New section added April 7, 2011; go to Original Post for the original post and earlier additions.
I lost a lot of sleep last night reconsidering scenarios and situations, to no good end.
And, eventually, found myself agreeing that–and this is the only way I can put it without getting a sudden headache:
In the Monty Hall puzzle, as discussed in this post (see below “Original Post”), in two-thirds of all possible scenarios, the contestant will gain by switching doors when the host makes the offer.
In other words, the post itself is wrong–which means I was wrong. Period.
The whole situation still gives me a headache. I’m not willing to use the wording that “the remaining door you didn’t choose has a 2/3 probability of being the right door”–that makes me want to scream. If a second contestant shows up, sees the two doors and is told which door the original contestant chose, I’d still assert that the second contestant is equally likely to win by choosing either door.
But I’m wrong as to the facts for the original contestant. Hell, I’ve been wrong before.
Note: I wrote this as the very first thing I did on my computer this morning–before reading two additional comments. Now, after pasting this same text into the original post, I’ll go read the comments and add another one, admitting that Seth (and others) were right and I was wrong.
And then maybe go take some aspirin.
Reading one of the last issues of Wired before my freebie print subscription expires, I ran across another use of the so-called Monty Hall Brainteaser, which I’ve previously seen in Marilyn Vos Savant’s column (I think) and probably elsewhere.
If you don’t know the thing, here’s a version:
You’re on Let’s Make a Deal. The host shows you three doors. One of the three has a car behind it. The other two have goats. You choose one of the three (let’s say Door 1). The host, who knows what’s behind each door, opens one of the other two (let’s say Door 3) and shows you a goat. Now he offers you a chance to change your choice (that is, to Door 2). Should you change your choice?
The supposedly-right answer is Yes–that the odds of Door 2 being the right one have increased from 1/3 (when no doors were open) to 2/3 (because one door is clearly wrong).
That seems counter-intuitive. That’s because it’s nonsense. It’s using trick mathematics to make something appear to be the case even though it isn’t.
Let’s reframe the situation:
- There are three doors, but you really only have two choices. Either you’re going to choose the door with the car or you’re not.
- Whatever door you choose, there will be a door with a goat behind it, a door that you didn’t choose. Therefore, that door–which could be either 1 or 3–really isn’t part of the equation.
- The odds of the door you chose being the one of the remaining two doors that has the car were 50% before the host opened the third door, and they’re 50% after the host opens that door.
- Therefore, there is no advantage to changing your choice.
Essentially, from an odds perspective, one door with a goat is simply irrelevant. There are three doors: One door-with-a-goat-that’s-going-to-be-opened, one door with a car, and one door with a goat that’s NOT going to be opened. You can’t choose the first door.
Yes, I know, some statisticians/mathematicians and The Woman With The Highest IQ (And The Tiniest Weekly Column) In The World will say I’m wrong. I don’t think so. Fact is, the host’s decision to show you a door–and there is always a door with a goat behind it that you didn’t choose–cannot change the location of the car, and thus cannot give one of the two remaining doors an advantage over the other one.
Even in WiredWorld.
Update 4/6: I’ve now been pointed to a lengthy Wikipedia essay explaining in excruciating detail just why I’m wrong and the counterintuitive answer is right. To which I can only say: I guess this is one of those cases where I must be dumber than a pigeon, because I’m still not buying it. If the game’s fair, with the car behind one door and the host knowing which door is which (and always opening one non-car door), then the fact that the host opens a non-car door cannot change where the car is. The car doesn’t move because the host opens a door. In which case it can’t change the actual probability of whether you’ve chosen the right door, which remains 50% after the host has opened a door, whether you switch or don’t switch.