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Monty Hall Problem

Has anyone read the OP and discussions following it? Monty opens the door and a goat is behind it. Switching results in you having two cards in the game rather than one. You have a 2/3 chance of winning.
 
lionking said:
Has anyone read the OP and discussions following it? Monty opens the door and a goat is behind it. Switching results in you having two cards in the game rather than one. You have a 2/3 chance of winning.
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Right. Per definition the opened door reveals a goat, as a matter of fact not condition. How this fact has been established, randomly or by choice doesn't matter obviously. It's given fact.

In case the opened door might reveal the car as well, that's a totally different story.
 
Switching will make you lose only if you had originally picked the door with the car. The probability that you originally picked the door with the car is 1/3. Therefore switching will only make you lose 1/3 of the time. Therefore switching makes you win 2/3 of the time. This is like spending 11 pages arguing about whether or not 1 + 2 = 3.
 
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Anti-Telharsic said:
Switching will make you lose only if you had originally picked the door with the car. The probability that you originally picked the door with the car is 1/3. Therefore switching will only make you lose 1/3 of the time. Therefore switching makes you win 2/3 of the time. This is like spending 11 pages arguing about whether or not 1 + 2 = 3.
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Very well put and welcome to the forum.
 
Anti-Telharsic said:
Switching will make you lose only if you had originally picked the door with the prize. The probability that you originally picked the door with the prize is 1/3. Therefore switching will only make you lose 1/3 of the time. Therefore switching makes you win 2/3 of the time. Does this really need 11 pages?
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I understand an ambiguity here. After the moderator has opened the door, where are we?

What is the probability P our choice wins the price

-1- and the moderator has revealed a goat
-2- under the condition the moderator has not revealed the car (but the goat) although he might have, because he has chosen the door randomly?

The second option makes the result of the moderator's choice conditional, and hence leads to a conditional probability of P=1/2, which is not my understanding of the problem.
 
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Yeah, the problem as usually stated has Monty revealing a goat always; much simpler than fussing around with all the possibilities otherwise.
 
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lionking said:
Has anyone read the OP and discussions following it? Monty opens the door and a goat is behind it. Switching results in you having two cards in the game rather than one. You have a 2/3 chance of winning.
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Herzblut said:
Right. Per definition the opened door reveals a goat, as a matter of fact not condition. How this fact has been established, randomly or by choice doesn't matter obviously. It's given fact.

In case the opened door might reveal the car as well, that's a totally different story.
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No, it does make a difference if Monty knows where the car is when he chooses the door.

- If Monty knows where the car is, and always chooses to show the goat, the chance that the car is behind the contestant's chosen door stays at 1/3 and the chance that the car is behind the other remaining door is 2/3.

- If Monty chooses a door at random, the game will stop if the door reveals a car. If the door reveals a goat, the situation for the two remaining doors becomes symmetrical: there is no reason to suppose that the car has more chances of being behind one or other of the two doors.
 
Michael C said:
- If Monty chooses a door at random, the game will stop if the door reveals a car.
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The door doesn't reveal a car per definition.

Edit: OK, now if you change the rules and assume, Monty is randomly choosing a door (but not yours) with two possible outcomes, then changing to another door (of your choice) still gives you 2/3 winning chance.
 
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GodMark2 said:
C: Monty may be evil, and thus only open a second door if he already knows you picked the car first, but not offering you the choice when you pick a goat.

Include that possibility in the mix, and switching (when offered) is not necessarily the best choice. With only a single instance from which to determine Monty's motive and the rules of the game, you don't have enough information.
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This is the only possible scenario in which switching would be a bad idea. Monty will only pull the stunt of opening another door to reveal a goat if he knows you have already picked the car. This is what I meant when I said, unless the game is rigged. Though whether you'd call that rigging, I'm not sure.

The rest is as Anti-Telharsic says.

I'm still trying to figure out what GreyICE means, and I think I understand. Would anyone else care to comment on his point that even if Monty is opening either of the other two doors at random, once you know that the door he did actually open has a goat, then the odds still favour switching?

