The Monty Hall Paradox

The Monty Hall 'problem' (dilemma? paradox?) is a probability puzzle, originally posed in 1975 by Steve Selvin in a letter to the American Statistician. The puzzle is based on the American television game show "Let's Make a Deal" and named after the show's original host, Monty Hall. It was popularized by Marilyn vos Savant in her "Ask Marilyn" column in 1990.

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?

Marilyn's answer was that you should switch doors, that in general you have a 2/3 chance of winning if you switch, and only a 1/3 chance if you stick with your original choice. This unleashed a torrent of critical responses (more than ten thousand in all, with about one thousand coming from PhDs) – and just a few defenses.

I have included a few of the responses, taken from vos Savant's magnificent – and highly recommended – book "The Power of Logical Thinking,"

"Your answer to the question is in error. But if it any consolation, many of my academic colleagues have also been stumped by this problem."

B. P., Ph.D., California Faculty Association
"You blew it and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I'll explain. After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the chances are the same. There is enough mathematical illiteracy in this country, and we don't need the world's highest IQ propagating more. Shame!"

S. S., Ph.D., California Faculty Association

At the time, around 1970, vos Savant had the highest recorded IQ in the world, with a score of 228. Nowadays, she is third overall, and the most intelligent female.

"Dear Marilyn:

Since you seem to enjoy coming straight to the point, I'll do the same. In the following question and answer, you blew it! Let me explain. If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2. As a professional mathematician, I'm very concerned with the general public's lack of mathematical skills. Please help by confessing your error and in future being more careful."

R. S., Ph.D., George Mason University

Wow! These Ph.D.s (all male, as it happens!) seem pretty sure of themselves. But they, and another thousand or so, Ph.D.s were wrong.

Justification

I have a degree in mathematics which is far below the giddy heights where doctoral academics reside. But I will try my best to explain.

Consider first, the don't switch strategy. You have a 1/3 chance, or probability, of guessing correctly (with one-third probability) which door the car is behind. Since you don't switch, you will win if you guessed correctly. So, the probability of winning is 1/3. And the probability of losing is 2/3.

Now, consider the switch strategy. You have a 1/3 chance of guessing correctly which door the car is behind. So, if you initially guessed correctly and then switched, you lose. With 1/3 probability. The probability that you will guess incorrectly is 2/3. And, NOTE that whenever you guess incorrectly, you will win when you switch!

Why do you always win when you switch after guessing incorrectly? Because then one of the two doors you didn't pick contains the car. And the other goat. The host will always reveal the goat, leaving the car behind the closed door you didn't initially choose. If you switch to the door that the host doesn't open, you win. (If this is confusing, just step through the simulation a few times, paying close attention to what happens when you switch!)

When you see it, you'll wonder how so many of these brainiacs ended up with egg on their faces!

Some Probability

There is a concept in probability theory called the expected value of a random variable. It is the sum of the probabilities of each possible outcome of the variable multiplied by the value of the outcome.

If XX represents a random variable (such as, in this case, the contestant winnings), then the expected value of XX is E[X]=∑xk⋅Pxk E[X] = \sum x_k\cdot P_{x_k} where xkx_k is the value of the kkth outcome of each 'experiment' (winning, losing, heads, tails, ...) and PxkP_{x_k} is the probability of that outcome (winning that prize, losing, getting a head, ...).

Expected winnings with a 'DON'T SWITCH' strategy:
Guessing the door with the car behind occurs with a one in three chance. By not switching, you win the car. You guess incorrectly with a two in three chance. By not switching, you win a goat. E[X]=Car⋅Pguessing correctly+Goat⋅Pguessing incorrectly=Car⋅13+Goat⋅23=Car3+2⋅Goat3 \begin{aligned} E[X] &= \text{Car}\cdot P_{\text{guessing correctly}} \\ &\quad\qquad + \text{Goat}\cdot P_{\text{guessing incorrectly}} \\ &= \text{Car}\cdot\frac 13 + \text{Goat}\cdot\frac 23 \\ &= \frac{\text{Car}}{3} + \frac{2\cdot\text{Goat}}{3} \end{aligned} You should expect (on average) to win the car one time in three.

Expected winnings with a 'SWITCH' strategy:
Guessing the door with the car behind still occurs with a one in three chance. But by switching, you do not win the car. However, if you guess incorrectly (which has a two in three chance) and switch, you win the car. E[X]=Goat⋅Pguessing correctly+Car⋅Pguessing incorrectly=Goat⋅13+Car⋅23=2⋅Car3+Goat3 \begin{aligned} E[X] &= \text{Goat}\cdot P_{\text{guessing correctly}} \\ &\quad\qquad + \text{Car}\cdot P_{\text{guessing incorrectly}} \\ &= \text{Goat}\cdot\frac 13 + \text{Car}\cdot\frac 23 \\ &= \frac{\mathrm{2 \cdot Car}}{3} + \frac{\text{Goat}}{3} \end{aligned} You should expect (on average) to win the car two times in three.