I want to add here that, looking around online, I see a lot of statements from people who believe that Vos Savant (and zut) are right and that if the host is clueless it makes no difference whether the player keeps the original door or switches. But I really cannot see that it matters how we get to the situation where a door is opened to reveal a goat. Maybe I’m the one being clueless here, but I don’t see how Monty’s state of mind changes the probability of the player having guessed correctly.
I also don’t see how a real player would know if Monty had intentionally opened a goat door or just opened it by chance, but I guess that’s taking this out of the realm of a logic puzzle.
One more time: list out the probabilities like I did above. The two situations (Monty is random vs. Monty only reveals goats) leave Monty with different decisions, which lead to different probabilities.
Here’s the point where you’re veering astray: seeing a goat may, or may not, give you new information, depending on the circumstances where you see it.
Let me restate a game I proposed above:
Suppose we have a new game, just the three of us: me, you, and Monty. In this game, you choose a door, then I choose another door, and Monty’s door is the one that’s left over.
Now, you should agree that, at the beginning, your chance of having the door with the car is 1/3, right? Likewise, the chance of my door having a car is 1/3, and Monty’s is 1/3. With me so far? Now, you should also agree that, at this point, you know that at least one of the other doors (mine and Monty’s) has a goat behind it, right?
So now, Monty opens his door. Turns out there’s a goat behind it. But, according to you, “seeing the goat doesn’t give [you] any new information about [your] own door.” Right?
So NOW Monty asks you, “do you want to switch doors with zut?” “Of course!” you say. “I have a 2/3 chance if I switch!” (I’m following your logic here.) Right?
So here’s a question: Would it be in my best interest to switch? Because, after all, “seeing the goat doesn’t give [me] any new information about [my] own door.” I’m in exactly the same boat as you.
As Zut has been saying, it all comes down to whether you learn new information from opening the first door. Whenever you learn new information, the game changes. If Monty knows where the prize is and always opens a goat door, you learn nothing. If he opens a random door that might have hidden the prize, you have learned something.
If Monty knows where the prize is, and only opens a door with a goat, then the problem is like this-
Chances of player picking prize door- 1 in 3
Chances of Monty opening prize door- 0 in 3
Chances of prize behind last door- 2 in 3
I think this is the problem as you see it. You learn nothing new by opening the first door. It is always a goat. It could not have been the prize. However,
If Monty does not know where the prize is, or opens a random door, then the problem is different-
Chances of player picking prize door- 1 in 3
Chances of Monty opening prize door- 1 in 3
Chances of prize behind last door- 1 in 3
In this case opening the first door does give you new information. The prize could have been behind the first opened door, but it was not. In this case, its like a new game. When you get new information, you have a new game with new odds.
Another thing to look at, is that when you add up the odds for all of the possibilities, they must add up to 1. In the above situation, 1/3+1/3+1/3=1.
You cannot have a situation with 2 possibilities that are both have a 1 in 3 chance.
I think what Lamia is saying is that when Monty asks if you want to switch, he is not just asking if you wish to switch to the door that is not opened, but rather, do you wish to switch to either of the other doors. So if Monty does not know he may very well show you the car and you may pick it.
This gives you the same odds, of winning 2/3, however it does not mean you should switch. If Monty reveals a car you switch, if he does not then it is 50/50 and there is no need to switch.
I agree it makes no difference whether the two players in this problem switch or not, but I think this is a significantly different problem than the one where there is only one player and a clueless host.
I’m willing to admit that maybe I’m just not getting it, and Google indicates that many others are on your side with this one, but once the single player has seen the goat I don’t see how he’s in a different situation with the clueless Monty problem then he would be in the original knowledgeable Monty problem. He has the same information either way. The player knows he had a 1/3 chance of guessing right at the beginning of the game. He knows that the already opened door definitely has a goat and thus a 0% chance of being the winning door.
Now, if the single player were offered the chance to switch after Monty chooses another door but BEFORE it is actually opened then I would agree it makes a difference whether Monty knows where the car is or not. If Monty is clueless then he has a 1/3 chance of having selected the door with the car, and the player has a 1/3 chance of winning whether he sticks with his original door or switches. If Monty is not clueless and the player knows Monty never chooses the car, then the player would also know that whichever door Monty picked definitely won’t have the car even before it is opened.
