} \right)^n \right) }$