Ivor the Engineer
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Monty Hall Problem
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billydkid
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This horse in quite completely decayed so - this above is what I said, right? The person who make the major point of saying I was EXACTLY WRONG, with oversized and bolded type - that it doesn't matter what Monty knows or doesn't know is wrong, right? This is all we know and all we need to know - you pick one of three doors. Monty opens another door to reveal a booby prize. And the only question is whether you odds improve by choosing the remaining door or they do not. They improve from 1/3 to 2/3. This is not about what Monty knows or doesn't know. The only things stipulated by the puzzle is what I have described above. In exactly the same way if you had a thousand doors and you picked one (your odds being 1/1000. Then Monty went on to reveal goats behind 998 of the other doors (regardless of why he chose those particular doors - it's not germaine to the problem). The odds of the prize being behind the unchosen door are 1000/1. This is correct isn't it?
Where I got really confused was with the Wikipedia page and they started talking about Monty's motives being critical. I just think that is wrong if you adhere to the stipulation of the problem - the he does, regardless of his motive or knowlege, pick a door with a goat. It is about statistics and nothing else.
No, you're failing to understand the nature of the problem.
Lets say Monty gets very, very luck with his thousand doors, and happens to open 998 of them at random without revealing the car.
The chances you now have are 1/2. Monty doesn't know where the car is, so he essentially won the lottery and happened to miss it. Now there's a 50/50 shot what door its behind. Think about it - there are a 999 outcomes on which door is remaining. 998 of those outcomes were eliminated, because he missed the car.
If he can't open the door with the car behind it, every possibility where he would open a door with a car becomes a possibility where he doesn't pick the car. The 998 outcomes where he picks a car aren't eliminated, in this case. They don't exist in the first place, in this scenario. Since no outcomes have been eliminated, the chances are overwhelming that its behind the final door.
I graphed this out (do it with paper) and its infinitely convincing once you see the paper in front of you.
The conceptual problem is Monty's motivations. Remove them, and the problem remains similar.
Flip two coins. But hide them, so you don't know the outcomes. Ask an impartial friend (A computer, if we want no humans in this) to reveal one if it came up heads (if both came up heads, he reveals just one). What are the chances the other coin is tails?
Now reveal one of the two coins at random. If its heads, whats the chances the other one is tails?
Finally program a computer to sometimes randomly reveal a coin, and sometimes reveal one of the coins if one is heads.
If the computer reveals heads, does it matter if its revealing coins at random, or doing the "reveal one if its heads" in terms of the possibility that the other coin is tails?
It seems odd, but which program the computer is running matters.
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billydkid
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No, no, no, no. Articulet has it exactly right and the talk about Monty "always having" to pick a certain door and so on is completely irrelevant. It doesn't matter how many doors you are talking about as long as there are more than two. We can leave Monty out altogether. There are 3 doors. When you pick a door you are essentially dividing these 3 doors into two groups with one door on one side and two on the other. Say, the door you picked on the right and the other two on the left. The odds of the the prize being on the right are 1 in 3. The odds of it being on the left are 2 in 3. A door without prize on the left is revealed (regardless of why). The odds of the prize being on the right remain 33% and the odds of it being on the left remain 67%. Take a billion doors with a single prize randomly places behind one of them. You pick a door. The prize is almost certainly behind one of the other doors. All of the other doors except one are revealed to be prizeless (regardless of why - it's not part of the puzzle). The prize is virtually guaranteed to be behind the remaining unchosen door. So you are saying there is a difference between whether he can't open the door with the car and he simply doesn't open that door? My argument would be that he can't by definition, because he doesn't. The definition of the problem is that he can't. How about he merely won't open the door with the car? Suppose this scenario - you choose one of a million doors and then Monty gives you the option of keeping your one door or switching to all of the remaining doors and you do that and then, for whatever reason, Monty reveals all of the remaining doors except the one you chose initially and just one of 999,998 on the other side. The car was a virtual certainty to be on the side with the umpteen doors before Monty revealed all of the remaining doors. Why would that change by him showing you all of the doors where the car is not?
