Exercise 2.42

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)

$$\begin{array}{llll}\hfill P(S)& =P(G_1)P(S|G_1)+P(G_1^c)P(S|G_1^c)& \hfill & \\ \hfill & =\frac{2}{3}p+0& \hfill & \\ \hfill & =\frac{2}{3}p.& \hfill & \end{array}$$

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)

$$\begin{array}{llll}\hfill P(G_1|D_2)& =\frac{P(G_1)P(D_2|G_1)}{P(D_2)}& \hfill & \\ \hfill & =\frac{P(G_1)P(D_2|G_1)}{P(G_1)P(D_2|G_1)+P(G_1^c)P(D_2|G_1^c)}& \hfill & \\ \hfill & =\frac{\frac{2}{6}p}{\frac{2}{6}p+\frac{1}{6}}& \hfill & \\ \hfill & =\frac{2p}{2p+1}.& \hfill & \end{array}$$

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.