This is wrong. Let’s label the goats A and B to simplify things (so we do not need to consider the positions of the doors). There are 3 cases:
1. You pick the right door. The other two doors have goats. The host may only choose a goat. Whether it is A or B does not matter.
2. You pick the door with goat A. The host may only choose goat B.
3. You pick the door with goat B. The host may only choose goat A.
The host’s intentions are irrelevant as far as the probability is concerned (unless the host is allowed to tell the contestant which door is correct, but I am not aware of that ever being the case). 2/3 of the time, you pick the wrong door. In each of those cases, the remaining door is correct.
The most strict argument is yet another statistics professor got basic statistics wrong.
> "The problem is not well-formed," Mr. Gardner said, "unless it makes clear that the host must always open an empty door and offer the switch. Otherwise, if the host is malevolent, he may open another door only when it's to his advantage to let the player switch, and the probability of being right by switching could be as low as zero." Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch.
The hosts’s intentions absolutely do matter, because the problem (as originally stated) doesn’t specify that the host always opens a door and offers a switch. Maybe he only offers a trade when you initially picked the good door.
Some of the comments aged fairly well, although not in the way that their authors intended:
> There is enough mathematical illiteracy in this country
> If all those Ph.D.’s were wrong, the country would be in some very serious trouble.
In the 1800s, Carl Friedrich Gauss lamented about the decline in mathematical ability in academia. Despite academia since having advanced mathematics farther, mathematical ability in academia still has evidence of decline. Professors tend to be good at extremely specialized things, yet they get the simple things wrong. I once had a Calculus professor who failed to perform basic arithmetic correctly, during his calculus class. All of the algebra was right, but his constants were wrong. This happened on multiple occasions.
> Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.
> “Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”
All you know is that in this particular instance the host has opened a door and offered a switch. You cannot conclude that the host always opens a door and offers a switch.
The problem as stated allows the host to offer switches only when the contestant picked the door with the prize, or only when the moon is gibbous, or only when the tide is going out. Diaconis and Gardner are completely correct to point out that the problem as stated is under specified and that the intent of the host matters.
That would be a rather convenient signal to the player.
That would make no sense. Also, Monty always opens a door with a goat. The problem is well-formed, but most people misrepresent it in order to object to it.
Nope! That’s you adding a constraint that does not exist in the original problem.
> Was Mr. Hall cheating? Not according to the rules of the show, because he did have the option of not offering the switch, and he usually did not offer it.
From [1].
Constraints matter. Don’t play fast and loose with them.
[1] https://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-do...
In a precise way, the reason the question is underspecified is because it doesn't say if the probability of the host offering you a chance to choose again is dependent on which choice you make. If the host offers the choice more twice as often when your pick right and when you pick wrong, then changing you pick is the incorrect choice.
Now, colloquially, it can makes sense to assume the host always offers the choice, but practically, if we're looking at how to use statistics in a real world situation, that isn't a safe to always assume that probabilities are independent.
This is like being presented with a nearly completed game of chess, asked if the loser can lose in 1 move and then arguing that the answer is more nuanced because there might have been other moves taken that produced a different end games rather than the ones that produced this particular end game. We do not care about those other end games, since we are only considering this particular one.
Whether the choice was already made by the host makes no difference, what matters is what information about the hidden state can be derived from that choice.
Let's say the rules of the game are modified to sat that the host never offers a re-selection when you already have selected a door with a goat. Then if the host has offered you a re-selection you should definitely not take it because you already have the good prize. You know this because the re-selection offer provides information about what is behind the door you selected.
In fact, any time your choice of door has amyy statistical effect on whether a re-selection is offered, then a re-selection offer (or lack) provides a small amount of information that modifies the expected value of choosing a new door.
> This is like being presented with a nearly completed game of chess, asked if the loser can lose in 1 move and then arguing that the answer is more nuanced because there might have been other moves taken that produced a different end games
It is absolutely nothing like that. That is not a question about statistics or probability.
I don't think you understand the concept of conditional probabilities correctly.
The fact that event B already happened doesn't make it any easier to compute P(A|B) nor it renders the P(B) useless.
On the contrary P(B) and P(AB) are key to solve this problem.
Is it really unreasonable to assume that the host would like to keep the car? As I see it, that's the economic intuition behind why most people don't switch.
To understand how the framing might change how you interpret the problem, consider the following scenario: You are in a game of poker, and you have a flush with king high. Your opponent reveals all but one card from their hand, which shows they have 4 hearts, and they also reveal that their last card is an ace, but they don't reveal its suit. It's your turn to bet. Do you bet, or do you fold?
Now you could treat this as a simple statistics problem -- there are four possible aces they could have in their hand, and only one is a heart, so only a 1/4 chance they will beat you. But is the solution to this problem that there is a 3/4 chance of winning the pot? In the problem text, we haven't specified under what conditions your opponent will reveal which cards in their hands. But somehow, by saying it's a game of poker makes you think that they probably are more likely to reveal their hand if they are bluffing, so the true probability is not 3/4.
We are primed by this description of this person as your "opponent" to think about them making the decision adversarially. What if instead we say that that game of poker is part of a game show and your opponent is the host of the game show? Depending on the assumptions you make about your opponent's motivations, you must calculate the odds differently, and simply saying "3/4" is not unambiguously correct.
> 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. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?
Diaconis is in fact correct that given just that information the problem cannot be solved. What is missing is a statement that the host will always reveal a goat and always offer you a chance to switch doors.
If the host can chose whether or not to make the offer then if you you happen to receive the offer when you are on the show you cannot say anything about whether or not switching is to your advantage.
For instance suppose the show has given away a lot of cars earlier in the season and the producers ask the host to try to reduce the number of cars given away during the rest of the season. The host might then only offer switching when he knows the contestant has picked the car door.
He will still open a goat door first because that's more dramatic. He just won't offer to let you switch before going on to open either your door or the remaining door.
