2.3.7 problem 47

(a)

Contestant wins under the "stay-stay" strategy if and only if the Car is behind Door $1$.

$$P\left(S\right)=\frac{1}{4}$$

(b)

If the Car is not behind Door $1$, Monty opens one of the two doors revealing a Goat. Contestant stays. Then, Monty opens the other door with a Goat behind it. Finally, contestant switches to the Door concealing the Car.

$$P\left(S\right)=P\left(C_1\right)P\left(S\right|C_1)+P(C_1^c\left)P\right(S\left|C_1^c\right)$$

$$P\left(S\right)=0+\frac{3}{4}\ast 1=\frac{3}{4}$$

(c)

Under the "switch-stay" strategy, if the Car is behind Door $1$ the contestant loses. Given that the Car is not behind Door $1$, Monty opens one of the Doors containing a Goat. The contestant will win if they switch to the Door containing the Car and will lose if they switch to the Door containing the last remaining Goat.

Thus,

$$P\left(S\right)=P\left(C_1\right)P\left(S\right|C_1)+P(C_1^c\left)P\right(S\left|C_1^c\right)=0+\frac{3}{4}\ast \frac{1}{2}=\frac{3}{8}$$

(d)

Under the "switch-switch" strategy, if the car is behind Door $1$, then Monty opens a door with a Goat behind it. The contestant switches to a door with a Goat behind it. Monty then opens the last door containing a Goat, at which point the contestant switches back to the door containing the Car.

If Door $1$ contains a Goat, Monty opens another Door containing a Goat and presents the contestant with a choice. If the contestant switches to the remaining door containing a Goat, then Monty is forced to open Door $1$, revealing the final Goat. The contestant switches to the one remaining Door which contains the Car. If, on the other hand, the contestant switches to the door containing the Car, then on the subsequent switch they lose the game.

Thus,

$$P\left(S\right)=\frac{1}{4}\ast 1+\frac{3}{4}\ast \frac{1}{2}=\frac{5}{8}$$

(e)

"Stay-Switch" is the best strategy.