The problem as stated does not give the host such an option. If it did, the host opening a door would imply that the player picked the right answer, and it would only happen 1/3 of the time.
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.
The problem as stated, in the article you pulled your quote from, puts no limits on how the host decides whether or not to offer a switch:
> 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.
The problem as stated has the host open an incorrect door and offer the player a chance to change his choice. Inferring that another possible variation might exist does not change the fact that we are discussing the variation that was presented. Both you and Diaconis are wrong.
> The problem as stated has the host open an incorrect door and offer the player a chance to change his choice.
Correct, in this one particular instance. You cannot conclude from this particular instance that the host always opens the door and offers a change.
> Inferring that another possible variation might exist
is totally reasonable, while denying the possibility that the host might be able to choose his actions specifically to benefit or screw you
over is an unwarranted leap.
The problem statement does not put constraints on the host. You cannot solve the problem by assuming that those constraints exist and then attack those like Diaconis who point out that those constraints don’t exist and that the thing that is unconstrained matters.
There is a genuine human language problem here, NOT A MATH PROBLEM, which accounts for two differing but self-consistent views. It is a legitimate difference in views because human language is genuinely ambiguous.
"You pick a door, the host opens another door and reveals a goat. Should you switch?"
Does this mean you are in one particular situation where the host opened a 2nd door, with a goat? Or does it mean the host always opens a 2nd door with a goat?
If the host always opens a 2nd door, showing a goat, you should switch to the third unopened door.
If all you know, is this time you picked a door, then the host revealed a goat, you don't know what to do. Maybe this host only opens goat doors after you pick the right door, in order to trick you into switching? In that case switching would be the worst thing you could do.
A host with that strategy is a special case, but special cases where a potential general solution (always switch) doesn't work, are all you need to disprove the general solution. It cannot be a general solution if their is even one special case it doesn't work.
Most people interpret the problem to mean the host always reveals goats.
But if the language isn't clear on that, then you do have a different problem, whose solution is really impossible to optimize for without some more information on general host behavior or strategies. Without that information, all you can do is flip a coin. Or always stay, or always switch. You have no means to improve your odds whatever you do.
The problem statement does put constraints on the host, by specifying that the host opened an unselected door with a goat behind it, only to ask if the player wants to change his choice. The answer to the question of whether the player should change the choice is well defined. Other variations are irrelevant since they are different problems.
Your argument is equivalent to denying that 2 + 2 = 4 is correct because the author had the option to write something other than a 2 as an operand.
Nope. The probability theory doesn't work like that. When you argue that 2+2=4 you assume 2 and 2 are known and they are not.
A=you picked the car at first
B=the host opened the door
P(A|B) can be anywhere between 0 and 1.
In your calculations you assume that P(A|B)=P(A) which is correct ONLY if A and B are independent. Independence of A and B is not in the problem statement, you invented this clause yourself.
> 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.
And in this particular instance, it makes sense to switch.
I'm sorry, but the problem is well-formed and well-specified.
> And in this particular instance, it makes sense to switch.
Are you are accepting that the host might be someone who only opens a door with a goat when your first choice was the door with a car behind it, and still arguing that you should switch?
The question asks what the best choice in this situation is, not a different situation. The answer does not depend on whether any other situations exist.
We don't really know the situation if all we were told is that we picked a door, and the host showed us a goat behind another door.
If we know that the host will always show us a goat behind another door, then yes, we should clearly switch.
If the host typically lets us just open the door, but will show us a goat before we open the door if the show is running too fast and they need to kill time, then we should switch if offered.
If the host typically lets us open the door, but will show us a goat if the show is running too fast, or if the prize budget is running low and we picked the car, then we should switch if we think the previous games went quickly, but not if there were some slow games already.
If the host only shows a goat when the contestent picks the car, then we should never switch.
Many problem statements include that the host always shows a goat; and if it doesn't you can kind of assume it, because it's a well-known problem, but if it's a novel problem and unsaid, then how are you supposed to know? I haven't watched enough Let's Make a Deal to know if they always give a second choice. Reading the NYT article linked elsewhere in the thread, I am reminded that Monty Hall could offer cash to not open doors too, so with the problem as stated and Let's Make a Deal being referenced, I have to assume an antagonistic host, unless provided with more information on their behavior.
As stated, assuming unknown behavior of the host, we can't put a number on the probability of switching.
Also, to address another point you made elsewhere in the thread. In addition to specifying the host behavior, it should also be specified in the problem statement that the car and goat positions were determined randomly, or at least the car was random, and the two goats are considered equal and assigned as convenient.
> So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!) Anything else is a different question.
That was how everyone interpreted the original column (according to thousands of letters sent to Marilyn vos Savant), and it was made explicit in a clarification made in a follow up to the column. What you are discussing is a different problem.
When you say "it makes sense to switch" you assume that P(A|B)=P(A) which is correct only if A and B are independent. Their independence is not given in the problem statement.
That they are independent is the only reasonable assumption, everything else is going out of your way to complicate the problem.
If they are NOT independent because Monty only shows you a goat behind the door if you've picked the wrong (or right) door, this is giving the game away. You don't need to guess, you always know what to pick with 100% certainty based on Monty's algorithm.
(Also, the game show doesn't work like this. And the text doesn't mention Monty's motivations, which in standard logic puzzle formulation must mean they are irrelevant, just as the phase of the Moon is also irrelevant and you must not take it into consideration)
If Monty picks randomly instead of always a goat, and he shows a car, the game has ended and no probabilities are involved, because you don't get to switch anymore; you've lost.
