Question

2b) Suppose you are playing “Let’s Make a Deal”, the “prize” is worth $10,000, and a goat is worth nothing. Further suppose that the initial pick of a door is not part of the game (since it’s random anyway). You’ve always picked door A initially before the game has begun. The game starts as nature randomly determines where the prize is (1/3 chance for any of the three doors); then Monty Hall selects one of doors B or C that has a goat (also a move by nature); then you choose to Stay at door A or to Switch to the unopened door. Finally, the outcome is resolved – either you find a goat or the prize behind the door you finally selected. If both B and C have a goat, Monty flips a coin (50/50 chance) to determine which one to open. If either B or C have the car, there is only one available door Monty can open, so he opens that one.

Draw the complete game tree, with Nature choosing the right door, then Nature revealing either door B or C after A has been selected (you choose whether or not to map out a degenerate lottery), then you Stay or Switch. Remember to include probabilities, nodes, payoffs, and so on?

Answer 1

It is easy to see that when the you makes your first choice of one of the three doors, without any prior information, that the probability that the prize is behind the chosen door is 1/3. Now the you makes a further choice between switching doors or changing doors. The probabilities can best be calculated with a tree diagram.

At the start of the game, you are asked to pick a single door. There is a 1/3 chance that you have picked correctly, and a 2/3 chance that you are wrong. This does not change when one of the two doors you did not pick is opened. The second time is that you are choosing between whether your first guess was right (which has probability 1/3) or wrong (probability 2/3). Clearly it is more likely that your first guess was wrong, so you switch doors .

Placement of Prize Door chosen Door opened   Path

by YOU     by Monty probabilities

The conditional probabilities can be expressed as follows:

  Placement of Prize Door chosen Door opened conditional probability

   by You by Monty probabilities

  

Now lets solve using Probability Theorem:

We will Bayes Theorem for this

Bayes’ theorem is a formula that describes how to update the probability that a hypothesis is correct, given evidence. In this case, it’s the probability that our initially picked door is the one with the prize behind it (that staying is right) given that Monty has opened a door showing a goat behind it: that Monty has shown us that one of the choices we didn’t pick was the wrong choice.

Let H be the hypothesis "door 1 has a prize behind it," and

E be the evidence that Monty has revealed a door with a goat behind it. Then the problem can be restated as calculating  P(H ∣ E ), the conditional probability of H given E.

Since every door either has a prize or a goat behind it, the hypothesis "NOT H " is the same as "door 1 has a goat behind it."

In this case, Bayes' theorem states that:

• P(H) is the prior probability that door A has a prize behind it, without knowing about the door that Monty reveals. This is 1/3​.
• P(not H) is the probability that we did not pick the door with the prize behind it. Since the door either has the prize behind it or not, P(not H) = 1 - P(H) =2/3
• P(E ∣ H) is the probability that Monty shows a door with a goat behind it, given that there is a car behind door A. Since Monty always shows a door with a goat, this is equal to 1.
• P(E ∣ notH) is the probability that Monty shows the goat, given that there is a goat behind door A. Again, since Monty always shows a door with a goat, this is equal to 1.

Combining all of this information gives:

The probability that the prize is behind door A is completely unchanged by the evidence. However, since the prize can only either be behind door A  or behind the door Monty didn't reveal, the probability it is behind the door that is not revealed is 2/3​. Therefore, switching is twice as likely to get you the prize as staying.