Lobstrosity

Well, that wasn't supposed to be a problem in that I tried to provide all the information you need to compute your best option... Eh, I just suck at this

Bumble Bee Tuna

well now we're talking real world here, and your new example also falls on its face in the real world. Of course Monty won't open the door with the car in it, that ruins all of the suspense, and serves no purpose- you know the other two are goats, so you would have no reason to switch from goat 1 to goat 2 and lose the $100. There would be no reason for Monty to open the car door.

Thanks! I guess the rumors are true: I am a goddamn math nerd. Oh well, I'm big on logic/puzzles. So sue me.

-B

Jesse

Actually, I think I calculated the wrong probability. Although it's true that if Monty picks one of the two remaining doors at random, then if this game is repeated many times we'll see that only in 1/3 of all cases does the remaining unchosen door have a car, if on the other hand we look at only the subset of cases where Monty opened a door with a goat behind it as opposed to a car we'll see that in 2/3 of these cases the remaining unchosen door will have a car. So if all I have is a single instance where I picked a door and then Monty opened another door and found a sheep behind it, it doesn't seem to matter whether I assume Monty was consciously avoiding the car or if he just happened to pick a door with a goat. I should have calculated p(door that neither you nor Monty picked has car given that Monty opened a door with a goat) rather than just p(door that neither you nor Monty picked has car) alone. So you're right Lobstrosity, as long as I assume Monty always opens at least one other door besides the one I picked, then I don't need to know anything about how he made the choice of which remaining door to open.

Ensign Steve

The "trick" of the doors is that while you will instinctually use your regular ideas of statistics and probability, there is a non-random aspect to the test. The host, Monty or Lobstrosity, knows the location of the car, so when he picks which door to open, he is doing so with that knowledge; he doesn't do so randomly. I can understand where the confusion comes from, however, since the question was phrased kind of ambiguously. I have seen Let's Make A Deal, though, so I got the idea of it. The $100 was a neat twist, though. I'd probably take it, as I almost always go with the sure thing, unless the odds are staggeringly in my favor. 1/3 or 2/3, it doesn't matter to me, if I've got a 1/1 chance at $100.

Did I get the balloon question right? I got to ponder that concept irl, as my mom and I schlepped not one, but forty-eight balloons in the back of her 4-Runner. Please do not ask why we didn't just rent a helium tank.

Jesse

As I said in my last post, I don't think this actually matters. Once you know he's opened a door with a goat, your calculation of the probabilities shouldn't depend on whether you assume he consciously avoided picking the car or whether he just picked a door with a goat by dumb luck. The only thing you have to assume is that the host always opens one other door, regardless of which door you picked.

Jabu Khan

switch. I trust Erdos.

Lobstrosity

The simplest way I can think to explain the doors problem is that the problem is isomorphic with this scenario:

You choose a door. Monte then makes this offer: you can keep your door and take the $100 or open BOTH remaining doors. If one of the two has the car, you get the car. If both are goats, you get a goat. When viewed this way, it's clear that by switching you improve your odds of getting the car from 1/3 to 2/3. I assert that this problem is identical to the one I posed initially.

Yeah, you did...damn you! I had hoped to have a few people give the obvious yet wrong answer, but people here are way to smart to fall for that. The balloon moves in the direction of the car's acceleration for the same reason it rises (moves in the opposite direction of gravitational acceleration). The surrounding air is heavier and is experiencing the same forces. Usually we forget that air has substance because it's mass is negligible. In this case, it's mass is key. The air rushes to the back of the car, setting up a pressure gradient that pushes the balloon to the front of the car with more force than the balloon's own inertia, which is trying to move it to the back of the car. Were the car a complete vacuum, the balloon would behave intuitively, moving to the back of the car. It would also no longer be floating in the air. It would also probably pop, but let's neglect that for now.

An analogous question that's more intuitively obvious would involve a car that's mostly full of water except for a small air bubble near the top. If the driver of the car accelerates forward, I think we can all see that the water would slosh to the back and thus the air bubble would be more towards the front. The air bubble in water is analagous to the helium balloon in air.

squiddy

from The Simpsons "Bart Gets a Goat" ....er Elephant

Bart: Well, all that money sounds mighty tempting, Marty, but I think I'm going to have to go with the goat.
Homer: [to Marge, happily] He's taking the goat instead of the
money.
Marty: [whispering] The kid wants the goat!
Bill: We don't have a damn goat.
Marty: Don't whisper into the mike!
Bill: Ahem, kid, the goat's a gag prize. Nobody takes the gag
prize. [nervous laughter] You want the cash.
Bart: [indignant] I want the goat!


I'd love to have a goat (lots of them!) if they were allowed within city limits!

Ensign Steve

Oh, I thought it was because the air in the car is invisible, but we can see the balloon.

Good analogy!

theyeti

There is one important difference: When Monty opens a door, and if he's not doing it at random, there is a 1/3 chance that he will open the one with the car behind it and the game ends there. (Actually, if the car is not behind the door you already picked, there is a 1/2 chance of him picking the one with the car. So your odds only improve when you switch if there are high odds of the game ending early.) Other than that, there can only be an overall advantage of switching doors if Monty opens the door non-randomly. Switching doors works because you choose to do it only after Monty has given you information about the two remaining doors, assuming you haven't lost. Here is a break-down of the possiblities:

Chances of winning if you don't switch: 1/3

Changes of losing automatically: 1/3

Chances of winning if you switch: 1/3

The reason why switching is advantageous is because you choose to do it only after you've seen whether or not you automatically lose. Only once you fold the chances of automatically losing into the odds for switching do your chances improve.

To illustrate what I mean, let's do things a different way: Let's say that you pick a door, and then Monty gives you the option of switching doors, but you have to decide on switching before he opens one of the two others. In other words he says, "You can keep what's behind door you picked. Or I will open one of the other two doors, and you will automatically get the one I don't open." In this case, if he's doing it randomly, you're just as good keeping the door you picked than you are switching (better, in fact, if he's giving you a bonus). Because if you haven't already picked the correct door, there's a 1/2 chance of losing anyway. But if he's going to open the door non-randomly, then you're better off switching.

theyeti