The Monty Hall Paradox
What Would Monty Do?
Posted May 31, 2018
Suppose I invited you to play a guessing game. It's a game of forced choice with incomplete information.
In my desk drawer I have three bills of US currency – two $1 bills and one $100 dollar bill. The drawer is closed, so you can’t see them. With colored pens I have placed a blue dot on one bill, a red dot on another bill, and a green dot on a third bill. You can have any bill you choose, but you must choose by color. Which color do you want?
Suppose you say, “red.” I reach in the drawer and remove a $1 bill which has a green dot. Now you know the $100 bill did not have the green dot. It has either the red dot (your guess) or the blue dot.
Before revealing whether you won the $1 or $100 dollar bill, I offer you this option: Do you want to stay with red, or would you like to change your choice to blue? What should you do? Does it matter?
This is the classic Monty Hall Paradox, named after the host of the 1960s game show, “Let’s Make a Deal.” At the end of the show, two contestants got to guess whether the grand prize was behind door number 1, door number 2, or door number 3. Behind one (and sometimes two) of the doors was a lesser prize. But sometimes one of the doors concealed a prize no one would want -- something like a pair of goats and 20 bales of hay. With two contestants and three doors, at least one guess must be wrong, and both guesses might be wrong.
Monty Hall always opened a wrong-guess contestant’s door first. Suppose, for example, that Bill and Betty are the two final contestants. Betty chooses door number 2, and Bill chooses door number 3. Monty opens door number 2 first. When he points to her door, Betty already knows she didn’t win the grand prize, because that prize is revealed last. Congratulations, lady. You won a dishwasher, a toaster oven, and $100 worth of Rice-A-Roni. Your life will never be the same.
Now Monty asks Bill the critical question: Do you want to keep door number 3, or would you rather exchange it for door number 1? What should Bill do?
Most people—including many PhDs with training in statistics and probability—get this wrong. Why change? The odds are the same, right? Fifty-fifty. Well…yes and no.
When Bill made his original choice, the odds were 1:3, not 1:2. In that situation, two out of every three contestants will choose the wrong door, just based on probability. However, once Betty’s door is eliminated and Bill is offered the chance to switch doors, he is being offered 1:2 odds instead of the original 1:3 odds. And since there is a 2:1 probability that his first guess was wrong, he should switch to door number 1. Here it is, point by point:
- If Bill chooses the grand-prize door on his first guess, he will win. But he won't know until the end of the game. (Forced choice with incomplete information.)
- After we learn that Betty didn't win, Bill still can't be sure of his original guess. His odds of guessing correctly were 1:3.
- But with Betty’s door eliminated, Bill's odds seemingly improve to fifty-fifty. So why change?
- Actually, Bill's odds at this point in the game are not fifty-fifty (1:2). His odds are still 1:3, because he made his choice before Betty's door was eliminated as the grand-prize door.
- When Monty offers Bill the choice of keeping door 3 or exchanging it for door 1, what Bill is really being offered is a new game with improved odds (i.e., 1:2), versus staying with the old game and original 1:3 odds.
- The essential issue isn't whether Bill should keep his door or exchange it. That's just a distraction. What's important is to recognize that Monty's offer to switch is actually an opportunity to play a different and better game.
Now, do you still want that red-dot bill?