The Wikipedia page (linked above) goes into great detail about how it's one of the most misunderstood problems in history. As cognitive psychologist Massimo Piatelli-Palmarini says, "...no other statistical puzzle comes so close to fooling all the people all the time... even Nobel physicists systematically give the wrong answer, and ...

*insist*on it, and they are ready to berate in print those who propose the right answer." This is a prime example of an extremely common failure to reason correctly, and a lesson that I think anyone could stand to learn from. Hell, I got it wrong at first, and was

*very*resistant to having my mind changed until it was shown in excruciating detail just where I went wrong.

So, to the point! For those who did not click the link or don't know about Monty Hall, the problem is: you're on a game show and have advanced to the final round. Before you are three doors, behind one of which is a car, and behind the other two of which are goats. You pick a door, and before opening it, the host stops and opens one of the remaining doors, revealing a goat. He then asks, "Do you want to keep what you've got behind door

*x*? Or risk it all to see what's behind door

*y*?" What should you do to maximize your chance of winning the car? Should you stand pat? Should you switch? Does it even matter?

Most people think it doesn't matter - a goat has been eliminated, meaning that of the two doors left, one has the car, the other has a goat, so it's 50/50. It will not make a difference whether you stand pat or switch, they say. This is incorrect, and in a big way. In truth, standing (as a strategy) will only win you the car 1/3 of the time, whereas switching will win you the car 2/3 of the time. Exploiting this common failure to reason correctly, and exacerbating the problem with some suggestive phrasing, Monty Hall saved his show a lot of money by handing out a lot fewer prizes than they otherwise might have.

The reason it is this way is that when you make your initial choice, you only have a 1/3 chance of selecting the car (the "right" door), and this probability

*does not change*when one of your choices is eliminated. Standing, as a strategy, stakes your bet on picking the right door

*of three*. Since there is a 2/3 chance that you picked the wrong door, there's a 2/3 chance that the car

*is*behind a door you did not pick, and thus Monty actually does you a

*favor*by eliminating a goat - which ought to make your choice all the more obvious.

Another way of looking at it is that there's (statistically) a 1/3 chance behind each door, and when you make your choice, you eliminate that door from the "opening pool," fixing its probability (since Monty will

*never*open the door you pick). Since Monty will

*also*never eliminate the car, but there's a 2/3 chance that it's behind the group of un-picked doors, eliminating one door "consolidates" the odds behind the un-picked and unopened door.

One of our corporation members proposed an apt re-phrasing of the problem which I will outright steal (with a little change). Suppose there are

*a billion*doors (and still only one car), so picking the right one off the bat is only a one-in-a-billion chance. You pick one anyway, and then Monty opens

*all the remaining doors*but one, showing goats behind all of them. The odds are

*overwhelmingly*likely that you picked a wrong door, and the car is behind one of the other nine-hundred ninety-nine million, nine-hundred ninety-nine thousand, nine-hundred ninety-nine doors (the number is more impressive when written out, I promise). When you are given your second chance, it's

*obvious*you should switch, since Monty's done most of the work for you. The same reasoning applies to the three-door scenario, it's just scaled down a bit and thus not so obvious, since only one door is picked by you and only one door is opened by Monty.

As an interesting side note, a less-obvious rephrasing could be that he eliminates

*one*of the remaining doors instead of

*all but one*, which just so happens to pan out as an identical behavior in the three-door scenario. Let's say there are four doors, so you have a 25% chance of picking correctly right off the bat. The remaining 75% is distributed evenly among the other three doors, until one is opened, at which point it's distributed among the two unopened and un-picked doors. Since we have no means by which to distinguish them, they're even, and both get half of the opened door's quarter, putting them at 37.5% each, so you should still switch. A five-door scenario, by these rules, yields a 20% chance for standing, and 26.7% behind each of three doors. As the number of doors rises, the margin of advantage to switching shrinks, but it's still non-zero and higher, thus making it always better to switch. The trend happens to have its peak in the three-door scenario.

I feel like this should be wrapped up somehow, but now all I can think of is other interesting problems in game theory, which is like the coolest thing ever.