Rolfe.
 
Rolfe said:
I'm still trying to figure out what GreyICE means, and I think I understand. Would anyone else care to comment on his point that even if Monty is opening either of the other two doors at random, once you know that the door he did actually open has a goat, then the odds still favour switching?
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If we play many times the version where Monty doesn't know where the car is and opens a door at random, it works like this:

In 1/3 of the games, the car is behind the contestant's chosen door. In 1/3 of the games, the car is behind the door that Monty chooses. In 1/3 of the games, the car is behind the door that neither of them chose. The important thing here is that when the car is revealed behind the door that Monty chooses, the game doesn't go on. We only get to the situation where the contestant has a choice between the two doors in 2/3 of the games.

This is the essential difference between this version of the game and the original one where Monty knows where the car is and chooses to reveal a goat: in the original case Monty can ensure that every game comes to the point where the contestant has a choice between the two doors. This means that if we play 300 games, all 300 of them will get to the point where the contestant has a choice between two doors. Since the contestant has chosen the door with the car in approximately 100 of the games, in the remaining 200 games the car will be behind the other closed door. Obviously he should swap doors.

In the version where Monty chooses at random, 1/3 of the games stop without the contestant getting the chance to choose between the two remaining doors. If we play the game 300 times, only 200 (approximately!) will get to the point where the contestant has the choice between the two doors. In 100 of these games, the contestant has already chosen the door with the car; in 100 of them the car is behind the other door. The contestant's chances of winning are 1/2: he can swap doors but it won't change his chances.
 
Michael C said:
In the version where Monty chooses at random, 1/3 of the games stop without the contestant getting the chance to choose between the two remaining doors. If we play the game 300 times, only 200 (approximately!) will get to the point where the contestant has the choice between the two doors. In 100 of these games, the contestant has already chosen the door with the car; in 100 of them the car is behind the other door. The contestant's chances of winning are 1/2: he can swap doors but it won't change his chances.
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That's wrong. You play the game only 200 times, 100 experiments are invalid and have to be disregarded because they don't correspond to the specified setup.
 
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Michael C said:
If we play many times the version where Monty doesn't know where the car is and opens a door at random, it works like this:

[....snip explanation of the two scenarios....]
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Yes, yes, yes, I understand all that.

GreyICE appeared to me to be saying that in the version where Monty opens one of the other two doors at random, on the occasions when he reveals a goat rather than the car, there is still some advantage to changing.

Here is what I said.

Rolfe said:
Now I may be remembering wrong, but I thought the conclusion I came to was, switch, whatever, given the problem as set out.

A. If Monty knows where the car is and is deliberately avoiding it, then you improve your chances.

B. If Monty is as ignorant as you are, then the chances stay at 50:50.

If you have no idea which scenario Monty is running it makes sense to switch. Because B is the worst case scenario (unless the game is truly bent), and in that one, while you don't gain by switching, you don't lose either.

By sticking, you are rejecting the possibility of improving your chances if A is in fact the scenario, and getting no possible benefit out of it.

By switching, you allow yourself to take advantage of the possibility of scenario A being the situation, without exposing yourself to a decrease in your chances.

Thus, if the rules are not specified, and A is within the bounds of possibility, then by definition switching is the sensible thing to do.
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I have already acknowledged GodMark2's additional point regarding the "bent game". However, here is what GreyICE said.

GreyICE said:
You ignore the fact that A and B are the same scenario. EXACTLY the same. If he opens it at random and happens to hit the goat, its exactly the same if he knew he hit the goat.

If he opens it at random and hits the car, you ought to know your chances of getting the car. DUH
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And again....

GreyICE said:
Rolfe, you're not getting it. Obviously you switch. But your scenario B isn't a scenario. It doesn't matter whether Monty opened the door at random or with foreknowledge, as long as he hit a goat.

