5.5 Monty Hall Problem

This problem was introduced in Section 2.7.

Example 1 Is it better to switch doors? - Monty Hall Problem revisited

You may recall that we were presented with a choice of 3 doors where a car is behind one and goats behind the others. Having picked one, the host opens up a door with a goat, and offers you the opportunity to change your answer. In order to assess the probabilities, we must remember that

the host never opens your door
the host always opens a door with a goat

We'll go through a specific example, that of you choosing door 1 and the host opening door 2. The analysis proceeds in identical ways for the other possibilities. We apply the Bayes' Recipe, where the models under consideration are

“car behind door 1”
“car behind door 2”
“car behind door 3”

The Bayes' Recipe proceeds as follows

Specify the prior probabilities for the models being considered
$\begin{eqnarray*} P\left(\text{car 1} \,|\, \text{you 1}\right) &=& 0.333 \\ P\left(\text{car 2} \,|\, \text{you 1}\right) &=& 0.333 \\ P\left(\text{car 3} \,|\, \text{you 1}\right) &=& 0.333 \end{eqnarray*}$
where, for example, $P\left(\text{car 1} \,|\, \text{you 1}\right)$ represents the probability that the door contains the car given that you chose door 1. Since your choice of door doesn't add any information about the location of the car, all of the probabilities are equal.
Write the top of Bayes' Rule for all models being considered
$\begin{eqnarray*} P\left(\text{car 1} \,|\, \text{you 1, host 2}\right) &\sim& P\left(\text{host 2} \,|\, \text{you 1, car 1}\right)P\left(\text{car 1} \,|\, \text{you 1}\right) \\ P\left(\text{car 2} \,|\, \text{you 1, host 2}\right) &\sim& P\left(\text{host 2} \,|\, \text{you 1, car 2}\right)P\left(\text{car 2} \,|\, \text{you 1}\right) \\ P\left(\text{car 3} \,|\, \text{you 1, host 2}\right) &\sim& P\left(\text{host 2} \,|\, \text{you 1, car 3}\right)P\left(\text{car 3} \,|\, \text{you 1}\right) \end{eqnarray*}$
Put in the likelihood and prior values
Due the restrictions on the host above, the host cannot open a door with a car, so $P\left(\text{host 2} \,|\, \text{you 1, car 2}\right)=0$ . In the case where you choose door 1 and the car is also behind door, the host has the freedom to choose either door 2 or door 3, so $P\left(\text{host 2} \,|\, \text{you 1, car 1}\right)=0.5$ . Where the information comes in is when the car is behind door 3 and you've chosen door 1. In that case, the host cannot open your door (door 1) or the door with the car (door 3) and must open door 2. Thus, $P\left(\text{host 2} \,|\, \text{you 1, car 3}\right)=1$ .
The final result of this step is
$\begin{eqnarray*} P\left(\text{car 1} \,|\, \text{you 1, host 2}\right) &\sim& 0.5 \cdot 0.333\\ P\left(\text{car 2} \,|\, \text{you 1, host 2}\right) &\sim& 0 \cdot 0.333 \\ P\left(\text{car 3} \,|\, \text{you 1, host 2}\right) &\sim& 1 \cdot 0.333 \end{eqnarray*}$
Add these values for all models
$\begin{eqnarray*} K=0.5 \cdot 0.333+ 1 \cdot 0.333 = 0.5 \end{eqnarray*}$
Divide each of the values by this sum, $K$ , to get the final probabilities
$\begin{eqnarray*} P\left(\text{car 1} \,|\, \text{you 1, host 2}\right) &=& \frac{0.5 \cdot 0.333}{0.5}=0.333\\ P\left(\text{car 2} \,|\, \text{you 1, host 2}\right) &=& \frac{0 \cdot 0.333}{0.5}= 0 \\ P\left(\text{car 3} \,|\, \text{you 1, host 2}\right) &=& \frac{1 \cdot 0.333}{0.5}= 0.666 \end{eqnarray*}$

Thus, in the case, given that you choose door 1 and the host chooses 2, the probability that the car is behind door 1 (your door) is 0.333 and the other door (door 3) is 0.666. Following the same steps through the other cases, we get in summary

Probability of...

Your Choice

Host Choice

Car Behind 1

Car Behind 2

Car Behind 3

(host can't open your door)

0.333

0.666

0.333

0.666

0.333

0.666

(host can't open your door)

0.666

0.333

0.666

0.333

0.666

0.333

(host can't open your door)

In summary, it is always better to switch to the remaining door, given these rules.

Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.