Sometimes counterintuitive is wrong

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.

Original Post

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.

13 Responses to “Sometimes counterintuitive is wrong”

  1. Seth Finkelstein Says:

    Here is a simple way to see the fallacy in the above post.

    Let’s say it’s a MILLION doors. 1,000,000 of them. One prize. 999,999 goats.
    You pick one.
    Monty opens *999,998* “goat” doors.

    Then asks if you want to switch your choice of doors.

    Are you really going to argue that you should not switch?

    What happened?

  2. walt Says:

    Yes, I read that response in the Wikipedia article.

    At the point that you’re offered a choice of doors, the odds of your door being the right one have improved from 1:1,000,000 to 1:2. Monty would never have opened the right door, so in effect there are 999,998 doors not really in play.

    Unless you can convince me that the car shifts doors because of which doors Monty opens, I’ll go with my original comment–not that you should *not* switch, but that there’s no gain in switching. (There’s no loss, either.) Call me obstinate and stupid; it won’t be the first time.

  3. Seth Finkelstein Says:

    Let me see if I understand you.

    There are a million doors. You have picked one at the start. Monty opens 999,998 wrong doors. There are now 2 out of a million doors remaining. Your original choice, and the one other door Monty has not opened.

    At this point, where you’re offered a choice in switching, are you really claiming your door now has a probability of being right of 1 in 2? There’s no gain in switching?

    Think of it this way: Monty is in effect saying: You can pick one door. Or you can pick 999,999 doors, and before I open the right door, I will open 999,998 wrong doors. Is that a 50/50 proposition?

  4. walt Says:

    Yes, that’s what I’m saying. Monty was never going to open the right door, so when all the door-opening is done, there are two doors, each of which has a 50-50 chance (for you, not for Monty) of being the right one. And that’s precisely as true no matter which of the million doors you choose…because Monty’s never going to open the right door.

    As to the final thought experiment, it becomes a 50/50 proposition, because 999,998 doors are irrelevant. No matter what door you pick, it won’t be one of the 999,998 that he opens.

    And I agree that I may be “mathematically wrong” or whatever. As for simulations and other testing, the only way that would matter is if it amounted to: I choose one. All but one of the “wrong” choices are removed. Now, I don’t honestly believe that whether I retain the one of two that I originally chose or switch to the other of two that hasn’t yet been revealed will matter–that each one has a 50/50 chance of being the right one.

  5. Seth Finkelstein Says:

    “… so when all the door-opening is done, there are two doors, each of which has a 50-50 chance (for you, not for Monty) ….”

    Why do you believe this (50-50 chance) is true? Forgive me, it sounds like the old joke “All events have a 50/50 probability of happening, because either they happen, or they do not”.

  6. walt Says:

    Let’s look at it another way:

    Suppose Guest B is invited onstage for this million-door unveiling, at the point that all but two doors have been opened. He’s told, correctly, that there’s a car behind one door. Would we agree that he has a one-in-two chance of choosing the right door?

    Effectively, that’s what “Do you want to switch doors?” does for Guest A: It abandons the previous game and begins a new one, one in which there are two doors, one of which has a car behind it. The guest has a one-in-two chance of choosing the right door. The other 999,998 doors are simply irrelevant at this point: Those affect the odds for the other game, the game that’s no longer in play.

    I don’t believe that the position of the car can be affected by the number of doors-without-a-car that are opened. That simple.

  7. laura Says:

    I get lost in logical arguments, but this seems like an opportune moment to remember this gem.

  8. Seth Finkelstein Says:

    There’s a big difference between what Guest B knows, and what Guest A knows. Guest B has a one-in-two chance of picking the right door (agreed) is not the same (agreed?) as the prize has an equal chance of being behind each door. Guest A has a much higher chance of picking it the right door, from knowing how the current state was created in the first place.

    Look at it this way: Suppose the game is viewed as follows – Monty says “There are a million doors. Pick one. This divides all the doors into two sets, Monty’s set of 999,999 doors, and Guest’s set of 1 door”. Now, would you argue that because there are two sets, and the prize is either in one set or the other, it has a 50% chance of being in either set? (see joke above).

