Question

In the problem, you are on a game show, being asked to choose between three doors. Behind each door, there is either a car or a goat. You choose a door. The host, Monty Hall, picks one of the other doors, which he knows has a goat behind it, and opens it, showing you the goat. (You know, by the rules of the game, that Monty will always reveal a goat.) Monty then asks whether you would like to switch your choice of door to the other remaining door.

(1) would you switch your choice or not? why?

(2) draw a normal form game. is there a dominating strategy?

(3) draw an extensive form game. support the choice in (1).

Answer 1

1.

Yes. Switching increases the probability of winning. In the first case when all the three doors are shown, the probability of the car bieng in behind any door is 1/3. However, after monty opens one of the doors revealing a goat, the probability of the car being in the door not chosen in the first round increases to 1/2, as monty did not open that door, but probability of the door chosen in the first round is still 1/3 . Adding up the probabilities, suppose you chose door A and Monty opened door C, and then you shifted to door B. The probability of the car being behind door A is 1/3 + 1/3 = 1/9, and the probabilitty of the car being behind door B is 1/3 + 1/2 = 5/6. Hence, by shifting, you increase your probability of winning by 5/6 - 1/9 = 13/18 or 72.22%.

2.

The normal form of the game can be drawn as follows -

Car is in box	You choose	Monty opens	You switch to	Result
A	A	B or C	A for B or C	loss
B	A	C	A for B	win
C	A	B	A for C	win
A	B	C	B for A	win
B	B	A or C	B for A or C	loss
C	B	A	B for C	win
A	C	B	C for A	win
B	C	A	C for B	win
C	C	A or B	A for A or B	loss

We can see that on switching, you can win 6 out of 9 times, which means the porbability of winning is 2/3 and that of losing is 1/3.

3.

The extensive form of the game can be drawn as follows -

Let us say that the car is in the door circled. In the first column, it is inside door A, and hence A is circled.

Consider the first row where you choose one of the three doors A, B or C with equal probability of 1/3. If you choose door A, Monty openes eiher door B or C with probability 1/2, and if you choose door B or C (which has goat behind), Monty is forced to open the other door with goat inside as he cannot open door A. Now consider the following -

Case 1: Your choose door A - Monty opens door B or C - You switch to B or C whichever is left unopened and you lose.

Case 2: Your choose door B - Monty has to open door C - You switch to A and you win.

Case 3: Your choose door C - Monty has to open door B - You switch to A and you win.

Hence, in 2 out of three cases, you win. Thus, your chance of winning is massively increased.

The same can be explained if you initially choose door B or C, as shown in the diagram.