Re^7: Marilyn Vos Savant's Monty Hall problem

I believe that you are stating the problem correctly, but then adding a big assumption before analyzing it. You assume that that, whether or not you have the right choice, the host will always put you in the current situation - you've picked a door, and another has been opened showing a goat. With that assumption then you should switch, and will have 2/3 odds of winning if you do.

But that is a big assumption. In analyzing probabilities you cannot just work with what has happened, you have to work with what could have happened instead. A probability problem is never fully specified until it includes both knowledge of what did happen and what might have happened.

In particular in this case there are models of the host's possible behaviour in which you would be an idiot to switch. (The host is trying to keep you from winning the car.) There are models of the host's behaviour in which it is a good idea to switch. (The host wants to draw the game out.) And the problem statement does not provide enough information to unambiguously decide which model of the host's behaviour is correct. (Saying that the host has enough knowledge to always draw the game out is not saying that the host will choose to do that.)

Therefore the problem is not fully specified. To come up with an answer you have to add an assumption of some sort.

If you want to make it truly unambiguous, you can state the problem like this, Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. The game works like this; you pick a door and the host, who knows what's behind the doors, will open another door and show you a goat. Then the host asks whether you want to switch your choice. You decide, then the host opens the other doors and you get the car if you've chosen it. When you play this game, is it to your advantage to switch your choice of doors?

The wording change is subtle but important. With this wording it is clear that, no matter what, the first time the host opens a door there will always be a goat. Not only have you been told what actually happened in the game, but you've also been told what would've happened no matter what your initial choice was.

Comment on Re^7: Marilyn Vos Savant's Monty Hall problem

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Re^8: Marilyn Vos Savant's Monty Hall problem by ysth (Canon) on Aug 24, 2004 at 17:26 UTC
In particular in this case there are models of the host's possible behaviour in which you would be an idiot to switch. (The host is trying to keep you from winning the car.) You've made this statement more than once now, and it completely loses me. Can you demonstrate it? I don't actually see where motivation comes into it; the stipulation is that you have a choice to switch or not switch. That implies that a goat door was opened by the host. (If the prize door was opened by the host, you no longer need choose.) And afaict, that a goat door was opened is enough to give you an advantage if you switch.	[reply]
Re^9: Marilyn Vos Savant's Monty Hall problem (motivation?) by tye (Sage) on Aug 24, 2004 at 18:31 UTC
The scenario is that the host (mostly) only opens the other door when s/he wants you to switch (because s/he knows that you have selected the "car" door). If you have selected a "goat" door, then the host (mostly) does something other than open one of the doors and offer you the chance to pick again. It's a bit of stretch to interpret the question this way, of course, and doing so makes the problem rather strange so I find it quite clear that this interpretation was not what was intended. Note that the question didn't clearly state that the car can't be moved after you've picked, that the goats aren't extremely valuable nor that the car isn't a cheap toy, that you actually "win" what is behind the door you select, or any number of other possibilities that make rather little sense. So I don't get why these alternate routes are so compelling to consider. - tye	[reply]
Re^9: Marilyn Vos Savant's Monty Hall problem by Aristotle (Chancellor) on Aug 24, 2004 at 18:38 UTC
Yes, if the host is trying to keep your from winning the car, he will open the car door if you haven't picked it. In that scenario, you should never switch. Of course, the only rationally plausible scenario in an actual game show is that in which the host will always pick a goat door to open. tilly is just pointing out that in purely formal terms, the problem is almost never stated unambiguously enough. Makeshifts last the longest.	[reply]
Re^8: Marilyn Vos Savant's Monty Hall problem by TrekNoid (Pilgrim) on Aug 24, 2004 at 18:58 UTC
But that is a big assumption. In analyzing probabilities you cannot just work with what has happened, you have to work with what could have happened instead. A probability problem is never fully specified until it includes both knowledge of what did happen and what might have happened. This is where I think we're in disagreement... Not with your statement, because I agree with it if you're trying to predict future behavior. But that's not what we're trying to do in this case. We're not 'about to go on the show'... and we're not 'about to begin the prize round'. The problem states that we're already in the prize round, chosen a door, and the host has revealed a goat door. That's where we start So we have to calculate the probabilities based on that set of variables. You can't examine the possibility that "The host might not have opened a goat door first" because this question started with that stated. Basically, I guess I'm stating that I agree with you that if we're talking about all possible future behavior, then we need to consider that the host might just open the prize door... or not offer a choice... etc... But for this specific problem, we already find ourselves faced with the host opening a goat door and offering a switch, so to decide from here what to do, we have to assume these as fixed variables in our analysis. You're beginning before the game started... I'm beginning from the point where two actions have already occured... Which changes things. It's like turning on the TV and finding a car race in it's last lap and wondering what odds are that the car in the back of the pack could win... You would base your conclusion on how many times you've seen the back car win from the last lap in the past... You wouldn't consider the possibility that he might have been in the front or the middle, because he isn't... If you were going to analyze the probability of his winning the next race, all that comes into play... but to analyze the probability that he's going to win this race, then you discard those options that can no longer occur. That's how I see the Monty Hall problem. You're considering the probabilities based on any version of the game... I'm considering the probabilities of this one game, where the goat door is already open, and a choice has been offered. And, since someone's mentioned 'arguing' with you... for the record, I've enjoyed this discussion, regardless of the outcome of it, or whether we ever agree on it :) Trek	[reply]
Re^9: Marilyn Vos Savant's Monty Hall problem by tilly (Archbishop) on Aug 24, 2004 at 21:54 UTC
What can I say? If you analyze a probability problem start to finish and add in the assumption that the outcome you saw is the only possible outcome that could have happened, then you'll usually get wrong answers. That isn't a matter of opinion - that's mathematical fact. In case you're curious, here's how the formal approach goes. You start with a priori estimates of various scenarios. You run the experiment and see what happened. You then compute the conditional probabilities of the scenarios of interest given what just happened using the well-known formula from Mr. Bayes, `P(A\|B)=P(A and B)/P(B)`. In English that says, "The probability of A given that B happened is the initial probability that A and B both happened divided by the initial probability that B happened." That gives you the right numbers every time. Any other approach can only be right when it winds up agreeing with that one. Look elsewhere in this thread for the problem that was misstated in tye's textbook about how likely the other child was a daughter. And try to understand the full (if informal) analysis that I gave. You should see how a correct analysis has to consider what might have happened (but didn't). Furthermore you'll see how very subtle differences in the stated problem change what was possible, and result in very different estimates. (Note that I didn't pull out the formal machinery. I would have needed to if I wanted to show how to adjust the answer to account for the fact that about 51% of children are male.) If you still think that I'm wrong, well that is your perogative. I've spent all of the energy on this thread that I'm willing to spend. I'm not here to present a course on elementary probability theory. You're not here to take one. So I'll let it drop here.	[reply] [d/l]