I see on preview that someone else has rung in. I appreciate everyone trying to explain this, because I’d like to understand the correct answer.
But you do learn something. The player already knew there was a 2/3 chance his original guess was wrong, and now he knows which of the other doors has the car if he was in fact wrong. If the player was right (1/3 chance) then it’s true that seeing the goat doesn’t help him, but if he was wrong (2/3 chance) then seeing the goat tells him where the car must be. Of course the player can’t be sure at this point whether he was right with his first guess or not, but he’s better off switching since the odds are against his having guessed right at first.
I agree that, as far as the player knew, before Monty opened his door there was a 1/3 chance of the car being there. But once the player sees the goat, there’s no longer a 1/3 chance that Monty chose the car. Once the door is open the car can’t 1/3 be there, it’s either 100% there or 100% not, and in this scenario it is not.
So the player has gained new information, but as far as I can tell this new information makes the new game exactly the same as the one where Monty intentionally chooses a goat: there’s a 0/3 chance that Monty’s open door actually has the car.
Now that I’ve figured out how to code this properly, I’ll present a table of my understanding of how the clueless Monty problem works.
Whether Monty is clueless or not, there are always three equally likely possibilities as the game begins:
Door A Door B Door C
Goat Goat Car (1/3)
Goat Car Goat (1/3)
Car Goat Goat (1/3)
In this case let’s say the player chooses Door A. As we can see, there are two possible outcomes where he was wrong, and one where he was right. The player doesn’t know if he was right or not though, because his door remains closed for now.
If at this point clueless Monty picks another door at random, he is equally likely to choose Door B or Door C. Whichever door he chooses, just as with the player’s original choice, there’s a 1/3 chance that Monty gets the car and a 2/3 chance that he gets a goat. Let’s say Monty chooses Door B.
If, before Door B is opened, the player is offered the chance to switch to Door C then he has no particular reason to do so. It wouldn’t matter either way. That’s because Monty doesn’t know where the car is, so the fact that he wants to open Door B doesn’t tell the player anything new. Nothing has changed since the player made his first choice: as far as anyone knows there’s a 1/3 chance that any given door has the car.
Player Monty
Door A Door B Door C
Goat Goat Car (1/3)
Goat Car Goat (1/3)
Car Goat Goat (1/3)
But if the player doesn’t have the chance to switch until AFTER Monty opens his door, as in the original problem, it does change things. If Monty’s door has the car then there’s no need to keep guessing where it is. Everyone will see that the car was behind Door B. So there’s a 1/3 chance that the game ends here without the player having to make another decision, and a 2/3 chance that it continues.
IF the game continues there are only two remaining possibilities, because Monty’s choice (Door B) must have had a goat:
Player Monty
Door A Door B Door C
Goat Goat Car
Car Goat Goat
The player now has the chance to switch. The big question here is what are the probabilities for his two possible choices? What are the odds that Door A has the car, and what are the odds that Door C does? It can’t be an equal 1/3 for both Door A and Door C anymore, because Door B is out of the running. There’s now a 0/3 chance that Door B has the car (everyone can see it doesn’t), so the odds for one or both of the remaining cars must have increased by a total of 1/3.
If I were unfamiliar with the classic Monty Hall problem then I probably would figure that the odds for both remaining doors have now increased equally by 1/6 (half of 1/3), giving Door A and Door C each a 1/2 chance of being the right door. In that case it wouldn’t matter if the player kept his first choice or switched.
Yet the original problem shows us that it’s possible to go from three equally likely choices to two choices that are NOT equally likely. The player in the classic Monty Hall problem would at this point would still have a 1/3 chance of winning the car if he sticks with Door A but a 2/3 chance with Door C.
