Jesse

As stated, I don't think the question is scenario is clear enough to give a definite answer--for example, it might be that you only open another door if I choose the door with the car to begin with, or if I choose the first door, or on a whim, etc. You have to add the condition that no matter which door I pick, you will always open up another door with a goat behind it and offer me the same deal.

Lobstrosity

This is what's stated, but I think I see where your confusion arises. Maybe you're interpreting the line "I then open one of the other two doors that has a goat" to mean that both remaining doors have a goat and I open one? Hopefully without ambiguity this time, the problem can be phrased as:

There are three doors. Behind one (randomly selected) is a car. behind the other two are goats. You choose a door (but it remains closed), I then open one of the two remaining doors to reveal a goat. I do this no matter what is behind the door you have chosen. It is now a complete mystery to all what is behind the door you have chosen. It is also a complete mystery to all what is behind the unchosen, unopened door. You simply know that one contains a goat and one contains a car. Do you stick with your door and take the $100 bonus or do you change doors?

enigma555

I dunno... a goat wouldn't be all that bad.

You'd never have to mow your lawn again.

Bumble Bee Tuna

I would assume that yes, you do want to switch. Imagine you tried this 18 times.

12 times, you'd get a goat.
6 times, you'd get a car.

If you switched every time,
12 times, you'd get a car.
6 times, you'd get a goat.

Seems simple enough. You go from 1/3 chance to 1/2 chance. Logically speaking, there is no difference between the two unopened doors. It's like Schrodinger's Cat, sort of. There's no reason not to change.

That is, unless the car was not worth much. If we say the goat is worth $50,, then the two ideas, switch vs. no switch, would average out like this:

If you don't switch, over 3 trials, you average ($2*150 + $Car+$100)/3.
If you do switch, over 3 trials, you average ($1*50 + $2*Car)/3
That's $133.3333 + $Car/3
vs $16.6666 + $2*Car/3
so if we subtract situation 2 from situation 1 we get $116.6666 - $Car/3.

So if the car is worth under $350, then you don't want to switch.

symbolically, if $300 + $Goat > $Car, you shouldn't switch.

-B

Jesse

No, I just meant that the way you originally stated it, it could be that you just happened to choose to open another door with a goat behind it on that one instance, for any number of possible reasons (like maybe you only opened another door if you saw that the door I picked was the one with the car behind it), rather than that you were operating according to a fixed rule that says "always follow the contestant's choice by opening a door that a) has a goat behind it, and b) is different from the one the contestant chose, and offer to let him pick the remaining unopened door instead of his original choice".

Lobstrosity

Well technically that one instance is all that matters, right? I mean you're on Let's Make a Deal and you've got your one shot at a car. You know for certain that it's two goats and a car behind those doors, and then you watch as Monte Hall (me) does his (my) thing. So the real question is, in that specific instance, if you were on that show witnessing what happened and knowing that there are two goats and a car, would you take Monte's $100 and stay or would you leave the $100 and switch. You know nothing about my motivations--I could be trying to trick you out of your car or maybe I'm trying to make you think I'm trying to trick you out of your car. You have no way of knowing. I still maintain that that you can probabilistic calculations to this one instance given the information you have, and in light of those calculations, you are still better off in switching (specifically because my motivations are given to be a complete unknown and hence would be inherently random for all intents and purposes--I think?). I guess then the question I'd ask you, Jesse, is what you'd do if you were on the game show and didn't know for sure why Monte showed you the goat?

And by the way, Bumble Bee, that's some nice work on the cost analysis!

Jesse

No, one instance is not all that matters. Suppose your secret rule is, "if the contestant picks the door with the car, open another door (which will have a goat behind it) and offer to let him choose the remaining unopened door; if the contestant picks a door with a goat, don't open any other doors, just give him the goat." However, I don't know that this is your rule. And let's say that in the one instance where I played this game, I chose the door with the car behind it. From my point of view, all I know is that I chose a door, then you opened another door and showed me a goat was behind it, then offered to let me switch my choice to the other unopened door. Obviously if I knew your rule I'd know that I should not switch my choice (because based on this rule there is a 100% chance that the door I chose had the car behind it), but I don't. Without knowing what internal rules you are obeying, there's no way to calculate the probability that the door I chose will have a car behind it vs. the probability that the other unopened door will have a car behind it, based on only a single instance where you opened a door with a goat behind it and offered to let me switch my choice.

