Consider the Monty Hall problem. Let's label the door with the car behind it a and the other two doors b and c. In the game the contestant chooses a door and then Monty chooses a door, so we can label each outcome as contestant followed by Monty, e.g. ab means the contestant chose a and Monty chose B

Question

(a) Make a 3 x 3 probability table showing probabilities for all possible outcomes.
(b) Make a probability tree showing all possible outcomes.
(c) Suppose the contestant strategy is to switch. List all the outcomes in the event the contestant wins a car. What is the probability the contestant wins?
(d) Redo part (c) with the strategy of not switching.

Answer 1

The Monty Hall problem is a well-known probability puzzle based on a game show scenario. Let's work through each part of this problem step by step.

(a) 3 x 3 Probability Table: The three doors are labeled as a (the car), b, and c. We'll analyze the probabilities for all the possible outcomes. The contestant can choose any of the three doors, and Monty, knowing where the car is, will always open a door that does not have the car behind it and was not initially chosen by the contestant.

Let's denote the contestant's choice by the first letter and Monty's choice by the second letter. The outcomes can be represented as follows:

• If the car is behind door a:
  • The contestant chooses a (a), Monty can choose b or c: outcomes are aa (choose a, Monty shows b) and ac (choose a, Monty shows c)
• If the car is behind door b:
  • The contestant chooses b (b), Monty can choose a or c: outcomes are ba (choose b, Monty shows a) and bc (choose b, Monty shows c)
• If the car is behind door c:
  • The contestant chooses c (c), Monty can choose a or b: outcomes are ca (choose c, Monty shows a) and cb (choose c, Monty shows b)

The overall probabilities for the contestant's choice and Monty's response depends on the car’s location. Each state (car behind a, b, or c) is equally likely, so each has a probability of \( \frac{1}{3} \).

Now, the table showing all possible outcomes (with their probabilities):

| Contestant | Monty | Probability | |------------|-------|-------------| | a | b | \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \) | | a | c | \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \) | | b | a | \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \) | | b | c | \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \) | | c | a | \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \) | | c | b | \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \) |

So the final table looks like this:

\[ \begin{array}{|c|c|c|} \hline \text{Contestant} & \text{Monty} & \text{Probability} \ \hline a & b & \frac{1}{6} \ a & c & \frac{1}{6} \ b & a & \frac{1}{6} \ b & c & \frac{1}{6} \ c & a & \frac{1}{6} \ c & b & \frac{1}{6} \ \hline \end{array} \]

(b) Probability Tree: A probability tree for this problem can be represented as:

1. The first level represents the contestant's choice (each branch has \( \frac{1}{3} \)):

   • \( A \) (Car behind a)
   • \( B \) (Car behind b)
   • \( C \) (Car behind c)

2. The second level represents Monty's choice (given the door that the contestant chose):

   • For \( A \): branches to \( b \) and \( c \)
   • For \( B \): branches to \( a \) and \( c \)
   • For \( C \): branches to \( a \) and \( b \)

Here's how it looks:

    (1/3) A (car behind a)
        /      \
       (1/2)  (1/2)
      b        c
     (1/6)   (1/6)
       
    (1/3) B (car behind b)
        /      \
       (1/2)  (1/2)
      a        c
     (1/6)   (1/6)
     
    (1/3) C (car behind c)
        /      \
       (1/2)  (1/2)
      a        b
     (1/6)   (1/6)

Each path will lead to the probability of that sequence of choices.

(c) Outcomes and Winning Probability for Switching Strategy: If the contestant always switches after Monty reveals a door, let's analyze each scenario:

1. Choose a (car behind a):

   • Monty reveals either b or c: contestant switches to the other unopened door (either b or c) — contestant loses.

2. Choose b (car behind b):

   • Monty will reveal a: contestant switches to c — contestant wins.

3. Choose c (car behind c):

   • Monty will reveal a: contestant switches to b — contestant wins.

The outcomes where the contestant wins by switching are:

• bA → switch to c (win)
• cA → switch to b (win)

The probability of winning by switching:

• Wins if the contestant first chooses b: \( \frac{1}{3} \)
• Wins if the contestant first chooses c: \( \frac{1}{3} \)

Total winning probability by switching becomes: \[ \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \]

(d) Outcomes and Winning Probability for Not Switching Strategy: If the contestant does not switch, then let's analyze:

1. Choose a (car behind a): contestant wins.
2. Choose b (car behind b): contestant loses.
3. Choose c (car behind c): contestant loses.

The outcomes where the contestant wins without switching are:

• aA → stay (win)

The probability of winning by not switching:

• Wins if the contestant first chooses a: \( \frac{1}{3} \).

Total winning probability by not switching becomes: \[ \frac{1}{3}. \]

To summarize:

• Switching gives a winning probability of \( \frac{2}{3} \).
• Not switching gives a winning probability of \( \frac{1}{3} \).