Why would these same odds not occur in the clueless Monty game? At this point in the game, what’s the difference whether Monty has a clue or not? He still picked a door with a goat, just as he would have if he’d known where the car was. There was a difference before clueless Monty opened the door, because he might have revealed the car and ended the game then. But once a goat is revealed, isn’t the player is facing the same final decision as he would have been if Monty had opened a goat door on purpose?
Nope. Same game. Since Monty’s clueless, it doesn’t matter which door he picks, so it may as well be the one neither you nor I chose.
No. When Monty acts differently, you glean different information, and that information allows you to modify the probabilities of different cases. Let’s go back to the beginning of your post:
That’s exactly correct. There are three possibilities; all of these three possibilities are equally likely (I’m using this phrase on purpose).
This is exactly correct: When Monty is clueless, if he reveals a goat, then what he’s done is eliminate one of three equally likely possibilities. Nothing more.
For example, suppose we had a game where I randomly grab a green, red, or blue marble out of a bag. At the beginning of the game, there are three equally likely scenarios:
-I pick green (1/3)
-I pick red (1/3)
-I pick blue (1/3)
Now, suppose I pick a marble, look at it, and tell you quite truthfully, “this marble is NOT red.” Now that I’ve eliminated one of three equally likely possibilities, what’s the chance the marble is green?
In both the marble game and the “clueless Monty” game, the additional information allows you to eliminate one of three equally likely possibilities, leaving two (still equally likely) possiblities. Nothing chages the relative probability of the two remaining posiblities.
In this case, the game again begins like so:
Door A Door B Door C
Goat Goat Car (1/3)
Goat Car Goat (1/3)
Car Goat Goat (1/3)
And we can again assume, without loss of generality, that the player chooses Door A, and the three possibilities are all equally likely. However, in this game, Monty never reveals the car, so none of these scenarios are invalidated. In fact, here’s what will happen:
Door A Door B Door C
Goat Goat Car (1/3) Monty OPENS door B, leaving door C (the car)
Goat Car Goat (1/3) Monty OPENS door C, leaving door B (the car)
Car Goat Goat (1/3) Monty OPENS either door B (1/6) or C (1/6), leaving a goat
It should be obvious here that, since all three possibilities are still equally likely, and in two of them you win by switching, that switching is the winning strategy.
For the same reason they have two different probabilities of being true in the classic Monty Hall problem. That’s my point. Once clueless Monty has revealed a goat, I think the two problems have the same solution: the player’s original 1/3 odds of guessing right on his first choice do not change, so there’s a 2/3 chance that the remaining door has the car. I can accept that I may be wrong about this, but I don’t understand why.
It’s some comfort to me that if I’m wrong, I at least wouldn’t be hurting my odds by always switching the way I think I should.
But if clueless Monty opens a door with a goat then how has he acted any differently than if he intentionally opened a door with a goat? His decision-making process may have been different, but what actually happens (and what the player observes) is exactly the same: Monty opens a door and reveals a goat. The player receives the same information whether Monty knew the goat was there or not.
*Isn’t this the same as asking which goat Monty picked, though? There’s a 1/2 chance that the marble is green, but if I’m after the red marble then I don’t care whether your marble is green or blue.
If we distinguish between the two non-winning options this way then the table of possibilities becomes more complicated but I don’t think it changes anything. Let’s say there’s a blue goat, a green goat, and a red car. Here are the possibilities:
Door A Door B Door C
Green Blue Red (1/6)
Blue Green Red (1/6)
Green Red Blue (1/6)
Blue Red Green (1/6)
Red Green Blue (1/6)
Red Blue Green (1/6)
The player picks Door A. The odds are an equal two out of six or 1/3 that the car is there, that the blue goat is there, or that the green goat is there.
Clueless Monty picks Door B at random and peeks behind it. As with the player, the odds are equal at this point that he will see the car, the blue goat, or the green goat. In two out of the six possibilities (1/3), Monty will see the car and the game will end. But if he can truthfully say “It’s not the car” then this is the same as the marble question, right? One of the equally likely possibilities has been eliminated (Monty didn’t get the car) so it’s now two out of four or 1/2 odds that Monty has the green goat. But this should be of no interest to the player. The important thing is that it was a goat.
