Given that the guest chooses door 1 and the host opens door 3, we apply Bayesian probability to find the probability that the car is behind door 1:

$P(1=\text{car}|3=\text{opened}) = \frac{P(3=\text{opened}|1=\text{car}) \cdot P(1=\text{car})}{P(3=\text{opened})}$

$P(3=\text{opened})$ is the total probability that the host opens door 3 when the guest picked door 1:

$P(3=\text{opened}) = P(3=\text{opened}|1=\text{car}) \cdot P(1=\text{car}) + P(3=\text{opened}|2=\text{car}) \cdot P(2=\text{car}) + P(3=\text{opened}|3=\text{car}) \cdot P(3=\text{car})$

$P(3=\text{opened}) = \frac{1}{2} \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} = \frac{1}{2}$

Substituting this back into the first equation:

$P(1=\text{car}|3=\text{opened}) = \frac{\frac{1}{2} \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$

Hence, the probability that the car is behind door 1 is $\frac{1}{3}$, given the guest selected door 1 and the host opened door 3.