6) (15 points) The Monte Hall Problem: You are on the game show Let\'s Make a De

Question

6) (15 points) The Monte Hall Problem: You are on the game show Let's Make a Deal, hosted by Monte Hall. A new car was placed behind one of three doors. You select a door, and for simplicity you choose door 1. Before opening your door, Monte shows you what was behind one of the other two doors, let's say door number 3 (nothing). Finally, Monte asks you if you would like to keep your original door or if you would like to switch your choice to door 2. What do you do? Use Bayes Theorem to justify your answer. Hint. Let C be the door that contains the car, so P(C = 1) = P (C = 2) = P(C = 3) = Let S be the door Monte shows you. We can assume that under no circumstances would he show you the door with the car. Hence, we can model S given which door the car is actually behind as follows: 1 PCS C = 1) PCs IC = 2) P(S |C = 3) 0 0.5 0.5 0 For example, the 1 in the bottom row of the table means that the probability Monte shows you what is behind door 2 (S) given that the car is actually behind door 3 (C) is 1. This is because Monte cannot show you what is behind door 1 since you picked it and he cannot show you door 3 either since the car is actually behind it

Explanation / Answer

This is famous Monte Hall problem

We should switch the door

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: Pr(AB)=Pr(BA)×Pr(A)Pr(B)Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty shows us a goat behind Door 2. Now let A be the event that the car is behind Door 1 and B be the event that Monty shows us a goat behind Door 2. Then
Pr(AB)=Pr(BA)×Pr(A)Pr(B)=1/2×1/31/3×1/2+1/3×0+1/3×1=1/3.The tricky calculation is Pr(B). Remember, we are assuming we initially chose Door 1. It follows that if the car is behind Door 1, Monty will show us a goat behind Door 2 half the time. If the car is behind Door 2, Monty never shows us a goat behind Door 2. Finally, if the car is behind Door 3, Monty shows us a goat behind Door 2 every time. Thus, Pr(B)=1/3×1/2+1/3×0+1/3×1=1/2.The car is either behind Door 1 or Door 3, and since the probability that it's behind Door 1 is 1/3 and the sum of the two probabilities must equal 1, the probability the car is behind Door 3 is 11/3=2/3. You could also apply Bayes' Theorem directly, but this is simpler.

So Bayes says we should switch, as our probability of winning the car jumps from 1/3 to 2/3.

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