problem 42

Let $G_{i}$ be the event that the $i$ -th door contains a goat, and let $D_{i}$ be the event that Monty opens door $i$ .

Let $S$ be the event of success under the specified strategy.

(a): $\begin{matrix} P (S) & = P (G_{1}) P (S ∣ ∣ G_{1}) + P (G_{1}^{c}) P (S ∣ ∣ G_{1}^{c}) = \frac{2}{3} p + 0 = \frac{2}{3} p . \end{matrix}$
Note that when $p = 1$ , the problem reduces to the basic Monty Hall problem, and we get the correct solution $\frac{2}{3}$ . In the case when $p = 0$ , Monty never gives the contestant a chance to switch their initial, incorrect choice to the correct one, resulting in a definite failure under the specified strategy.
(b): $\begin{matrix} P (G_{1} | D_{2}) & = \frac{P (G_{1}) P (D_{2} | G_{1})}{P (D_{2})} = \frac{P (G_{1}) P (D_{2} | G_{1})}{P (G_{1}) P (D_{2} ∣ ∣ G_{1}) + P (G_{1}^{c}) P (D_{2} ∣ ∣ G_{1}^{c})} = \frac{\frac{2}{6} p}{\frac{2}{6} p + \frac{1}{6}} = \frac{2 p}{2 p + 1} . \end{matrix}$
Note that if $p = 1$ , the problem reduces to the basic Monty Hall problem, and the solution matches that of the basic, conditional Monty Hall problem. If $p = 0$ on the other hand, then the reason Monty has opened a door is because the contestant’s initial guess (Door $1$ ) is correct. By choosing the strategy to switch, the contestant always loses.