If Monty knows where the car is then he also has a 1/2 chance of revealing the green goat, provided he doesn’t favor one color of goat over the other. If the green goat is behind Door A, Monty must reveal the blue goat (1/3). If the blue goat is behind Door A, Monty must reveal the green goat (1/3). If the player got lucky and picked the car, Monty may randomly choose either goat (1/6 to blue, 1/6 to green). So that’s 1/3 + 1/6 = 1/2 odds that a particular goat (blue or green) will be revealed.
Once clueless Monty has already opened a door and found a goat, I don’t see any way of distinguishing this situation from the one where Monty intentionally chose a door with a goat. In either case Monty chooses a goat, and it’s equally likely (1/2 either way) that it was blue or green.
*What’s the difference between a game where Monty can’t reveal the car and a game where he doesn’t reveal the car? If Monty doesn’t know where the car is then he might reveal it by chance, but if that happens the player has no need to weigh the probability of switching. The solution is obvious. If he’s allowed to switch to Monty’s door then that’s a 100% chance of winning for the player – he knows where the car is. If he can only choose one of the two doors Monty didn’t pick then he has no chance of winning either way because Monty got the car. The odds of switching only present a puzzle if Monty doesn’t choose the car, in which case the player has the same information he would if Monty couldn’t choose the car: he saw one of the doors he didn’t choose open to reveal a goat.
I’ve read barely any of this thread. But this stood out to me:
The player receives more information than just knowing what’s behind a certain door. They also receive the information that Monty chose to open that door. [Note, if a third party told you “There’s a goat behind Door B”, and then told you “Monty is going to reveal Door B”, the second statement would be giving you new information you didn’t have before]. How exactly this second morsel of information affects probabilities depends on what Monty’s decision-making process is.
That there is this additional information being received is the key to the whole thing. If the only information the player received was “There’s a goat behind Door B”, then they would be correct in stating that Door A and Door C have equal probability of winning. It’s the new information “Monty chose to reveal Door B, instead of Door C” that shifts the probabilities.
But as far as the player knows, Monty always freely chooses the door that is revealed.
It’s true that Monty’s actual choice is forced if he both knows where the car is and the car is NOT behind Door A. Monty can only open the door that doesn’t have the car, whichever one that is. If the car is behind Door B he must open Door C, and vice versa. If the car is behind Door A then Monty has an actual choice and can freely choose Door B or Door C. If the game is fair this choice will be random.
If Monty doesn’t know where the car is then he can only ever make a random choice between Door B and Door C. But how does this look any different to the player than in the game where Monty knows where the car is all along?
If the player can somehow tell the difference between Monty making a forced choice and Monty making a random choice, that changes the solution to the original problem. Monty only makes a random choice when the player was right with his first guess of Door A, so when the player can tell that Monty’s choice was random he shouldn’t switch. If he can tell the choice was forced then he should switch to whichever door Monty didn’t open because that’s sure to be the right one.
The player is not able to engage in any probabilistic reasoning without some model of how the game goes. In addition to including some model of how the original winning door is chosen (presumably, with all three equally probable), this will include some model of how Monty chooses which door to reveal. The details of the latter highly influence the probabilities the player calculates.
If the model is “Monty chooses a door in a manner probabilistically independent of its contents (i.e., Monty has no clue)”, the information “Monty chose door B” will tell the player nothing about which door is the winning one. The probabilities of each door winning will remain the same after learning this information as before.
If the model is “Monty chooses a door in a manner influenced to some degree by its contents (i.e., Monty has some clue)”, then the information “Monty chose door B” will tell the player something about which door is the winning one. The probabilities of each door winning will potentially change after learning this information.
It’s up to the player to choose a model with which to reason. Probabilities always only exist relative to such a model. That having been said, there may be good considerations for picking one model rather than another.
I lost sleep over this last night, but I think I have it now.
What was helpful to me was to go to the ten doors/“I’m thinking of a number between 1 and 10” example that was used to teach me the original Monty Hall problem. With the “I’m thinking of a number” game, it’s especially obvious (to me a least) why the host’s knowledge makes a big difference when the player is offered the chance to switch.
