The Monty Hall Puzzle

The Monty Hall problem is one of the best brainteasers I've come across; it's also one of the most debated. I first ran into this problem while taking a statistics class in college. It goes something like this:

Suppose you are on a game show, similar to Let's Make a Deal. Monty Hall presents 3 doors, and asks you to select one. Behind one of the doors is a big prize; behind the other two are fake (or very low value) prizes. You select a door. Before opening your chosen door, Monty says, "Before I show you what's behind that door, let me show you what you didn't pick." He then opens one of the other two doors that you did not choose, revealing one of the fake prizes. "Now, would you like to stick with your original choice, or switch to the one remaining unopened door? Or does it not make any statistical difference?"

So, statistically speaking, should you stick with your original choice, switch doors, or does it matter at all?

There's a host of answers for all different reasons -- my favorites are the psychological answers, such as, "You stick with your original choice. After all, if you switch and end up being wrong, you'd never forgive yourself." The next most common answer is that it doesn't matter, and it's often argued that the choice is simply a 50/50.

The psychological answers are intriguing answer (and true), but it's wrong. Statistically speaking, you're better off switching. Your odds of winning actually increase from 1/3 to 2/3. You can draw out a truth table for all scenarios; but the best way to describe why, in my experience, goes something like this: First, look at the problem from the other way around -- what are your odds of losing? Pretty simple, right? 2/3 -- there are 2 wrong doors and 1 correct door, so you'd lose two-thirds of the time. It's a pretty safe bet, then, that your initial choice is a losing door. When a losing door is revealed to you, you know with 100% certainty that it's a losing door. I know that sounds painfully obvious, but go with me here. This doesn't change the fact that the door you picked, two-thirds of the time, is still a losing door. If you decide to stay with this door, how often will you win? 1/3. Moreover, if you're two-thirds certain you've got the wrong door to begin with, why wouldn't you switch?

Once explained this way, most people "get" the puzzle and it's no longer difficult. The disjunction is getting past the emotional involvement. As an aside, this reminds me of a grade school problem, "If you flipped a coin ten times and got heads on each flip, what are your odds of getting a head on the eleventh flip?" The answer is simple: 50/50. The emotional involvement here makes us think, "What are the odds of getting eleven heads in a row?" -- and that leads to the incorrect answer of 1/2ˆ11.

I decided to build a page dedicated to simulating this puzzle (you can see it here). Over time, these statistics should stabilize. Comments on this puzzle? Post them in the comments section.
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