Michael C
Graduate Poster
Here we go again!
Billy, what is you answer to this slightly different problem:
After you have chosen a door, Monty lets you designate a door to be opened. The door reveals a goat. Now you are offered the choice to switch from your chosen door to the other remaining one. What are now your chances of winning if you stay? If you switch?
ThatSoundAgain
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I think that this is easier to explain if you take care to remind people of the "hidden branch" - the possibility that the host unwittingly opens the door on the prize in step 2. In the latest scenario from your post, you'll have to take into account that the player could have revealed the prize with their second choice, but didn't.
CurtC
Illuminator
Michael C - I really like that variation.
Billy, I thought this was a settled issue, with everyone finally agreeing, until you showed up again. Please think carefully to realize why everyone else here has agreed that you need to know Monty's methods and not just what has happened on this one trial.
Billy, I thought this was a settled issue, with everyone finally agreeing, until you showed up again. Please think carefully to realize why everyone else here has agreed that you need to know Monty's methods and not just what has happened on this one trial.
I think that earlier on, when I was saying that Monty's motivations 'don't matter', people may have misunderstood what I meant. When we look at the problem, we as human beings with motives see that there is a person in the problem acting in a particular way. Because of this, we automatically assign him a motive to explain his behaviour. More than this, because the difference between the 'classic' Monty Hall problem and the 'random' Monty Hall alternative is that either a door is opened to always show a goat or a door is opened at random, we have to explain how this gels with what is going on inside the head of the person opening the doors - in this case Monty.
If Monty is always going to reveal a goat, then he has to know what is behind each door - otherwise it is improbable to the point of impossibility that he will never accidentally reveal a car.
If, however, Monty opens a door at random, we need to explain that too. The easiest way to do this is to have Monty not know where the prize is (so that it will not influence his door choice). There are other ways to randomize this, but the most intuitive is simply that he doesn't know where the prizes are.
This is where things start to get a bit back-asswards. In real life, motive precedes behaviour. This leads people to think that Monty's motive is the important part of the setup, as it influences his behaviour (and thus what game you are playing). Were this real life, that'd be the right way to do things...but in the context of the Monty Hall problem, it's getting the cart before the horse. Monty doesn't act in a particular fashion because of his motives, his motives have been assigned to match the setup in the scenario. It is entirely possible to set the problem up without including Monty at all, and to simply describe the rules of the scenario - but such a setup is dry, and not as 'fun' (in a manner) as one where you are on a gameshow with a host.
So while it may be fun to assign motives to Monty, they don't actually matter - the motives are there to serve the scenario, rather than define it.
If Monty is always going to reveal a goat, then he has to know what is behind each door - otherwise it is improbable to the point of impossibility that he will never accidentally reveal a car.
If, however, Monty opens a door at random, we need to explain that too. The easiest way to do this is to have Monty not know where the prize is (so that it will not influence his door choice). There are other ways to randomize this, but the most intuitive is simply that he doesn't know where the prizes are.
This is where things start to get a bit back-asswards. In real life, motive precedes behaviour. This leads people to think that Monty's motive is the important part of the setup, as it influences his behaviour (and thus what game you are playing). Were this real life, that'd be the right way to do things...but in the context of the Monty Hall problem, it's getting the cart before the horse. Monty doesn't act in a particular fashion because of his motives, his motives have been assigned to match the setup in the scenario. It is entirely possible to set the problem up without including Monty at all, and to simply describe the rules of the scenario - but such a setup is dry, and not as 'fun' (in a manner) as one where you are on a gameshow with a host.
So while it may be fun to assign motives to Monty, they don't actually matter - the motives are there to serve the scenario, rather than define it.
Michael C
Graduate Poster
You're right: I agree that what makes a difference is Monty's behaviour. We have to be careful when we're saying this, though: what matters is not just his behaviour this time, it's his behaviour pattern. To say that he opens a door revealing a goat is not enough information: we need to know if this always happens, or if it only happens when there is a car behind the contestant's door, or if it happens at random, or if some other conditions apply.
