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

P( Not winning a car)= P(NWC)=2/3; and P(WC)=1/3

a) If door one picked P(WC)=1/3

b) since the host doesnt open door 1 then there are only two chances either open door 2 or door3

the probabiity of opening door2 is=1/2 and probability of opening door 3 is =1/2

c) given that O₃ has happened probability of (O₁/ O₃)=P(O₁)=1/2 (beacuse after opening door three the chances of winning the car increases and O₁ and O₃ are independant

similar argument holds if he decied to switch so P(O₂/O₃)=1/2