Scenario 1: Monty chooses a number himself and the player makes a guess, let’s say 7. Monty says to the player, “You may stick with 7, or switch to 3.” At this point the player knows that either his first guess was correct (1/10 odds) or Monty has just told him the correct answer (9/10). The player is better off switching.
Scenario 2: Monty does NOT choose the number and the player knows this. It’s selected randomly in some way that neither Monty nor the player can see, such as one of the spokesmodels spinning a number wheel behind a curtain. The player guessed 7. Monty says to the player, “You may stick with 7, or switch to 3.” This is no more helpful to the player than if anybody else had suggested an alternate number. It might as well be a two-player game where they’re offered the chance to switch with each other once both have selected a number. Monty doesn’t know what the right number is, so he has the same 1/10 chance of guessing correctly as the player did. It makes no difference if the player switches, because both 7 and 3 have an equal 1/10 chance of being correct. 80% of the time neither number will have been correct.
In Scenario 1, Monty’s choice of a number is a lot more meaningful than in Scenario 2. In Scenario 1 it wouldn’t really matter if the other eight possible numbers were proven to be incorrect (their doors opened to reveal goats) or not, as long as the player trusts that Monty is honest and that he really knows where the car is. Even before seeing the goats, the player knows that either #7 or #3 is correct, and that Monty’s suggestion is much more likely to be the right one.
In Scenario 2, unless the doors are opened then as far as the player knows both his choice and Monty’s choice are incorrect. This is the most likely scenario by a long shot, since there’s an 80% chance that the car is behind some door other than #7 or #3. So if the eight remaining doors are opened then the majority of the time this will end the game without the player needing to make a decision to switch or not. The other 20% of the time there will be goats behind all the other doors and the game will continue. But since the player knows that Monty guessed at random, he knows that Monty is no more likely to have guessed correctly than the player was. The player has equal odds of winning at this point if he keeps his original guess or he switches.
Part of what’s tricky here is that revealing the goats doesn’t really change anything for the player. I was hung up on this before. I said above that “if the single player were offered the chance to switch after Monty chooses another door but BEFORE it is actually opened then I would agree it makes a difference whether Monty knows where the car is or not”, but I didn’t understand that opening the doors provides the player with little new information. The only real difference opening the doors can make is that in Scenario B it’s possible this will automatically end the game.
Another tricky part is that, as Indistinguishable points out, it makes a difference what the player understands Monty’s decision-making process to be.
In a fair two-player game, it should be fairly obvious to each player that they were each equally likely to guess correctly and that nothing is to be gained by switching. It’s a lot less obvious that the host would be no more likely to guess correctly than the player. Unless the player KNOWS that Monty is just guessing, if the doors are opened in Scenario B (clueless Monty) and the car is still in play then it looks exactly the same as Scenario A (knowledgeable Monty). If the player doesn’t know how Monty made his decision, at this point it should seem a lot more likely that Monty really knows where the car is – he’s the host and plausibly has access to special knowledge, and there was only a 20% chance of getting to this point through sheer luck with a clueless Monty.
The way the clueless Monty scenario was first presented here (quoted from Wikipedia by aldiboronti in post #29), the player does NOT know that Monty is just guessing: “In this version, Monty Hall forgets which door hides the car. He opens one of the doors at random and is relieved when a goat is revealed.” This indicates a game where Monty normally does know where the car is. He just happens to have forgotten this time. Monty manages to guess a goat door as usual, so the player has no reason to believe things have changed for this round.
Although in reality it makes no difference whether the player switches or not here, the player has no way of reaching this conclusion in the game described. This game looks exactly the same to the player as the usual version where Monty knows where the car is. Luckily, it is safe for the player to go ahead and behave as though this is the classic Monty Hall problem. If there’s any doubt as to how the game is being run then switching is the smarter choice, because it’s more likely to be correct if Monty knows where the car is and equally likely to be correct if he doesn’t.
