Difference between revisions of "Bayes theorem eg3 S13" - Rhea

Revision as of 21:00, 16 March 2013

Example 3: Monty Hall

by Maliha Hossain

keyword: probability, Monty Hall, Bayes' Theorem, Bayes' Rule 

The Monty Hall problem is based on the television game show Let's Make a Deal. The problem is named after the show's original host, Monty Hall. The problem was published by Marilyn vos Savant in Parade Magazine as follows:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Many readers, including scholars holding PhD's, refused to believe that switching would be to the contestant's advantage even though it can be demonstrated using proofs and computer simulation.

We will use Bayesian inference to show that contestants have a 2/3 chance of winning the car by switching whereas contestants who do not switch only have a 1/3 chance of winning.

Let's assume you are the contestant on Let's Make a Deal and you choose door number 1.

Let $ a $ be the event that the car is behind door number 1.

Let $ b $ be the event that the car is behind door number 2.

Let $ c $ be the event that the car is behind door number 3.

The prize is equally likely to behind any one of the three doors, so

$ P[a] = P[b] = P[c] = \frac{1}{3} $

Let C be the event that Monty Hall opens door number 3 after you have chosen door number 1.

Given that the car is behind door number 1, the door you have chosen, Monty has a 1/2 chance of picking door 3 from the remaining two doors to reveal a goat. So we have that

$ P[C|a] = \frac{1}{2} $

Given that the car is behind door number 2, and you have chosen door 1, that leaves Monty with only one choice. He can't reveal the door with the car and he also can't reveal what's behind the door you've chosen. So he has to pick door number 3.

$ P[C|b] = 1 $

Given that the car is behind door number 3, and you have chosen door 1, Monty will not choose door 3 since in doing so, he will reveal the car. Monty can not reveal the car before he's given you a chance to revise your decision. So

$ P[C|c] = 0 $

By the theorem of total probability, we have that

$ \begin{align} P[C] &= P[C|a]P[a] + P[C|b]P[b] + P[C|c]P[c] \\ &= \frac{1}{2} \times \frac{1}{3} + 1\times\frac{1}{3} + 0\times\frac{1}{3}\\ &= \frac{1}{2} \end{align} $

What is the probability that you've lost by switching to door number 2? In other words, what is $ P[a|C] $? By Bayes' Theorem, we have that

$ \begin{align} P[a|C] &= \frac{P[C|a]P[a]}{P[C]} \\ &= \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{1}{2}}\\ &= \frac{1}{3} \end{align} $

Now what is the probability that you've won by switching? That is to say, what is $ P[b|C] $? By Bayes' Theorem, we have that

$ \begin{align} P[b|C] &= \frac{P[C|b]P[b]}{P[C]} \\ &= \frac{1 \times \frac{1}{3}}{\frac{1}{2}}\\ &= \frac{2}{3} \end{align} $

Since the assignment of door numbers that the you pick and the event C are arbitrary, you will arrive at the same results regardless of which door you pick first and which Monty opens later.

References

• Mark Haddon, The Curious Incident of the Dog in the Night-Time

Questions and comments

If you have any questions, comments, etc. please post them below:

• Comment / question 1

Back to Math Squad page