Question

(Monty Hall problem) Suppose you’re on a game show, and you’re given the choice of three doors, say Door 1, Door 2, and Door 3. Behind one door there is a car; behind the others, goats. Assume it is equally likely that the car is behind any door, i.e., P(D1) = P(D2) = P(D3). You will win whatever is behind the door you choose.

(a) If you pick Door 1, what is your probability of winning the car? [2 point]

(b) Now the host, who knows what’s behind the doors, decides to give you more information. He will open another door that does not have the car behind it. He will not open Door 1 since you have picked it. What are the probabilities of the host opening Door 2 and Door 3. You may use O1, O2, or O3 to denote the events that he opens Door 1, Door 2, or Door 3, respectively. [4 points]

(c) Suppose that the host opens Door 3 (O3), which has a goat. He then gives you another opportunity to choose. Given this event, calculate the probabilities of winning the car if you stick to your original choice, and if you switch to pick Door 2. Based on this, should you stick to your original choice or switch? [4 points]

Answer 1

(a) Probability of winning a car if you pick Door 1

It is given that the car can be behind any of the three doors and also assumned that it is equally likely that the car is behind any door, i.e., P(D1) = P(D2) = P(D3). So, there are 3 possible outcomes for car which is Door 1 or Door 2 or Door 3, but there is only 1 favourable outcome which is Door 1.

Therefore, P(D1) = Number of favourable outcomes / No of total outcomes = 1/3

(b) Probabilities that the host opens Door 2 and Door 3.

The probability will depend upon which of the 3 doors actually has car behind it. Since you have picked the Door 1, host will not open it. Also, note that the host always know what is behind each door. Considering these, there can be 3 cases which are defined below and probability will depend on the cases. Let O1, O2, or O3 to denote the events that he opens Door 1, Door 2, or Door 3, respectively

Case1: If the car is actually behind Door 1.

Since the car is actually behind Door 1 and host knows that, so he can open any of the Door 2 or Doo3. This is equally likely event and thus probabilities will be same. Since there are 2 doors and anyone can be opened by host probaility will be 1/2.

Therefore P(O2) = P(O3) = 1/2

Case2: If the car is actually behind Door 2.

Since the Car is actually behind Door 2 and the host know that, so he is not going to open Door 2. So, he has now no other option than to open door 3.

Therefore, P(O2) = 0 and P(O3) = 1

Case3: If the car is actually behind Door 3.

Since the Car is actually behind Door 3 and the host know that, so he is not going to open Door 3. So, he has now no other option than to open door 2.

Therefore, P(O2) = 1 and P(O3) = 0

(c)

Given that the hosts opens Door 3 and there is a goat behind it. From the part (a) of the question we know that probability of a winning a car if you pick Door 1 is 1/3. It is same for you had picked any other door. If probablity of winning a car for each Door is 1/3 then, we can probability of the car being behind Door 2 or Door 3 is 1/3 + 1/3 = 2/3

So, lets say, Event 1 is P(D1) = 1/3

and Event 2 is P(D2 or D3) = 2/3.

Since, Door 3 is opened and has a goat, so the entire probability of the second (P(D2 or D3)) event will now be concentrated on Door 2 so now the Event 2 will look like P(Door 2) = 2/3 as Door 3 is now ruled out.

So you can see that P(D1) = 1/3 and P(D2) = 2/3.

Therefore P(Winning a car if you stick to Door 1) = 1/2 and

P(Winning a car if you switch to Door 2) = 2/3

Since, probability of switching to Door 2 is more than sticking to Door 1 so, based on this you should Swithch