I suppose I'd try to make some sort of educated guess about the meta-probability that the host was obeying different possible internal rules, based on my beliefs about his psychology or motivations, then use that to weigh the probabilities of either door having a car behind it which derive from each possible rule.

Lobstrosity

First of all, on Let's Make a Deal Monte never just gives you one choice. That'd be no fun. What's fun is knowing that you screwed yourself out of a car by changing your mind (or conversely winning yourself a car by making the smart move and changing your mind). So it's pretty much a given that on that particular show you're going to get more than one choice rather than find yourself saddled with a goat with your first pick.

Eh, but that's all besides the point. The point is that you have more information after he opens the door than before. Couple this with the fact that you honestly have zero information about his motivations. If you try to convince yourself that you can figure out his psychology, you're just going to fall victim to the old "well, I have a hunch it's probably behind door X because it feels right." It simply doesn't make sense that a game show host would obey the simple algorithm of "if they pick the good prize, allow them to change." After the first three shows everyone would know to never change if the offer came up. The whole reason the show worked was because you never knew if the change would be good or bad (i.e. it was approximately random). So could there be an algorithm at work behind the scenes you don't know about? Sure. Do you know anything about this possible algorithm? No. Therefore trying to make a decision that takes this complete unknown and gives it weight would probably not be the best way to go. Anyway that's my opinion on how I would approach this matter were it a real-world experience rather than an ideal statistical case. You know, it's kind of interesting debating how the transition from the ideal statistical case to the real world could skew the solution to the problem--I hadn't considered that before.

Jackalope

I think part of the problem is that half our readers here haven't ever seen "Let's Make a Deal." When that puzzle was first posed, it was a part of US culture and even if you'd never watched the show, you had some idea what the problem was talking about. Now, a good many folks on the internet were born after the show went off the air! Not to mention all the non-US readers.

Jesse

I think you're contradicting yourself a bit here--first you said we have zero information about his motivation, then you make an argument about why he'd be silly to use the algorithm "only offer the contestant another choice if his first choice is the car."

OK, but there are other rules he could be using that would also be "approximately random", and would allow him to always offer the contestant a second choice like you stipulated above. For example, he could obey the rule "after the contestant picks a door, randomly select one of the two remaining doors and open it to show what's behind it, then allow the contestant to switch his choice to the other unopened door." In this case, if the door you chose had a goat behind it, there'd be a 50% chance Monty would open the door with the car and a 50% chance he'd open the door with the other goat (if your door had the car behind it, then of course the door he will open will be guaranteed to have a goat behind it). If you thought he was using this rule, then if you pick a door and Monty opens another door with a goat behind it, then you would not have any reason to switch--if you work out the probabilities it's equally likely that your original choice had the car behind it as it is that the remaining unopened door has the car behind it (see below).

It's only if you assume he will consciously avoid opening the door with the car that his opening one door you didn't choose increases the likelihood that the other door you didn't choose has a car behind it. In that case you know that if your door had a goat behind it (and there's a 2/3 probability that's true), then his choice was "forced", so there's a 100% chance the door he didn't pick has the car behind it. On the other hand, if you chose the door with the car, there's a 100% chance that both remaining doors have goats behind them. So:

p(door that neither you nor Monty picked has car) = (2/3 chance you chose goat)(100% chance door that neither you nor Monty picked has car, given you choosing goat) + (1/3 chance you chose car)(0% chance door that neither you nor Monty picked has car, given you choosing car) = 2/3

On the other hand, if Monty wasn't consciously avoiding the door with the car and just randomly selected one of the two remaining doors, you'd have:

p(door that neither you nor Monty picked has car) = (2/3 chance you chose goat)(50% chance door that neither you nor Monty picked has car, given you choosing goat) + (1/3 chance you chose car)(0% chance that door neither you nor Monty picked has car, given you choosing car) = 1/3

...which is the same as the probability that your original choice had the car behind it. So again, your guess about Monty's "rule" plays a critical role in deciding whether you have a better chance of getting the car if you switch.