Let $A_{i}$ be the event "the prize is in th $i$ - th door". Let's consider $P(A_{1} | A_{2}^c)$, that is, the probability that the prize is in the first door, given that it is not in the second door (Monty Hall has revealed us the door). Using Bayes' theorem, we get
$$P(A_{1} | A_{2}^c) = \dfrac{P(A_{2}^c | A_{1}) \cdot P(A_{1})}{P(A_{2}^c)}.$$
The probability that the prize is not in the second door, given that it is in the first one is clearly 1. Hence, $P(A_{2}^c | A_{1}) = 1$. Because the prize is put at random, $P(A_{1}) = \dfrac{1}{3}$. Finally, the probability that the prize is not in the second door is $\dfrac{2}{3}$. In consecuence, $P(A_{2}^c) = \dfrac{2}{3}$. Substituing the values, we get
$$P(A_{1} | A_{2}^c) = \dfrac{1}{2}.$$
However, from what I have found, the correct answer is $\dfrac{1}{3}$. What am I doing wrong?