It's true that we don't have to ascribe motives to Monty in order to describe his behaviour pattern: we could just as well say that there is a machine that opens a certain door depending on certain conditions. But if we just say "a door is opened, revealing a goat", there is not enough information to give a unique answer. Equally, stating that the function of Monty is to open a door with a goat, or that the problem is so defined that Monty opens a door with a goat, does not provide enough information for a unique answer. We need to define the function of the "Monty machine" more precisely. Does it always open a door with a goat, and, equally important, does it always give the contestant the option of switching doors?
So much hinges on that word "always". Does the wording of the OP imply everything always happens the same way? I say it doesn't, but many posters here were adamantly insisting that it does. Apparently there is something ambiguous there, otherwise we wouldn't have so much disagreement (though I do wonder if a few people weren't just being pig-headed, refusing to even consider alternative interpretations, either because they were too attached to a mythical one and only "real" Monty Hall problem, because they were too attached to what they had first said or because they just wanted to annoy).
For those who insist that the notion that Monty always reveals a goat is implicit in the OP, I offer the following problem:
Minty Hill opens the front door of her home, revealing a goat on her doorstep. What should she do?
Rolfe
Adult human female
I'm going on from this, because I'm fed up repeating the same ground.
Which is.
We know that if Monty will always open another door at step 2, and will always reveal a goat, irrespective of your initial choice, your chances of winning are doubled if you switch. End of.
We know that if Monty has opened a door at random at step 2, switching doesn't alter your chances of winning. End of, again.
We've gone round this so often I'm getting dizzy.
And then there are the other scenarios where is decision as to whether or not to open a door at all is dependent on whether or not you already have the right door. Once you start allowing for these (that is, scenarios where Monty may not offer you the choice at all), there really is no way to answer the conundrum, because you can't exclude the possibility that Monty is deliberately trying to trap you into switching away from the correct choice.
So, can I stand the problem on its head? Just for a bit of variety? Let's look at it from Monty's point of view, assuming he knows where the car is (as would certainly have been the case in the real show), and he's an active participant rather than a computer model.
Step 1, you choose a door. There are only two possibilities here - you've either picked the prize door or you haven't.
Monty now has one decision to make - will he give you a chance to switch, or won't he?
So there are really only four possibilities.
- You already got the prize, and Monty decides to let you have it. He would do this by opening the door you already chose.
- You don't have the prize and Monty decides not to give you a second chance. He could do this either by opening the door you chose, or by opening to door with the prize to show that you don't have it.
- You already got the prize, and Monty decides to try to entice you away from it. He opens one of the two goat doors and sweetly invites you to change.
- You don't have the prize, and Monty decides to give you a second chance. He opens the other goat door, and sweetly invites you to change.
Looked at this way, it's not "do I feel lucky", but "do I think Monty likes me?"
Really, your chances of a switch being beneficial depend on whether Monty is more likely to offer someone who has initially chosen wrong a second chance, or whether he's more likely to try to snooker someone who chose right initially.
No doubt some information could be gained on this by observing the actual outcomes of a sufficient number of previous games. However, since Monty probably isn't tossing mental coins here, other factors will come into play such as audience reaction, how stubborn he thinks you are, and indeed whether or not he likes you!
Rolfe.
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TrueSceptic
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Once you start thinking about this and then consider Monty's motives, there's no end to it.
One more possibility, and I apologise if it's been mentioned already: what if Monty needs someone to win the prize every nth time, to maintain interest and viewer levels? On these occasions he steers the game towards you winning, changing from the default strategy of steering the game towards you losing. How often would he do this, i.e. what are your chances of being favoured?
Is there any way this puzzle can be resolved without more information?
hi Rolfe
your terms here appear unclear. please just point me to the relevant post if this was already clarified or if you are interested in a nonstandard definition of the problem (like opening the chosen door).
if monty opens one of the other two doors at random, there are two options:
a) the door he opens is the prize door: your chances of winning are now zero and you know it
b) the door he opens is a goat: you can increase your chances of winning by switching (the standard "two doors versus one door" argument holds regardless of Monty's ignorance).
if he opens the door you have, you know you have won, or you know you should switch; even in this nonstandard verison you gain information and can improve your "chance of winning".
so even opening a door at random introduces more information and thus alters your "chances of winning".