    Monty then gives information about the wrong doors in his set. Did the OVERALL PROBABILITY of the prize being in his *set* of doors change, because he gave information about *specific* wrong doors within it?

    That is, Monty’s set was originally 999,999 doors. Using his knowledge of which were wrong, he threw out 999,998 of the wrong ones. Did that change the probability of the prize being within his original set of 999,999 doors?

    Guest B doesn’t know this. Guest A does.

  9. walt Says:

    Laura: Thanks.

    Seth: You’re saying that the mental state of the person making the choice affects the results?

    Guest B’s knowledge does not, as far as I can tell, affect the actual situation at the point at which she must choose: That is, either the car is behind door X or door Y.

    At that point, assuming a fair setup, the odds are 50% in each case. If they’re 50/50 for Guest B, they must be 50/50 for Guest A…even as Guest A knows that another 999,998 doors were discarded during the process. It’s a new game at the point where “change your choice?” is offered; I don’t believe any of the previous odds are, at that point, relevant. Or, rather, I don’t think they are: it’s not a matter of belief.

    The current state in this problem is always: There was some number of doors N greater than two (the number doesn’t matter). The guest chose one door. The host revealed some number N-2 of non-car doors. At that point, the guest has a choice: Which of two doors to choose? The odds, at that point, are 50/50 for both doors. The remainder (N-2) of the doors are no longer part of the game.

  10. walt Says:

    And, damn, it sure is interesting to see which posts draw lots of comments. Sometimes I wish it was the ones I wanted comments on, but my desires have no real effects on the outcome. If they did, I’d have won SuperLotto by now.

  11. Seth Finkelstein Says:

    > “Seth: You’re saying that the mental state of the person making the choice affects the results?”

    Information possessed != “mental state”

    If a stagehand slipped the person the location of the prize beforehand, that would be information, not “mental state”. Suppose that slipped information was not a certainty, but a probability – e.g “There are two different assistants who set up the car prize, Goofus and Gallant. If Goofus does it, it’ll always be on the left side, since Goofus is lazy and won’t push it further than necessary. Gallant is conscientious and will put the car anywhere on stage. But I don’t know who is on duty tonight”. That’s still useful even if not definitive.

    “That is, either the car is behind door X or door Y.” – are you saying this means it has a 50-50 chance of being behind door X or door Y? That’s not the same thing.

    “At that point, assuming a fair setup, the odds are 50% in each case”

    But that’s the point – the setup is not 50/50, and Guest A knows that, while Guest B doesn’t (note, but Guest B still has 50/50 chance of guessing right, even if the setup is NOT 50/50 chance for each door).

    “If they’re 50/50 for Guest B, they must be 50/50 for Guest A”

    Umm, why? This is quite false. If Guest A knows that the prize is almost always in one door, say due to the way the stagehands use it more often because it’s easier to get the car into it, it’s still a 50/50 chance for Guest B to pick the right door (not knowing that), but Guest A has a much higher chance.

    That is, the previous game has given Guest A information that the setup for the current game is not 50/50 for each door.

  12. Terry (@GameCouch) Says:

    Since Marilyn Vos Savant has said this one issue brings in the most mail, I’m not surprised you’re getting so many comments. May I recommend going here and actually testing your theory: http://www.philosophyexperiments.com/montyhall ? I think you’ll see, in the long run, that switching doors makes sense.

  13. walt Says:

    Seth, Terry, others:
    See the revised post–only the first section, since to revise anything else would be to hide my mistake.

    I was wrong. I lost a couple hours sleep last night realizing that I was wrong. I will say that the Wikipedia article not only didn’t convince me, it totally confounded me. I won’t bother to test my theory; I’m now convinced that, in two-thirds of all scenarios for Guest A, Guest A will gain by switching doors when that switch is offered. Even if that does give me a headache.

    Oh, and Seth: Thanks for continuing to push at this. Even the thickest skulls, like mine on occasion, can occasionally be penetrated.

    Note that, to reach those who read the first post but are unlikely to return, the new text at the top of this post also appears as a separate post.


This blog is protected by dr Dave\\\\\\\'s Spam Karma 2: 103742 Spams eaten and counting...