91Ó°ÊÓ

In this exercise we will use Bayes’ theorem to solve the Monty Hall puzzle (Example 10 in Section 7.1). Recall that in this puzzle you are asked to select one of three doors to open. There is a large prize behind one of the three doors and the other two doors are losers. After you select a door, Monty Hall opens one of the two doors you did not select that he knows is a losing door, selecting at random if both are losing doors. Monty asks you whether you would like to switch doors. Suppose that the three doors in the puzzle are labeled 1, 2, and 3. Let W be the random variable whose value is the number of the winning door; assume that p(W = k) = 1?3 for k = 1, 2, 3. Let M denote the random variable whose value is the number of the door that Monty opens. Suppose you choose door i. a) What is the probability that you will win the prize if the game ends without Monty asking you whether you want to change doors? b) Find p(M = j ? W = k) for j = 1, 2, 3 and k = 1, 2, 3. c) Use Bayes’ theorem to find p(W = j ? M = k) where i and j and k are distinct values. d) Explain why the answer to part (c) tells you whether you should change doors when Monty gives you the chance to do so.

Step by step solution

01

- Probability of Winning Without Switching

When the game ends without Monty asking you to switch doors, you essentially have a 1 in 3 chance of picking the correct door since there are three doors and only one has the prize behind it. Therefore, the probability of winning without switching is \(\frac{1}{3}\).

02

- Condition Probability p(M = j | W = k)

To find p(M = j | W = k), consider the scenarios where Monty opens a door: - If W = 1 (prize behind door 1), M can be 2 or 3 with equal probability, so p(M = 2 | W = 1) = \(\frac{1}{2}\) and p(M = 3 | W = 1) = \(\frac{1}{2}\).- If W = 2 (prize behind door 2), M can be 1 or 3 with equal probability, so p(M = 1 | W = 2) = \(\frac{1}{2}\) and p(M = 3 | W = 2) = \(\frac{1}{2}\).- If W = 3 (prize behind door 3), M can be 1 or 2 with equal probability, so p(M = 1 | W = 3) = \(\frac{1}{2}\) and p(M = 2 | W = 3) = \(\frac{1}{2}\).

03

- Using Bayes’ Theorem

To find p(W = j | M = k) using Bayes’ theorem where i, j, and k are distinct values, we have\[ p(W = j | M = k) = \frac{p(M = k | W = j) \, p(W = j)}{p(M = k)} \]Since p(W = j) = \(\frac{1}{3}\) and p(M = k) = \(\frac{1}{3}\), and using the values from Step 2:\[ p(W = j | M = k) = \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{1}{3}} = \frac{1}{2} \]

04

- Decision on Switching Doors

If Monty opens a door, the probabilities shift. Using Bayes’ theorem finding that p(W = j | M = k) is \(\frac{1}{2}\) indicates that there is a higher probability of winning if you switch doors. This leads to the conclusion that you should switch doors when Monty gives you the chance to maximize your probability of winning from \(\frac{1}{3}\) to \(\frac{2}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bayes' theorem

Bayes' theorem provides a way to update the probability of a hypothesis based on new evidence. In the context of the Monty Hall Problem, Bayes' theorem helps us calculate the probability that a certain door hides the prize after Monty reveals a losing door. The formula for Bayes' theorem is \( p(A | B) = \frac{p(B | A) \, p(A)}{p(B)} \). Here, p(A | B) is the updated probability we're looking for, p(B | A) is the probability of the new evidence given the initial hypothesis, p(A) is the initial probability of the hypothesis, and p(B) is the overall probability of the new evidence.

conditional probability

Conditional probability captures the idea that the likelihood of an event happening may depend on another event. For instance, in the Monty Hall Problem, we want to calculate the probability that the prize is behind door j given that Monty opened door k (distinct from our initial choice). Denoted as p(W = j | M = k), it tells us how our chances change after Monty's action. By calculating p(M = k | W = j) for each possible scenario, we use this to find our updated probabilities of winning after Monty reveals a door.

probability theory

Probability theory deals with the analysis of random events. In the Monty Hall Problem, basic concepts from probability theory are used, such as the probability of a winning door being 1/3 initially. When we use probability calculations, we're exploring different outcomes and their likelihoods. Key ideas include:
- Probability of an event, like selecting the winning door, is \( \frac{1}{3} \) given there are 3 doors.
- Equal likelihood for Monty to open any losing door when multiple losing doors are available.
- Combining such probabilities helps us arrive at informed decisions.

decision making

Decision making involves choosing an optimal action based on given information. When applied to the Monty Hall Problem, understanding conditional probabilities guides our choices. Initially, the probability of picking the winning door is 1/3. After Monty reveals a losing door, this probability changes. By applying Bayes' theorem, we find that switching doors increases the probability of winning to 2/3. The decision to switch is thus backed by mathematical probability, showing that shifting your choice improves your odds significantly. This move from intuition to data-driven decision making is a powerful application of probability theory.