Hmm…let’s suppose I choose door 1, you choose door 2. I rudely ignore you for the moment, as if you were never there. Monty opens door 3 to reveal a goat. It’s to my advantage to switch–that’s been established, right? Well, is it to your advantage to swtich to door #1? It’s better for both of us to switch?
Worded differently: It’s just me, and I choose 1. Monty shows door number 3–it’s a goat. I should switch to door #2. Same scenario, except I originally chose 2. Still to my advantage to switch to #1?
Well, I can in the instance I’m describing, because he has revealed a goat, not a car. As I said originally, I understand that a random selection could show a car. But if it didn’t, why would my decision to switch be affected by whether or not the “reveal” was deliberate or lucky.
I’m suggesting that in instances where Monty reveals a goat, I should switch, because based on the subset of instances where he shows a goat, the odds are in my favor to switch, and that doesn’t change if he did it knowingly, or if the wind blows the door open. 2/3 of the time, it would be better for me to trade my door for the remaining two. Even a random “reveal” effectively permits me to do that.
Not to belabor, but I get that a random opening of a door could well reveal a car, and a large number of such events would do so a third of the time. But we’re not analyzing an endless series of events–it’s a single instance, where Monty opens the door and reveals a goat. Unless I’m missing something, whether or not Monty knew what was behind the door or not does not change the odds affecting my decision. Why would it?
Exactly. So…is it? Can you reconcile your answer to what you believe about the problem?
Take a look at what Indistinguishable wrote. Whether Monty chooses a door randomly, or to intentionally reveal a goat, gives you different information. Make a chart and run through the probabilities for each case, and you’ll see they’re different.
Yes, I believe it’s advantageous, because in either instance, 2/3 of the time I would be switching to the car. In that specific instance, obviously, it is only actually advantageous for one of those choices.
I already did, perhaps incorrectly.
If I chose door 1, and Monty reveals door 3 as a goat, knowing ahead of time it was a goat, I now know that if I chose incorrectly, the car is behind door #2. I also know that 2/3 of the time I will have chosen incorrectly, so I ought to switch doors.
If I chose door 1, and Monty reveals door 3 as a goat, NOT knowing ahead of time it was a goat, I STILL now know that if I chose incorrectly, the car is behind door #2. I also know that 2/3 of the time I will have chosen incorrectly, so I ought to switch doors.
Indistinguishable’s point seems to be that if Monty selects a door randomly, I can infer nothing about what will be behind the door ahead of time, a notion I concede (or perhaps misunderstand). That does not change the fact that a randomly opened door that reveals a goat does give me the same information—exactly the same–as one made with knowledge, when it happens to reveal a goat. In both instances it tells me which of the remaining doors is definitely a loser. That’s all I need to know.
Again, I’m not plotting out all the possibilities of what might occur with a randomly selected door, since those possibilities ignore an important piece of info–this door was opened, and I’m looking at a goat. I need only concern myself with the probabilities associated with that circumstance.
By the way, I’m not certain I’m right–just not seeing the light, though.
If Monty opens Door 2, and it reveals a goat, you have learned not just one, but two pieces of information:
A) Door 2 contains a goat.
B) Monty chooses to open Door 2
If you started off thinking all three doors were equally likely to be winners, then after learning A), you will switch to thinking Door 1 and Door 3 are equally likely to be winners, while Door 2 definitely is not.
But you’re not done after learning A). You also have to factor in the effects of learning B). What does that do for you?
Well, it depends… if you model Monty’s choice of a door as probabilistically independent of which one wins, then B) gives you no relevant information. You’ll continue to think of Door 1 and Door 3 as equally likely to be winners, while Door 2 is definitely not.
On the other hand, if you model Monty’s choice of a door as not probabilistically independent of which one wins, then the information given by B) is relevant, and will potentially shift your probabilities. Exactly how it does so depends on the details of the probabilistic connection between Monty’s choice of a door and the winning door in your model. [In the commonly intended setup of the problem (where Monty is equally likely to choose any goat-door other than the one you’ve selected, and guaranteed to choose no other), this is where the probabilities shift to your selected door have probability 1/3 of winning, while the remaining one has probability 2/3. On other setups, the probabilities we calculate can be different.]