(or: which of the 700 other posts should i have read?)
agreed.
your terms here appear unclear. please just point me to the relevant post if this was already clarified or if you are interested in a nonstandard definition of the problem (like opening the chosen door).
if monty opens one of the other two doors at random, there are two options:
a) the door he opens is the prize door: your chances of winning are now zero and you know it
b) the door he opens is a goat: you can increase your chances of winning by switching (the standard "two doors versus one door" argument holds regardless of Monty's ignorance).
if he opens the door you have, you know you have won, or you know you should switch; even in this nonstandard verison you gain information and can improve your "chance of winning".
so even opening a door at random introduces more information and thus alters your "chances of winning".
(or: which of the 700 other posts should i have read?)
Paul C. Anagnostopoulos
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Does it change anything if two doors contain goats and the other door contains free will?
~~ Paul
~~ Paul
Paul C. Anagnostopoulos
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If you're saying what it sounds like you're saying, then no.lenny said:
Make a table and fill it in.
~~ Paul
articulett
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Right. If you are shown the goat and you are asked to make a choice and you have no other knowledge on the subject (such that Monty always asks a person if they want to change)-- then it doesn't really matter what you do-- it's roughly 50-50.
But if it's stipulated as it is in the classic Monty Hall problem-- Monty always shows a goat (so Monty can't be choosing blindly) and you are always offered a choice (so Monty can't be offering you the choice for nefarious purposes)-- then you double your odds from 1/3 to 2/3 by switching.
And if there is free will behind on door and a dead horse behind the other two...
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agreed.
ThatSoundAgain
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Ruling out the possibility of a) in the standard version is exactly what increases the players' odds. You can think of the probablilties of a) and b) as merged in the standard version: "Do you want the single door you first picked or whatever prize is larger of the other two?"
This only occurs if a) is - per problem definition - an impossibility. In the random Monty scenario, you'll have 1/3 chance of winning the prize by staying, 1/3 by switching, and a 1/3 risk of having Monty end the game by picking the prize. This is as viewed from the start of the game - if you get to the point in the game where Monty has opened a non-prize door, each remaining door will have a 1/2 chance of being the winner.
Since I mapped it out anyway:
Code:
Doors | Outcome
[B]1)Car[/B] [B]2)Goat[/B] [B]3)Goat[/B] | [B]Stay[/B] [B]Switch[/B] [B]Game over[/B]
p m | 1 0 0
p m | 1 0 0
m p | 0 0 1
p m | 0 1 0
m p | 0 0 1
m p | 0 1 0
p: Players' choice
m: Monty's choiceETA: The above is, of course, a table of the non-standard "random Monty" variation. In the classic version, I agree that you increase your odds by switching.
ETA ETA: Your odds increase in one version and not the other, not by magic but by Monty essentially telling you something you don't know. As much of this thread has been about, only if the rules are strictly as in the classic version (and known to you) can you decipher his message.
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OK: i will make a table tomorrow, but i expect it to look rather like the standard table; given additional information, i can exploit this information within the rules to change "my chances" or i know immediately that the proability of success is zero, that this also changes "my chances".
can i assume you only object to (b)? (if
(a) monty opens the door with the prize, we agree that my chances of winning have changed (they are zero)
(c, nonstandard) monty open my door, i have the option to change if i have not won, and "my chances" of winning are increased from zero (if i do not change).
if i learn one of the doors i did not pick was not the winning door, this information is of value. full stop.
ThatSoundAgain
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lenny, I hadn't thought of your option c. Goes to show that strict definitions are necessary here - my table above operates under the rule that Monty will always choose another door than the player.
ETA: And now I see you wrote about that above. Sorry.
ETA: And now I see you wrote about that above. Sorry.