If Monty opens a door and there's a goat, we're within the parameters of the problem as stated (and you should switch!).
> That they are independent is the only reasonable assumption
No, this is not true. From the mathematical viewpoint Monty can have any strategy as long as it satisfies the problem statement. Which is, he DID open the door for whatever reason, the rest is uncertain. This literally what Diaconis means when he says "the strict argument would be that the question cannot be answered without knowing the motivation of the host" -- yes, in the strict sense he is indeed correct. This thread started because ryao stated that Diaconis is wrong [1].
Now even if you try to play the card of "reasonable assumptions" and rule out "boring" strategies because they are "giving the game away" this still won't eliminate all "non-independent" cases. The space of possible probability distributions here is way bigger than your list above. I can come up with an infinite number of "reasonable non-independent" strategies for Monty.
For example:
1) He rolls a dice before the game in his dressing room, secretly from the audience.
2) If he gets 6: he will open a door if you guessed incorrectly. If you guessed correctly he won't open the door.
3) If he gets 1-5: he will open a door if you guessed correctly. If you guessed incorrectly he won't open the door.
The situation is still the same: you've made your guess, then Monty opened the door with a goat and now you need to figure out whether to switch or not. It matches the problem definition stated above: the door was opened but we don't know why.
Let's see your chances if we assume Monty follow the dice approach:
event A: you've guessed correctly from the first try
event B: Monty opened the door
P(A|B): probability that you've guessed correctly given that Monty opened the door -- if it's less than 50% you should switch
P(A) = 1/3
P(B) = (1/6)x(2/3) + (5/6)x(1/3) = 7/18
P(AB) = (5/6)x(1/3) = 5/18
P(A|B) = P(AB)/P(B) = 5/7
So, in this case Monty doesn't "give up the game" -- there's still a significant random aspect to it. However in this setup for you it's better for you to stay (5/7 of winning) rather than switch (2/7).
You're right: I was focused on Monty picking a door with a goat depending on whether you had picked the right door. That would certainly give the game away, but indeed is not the only option.
However,
> Now even if you try to play the card of "reasonable assumptions" and rule out "boring" strategies because they are "giving the game away" this still won't eliminate all "non-independent" cases. The space of possible probability distributions here is way bigger than your list above. I can come up with an infinite number of "reasonable non-independent" strategies for Monty.
None of the assumptions you proceed to list are "reasonable". They introduce enough to the puzzle that they ought to be stated as part of the problem. Since they aren't, it's safe to assume none of those are how Monty picks the door.
Your "dice rolling" formulation of the puzzle is nonstandard. If you want to go with it, you must make it clear in the presentation of the puzzle. There are infinite such considerations; maybe Monty observes the phase of the Moon, maybe Monty likes the contestant, and so on... it wouldn't work as a puzzle!
Given no additional information or context, all we're left with is assuming Monty always opens a door with a goat behind it.
If we want to introduce psychology: I bet you almost all of the naysayers to vos Savant's solution to the puzzle are a posteriori rationalizing their disbelief: they initially disbelieve the solution to the standard puzzle, then when shown it actually works, they stubbornly go "oh, but the problem is underspecified"... trying to salvage their initial skepticism. But that wasn't why they reacted so strongly against it -- it was because their intuition failed them! I cannot prove this, but... I'm almost certain of it. Alas! Unlike with probabilities, there can be no formal proofs of psychological phenomena!
> Given no additional information or context, all we're left with is assuming Monty always opens a door with a goat behind it.
If you're playing against an opponent and trying to devise a winning strategy against him you can't just say "given no additional information or context, all we're left with is assuming his strategy is to always do X" and viola: present a strategy Y that beats X.
In this case X is "always opens a door with a goat behind it" and Y is "always switch doors". This is fascinating but simply incorrect from the math standpoint.
> Your "dice rolling" formulation of the puzzle is nonstandard. If you want to go with it, you must make it clear in the presentation of the puzzle. There are infinite such considerations; maybe Monty observes the phase of the Moon, maybe Monty likes the contestant, and so on... it wouldn't work as a puzzle!
The "dice rolling" it's not a problem formulation, it's one of the solutions to that problem i.e. specific values of X and Y that satisfy all the requirements. I present it to prove that more than one solution exist and furthermore not all solutions have Y="always switch", so you can't establish Y independent of X.
They key difference here is that I don't consider it as a "puzzle", whatever that means. I consider it to be a math problem. Problems of this kind are often encountered in both Game Theory and Probability Theory. It's perfectly fine to reason about your opponents strategies and either try to beat them all or find an equilibrium: this is still math and not psychology.
You can argue that it's a puzzle instead and I don't mind. What I do mind however is saying that Diaconis was wrong. He specifically said "the strict argument would be..." meaning that his conclusions hold when you consider it as a math problem, not as a "puzzle". My whole point is to demonstrate that.
> They key difference here is that I don't consider it as a "puzzle", whatever that means. I consider it to be a math problem. Problems of this kind are often encountered in both Game Theory and Probability Theory. It's perfectly fine to reason about your opponents strategies and either try to beat them all or find an equilibrium: this is still math and not psychology.
This was quite obviously a puzzle, of the "math problem" kind. It admits a pretty straightforward -- but counterintuitive -- solution, which made some admittedly smart people upset.
Everything else is smoke and mirrors.
> this is still math and not psychology.
If you read the responses to vos Savant's column, they are quite emotional. There was quite obviously an emotional response to it, of the "stubborn" and/or "must attack vos Savant's credentials" kind, too.
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.