He reveals the same information either way. It's your lack of understanding that makes you think that they're different scenarios.

The only way it doesn't influence the outcome is if the event is 50:50 going in, and you lose if it hits the car.
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I'm trying to understand if he has a point here. I'm seeing "DUH" and "you're not getting it", and "your lack of understanding", so I'm trying real hard to figure out if he might possibly be right.

I mean, arrogant posturing doesn't necessarily mean a poster is wrong, no?

Rolfe.
 
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I'm going to stretch this out even just a little further just to irritate Gnome. It's not about assumptions or anything like that. It doesn't matter what Monty knows or doesn't know. The description is this - there are 3 closed doors - 1 with a prize and 2 with crap. You pick one of the doors. Monty (and it doesn't even have to Monty) opens one of the other doors to reveal crap. And the question is are you better off switching to the remaining unchosen door or keeping your original door. You are better off switching. People bring in all sorts of irrelevancies which do nothing but confuse the issue. At first blush it seems counterintuitive, but it is not.
 
Herzblut said:
That's wrong. You play the game only 200 times, 100 experiments are invalid and have to be disregarded because they don't correspond to the specified setup.
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Yes, you can consider it like that if you wish. It doesn't change the result: of the 200 times you actually play the game, in 100 of them the car is behind the door you chose and in 100 of them the car is behind the other closed door. The probability of winning is 1/2.
 
billydkid said:
I'm going to stretch this out even just a little further just to irritate Gnome. It's not about assumptions or anything like that. It doesn't matter what Monty knows or doesn't know. The description is this - there are 3 closed doors - 1 with a prize and 2 with crap. You pick one of the doors. Monty (and it doesn't even have to Monty) opens one of the other doors to reveal crap. And the question is are you better off switching to the remaining unchosen door or keeping your original door. You are better off switching. People bring in all sorts of irrelevancies which do nothing but confuse the issue. At first blush it seems counterintuitive, but it is not.
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No. Consider:

There are three doors, A, B and C. One door has a prize, the other two have crap.

The probability of the prize being behind door A is 1/3. So is the probability of it being behind door B or door C.

At random, you choose door A. You open it and find crap. What is the probability of the prize being behind door B now? What is the probability of the prize being behind door C?
 
billydkid said:
I'm going to stretch this out even just a little further just to irritate Gnome. It's not about assumptions or anything like that. It doesn't matter what Monty knows or doesn't know. The description is this - there are 3 closed doors - 1 with a prize and 2 with crap. You pick one of the doors. Monty (and it doesn't even have to Monty) opens one of the other doors to reveal crap. And the question is are you better off switching to the remaining unchosen door or keeping your original door. You are better off switching. People bring in all sorts of irrelevancies which do nothing but confuse the issue. At first blush it seems counterintuitive, but it is not.
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This is exactly wrong. There's all kinds of misinformation swirling around here, so
Everyone please pause and read the following.

Sometimes the problem is stated such that Monty always shows you a goat before giving you the choice to switch. If it's stated that way, there is no argument that you'll win 2/3 of the time by switching.

However, usually when I see it stated, it's not explicit in constraining Monty to always offer the switch, all you know is that in this case, your only data point, he did offer the switch.

IN THIS CASE THERE IS NOT ENOUGH INFORMATION TO ARRIVE AT AN ANSWER WITHOUT KNOWING MONTY'S REASONS FOR OFFERING THE SWITCH.

Let's take three assumptions that we can make:

A: He knows where the prize is, will always reveal a goat door, and offer you the choice to switch. We've already determined that you'll win 2/3 of the time by switching.

B. Monty doesn't know where the prize is, he just happened to pick one door which was a goat, and now he's giving you the choice. In this case, if you simulate it yourself with pen and paper, you'll quickly see that your chances are not improved by switching: it's a 50/50 proposition.

C. Monty knows where the prize is, but wants you to lose the game, therefore he offers the choice to switch only for contestants who picked right the first time. In this case, obviously, you need to stick with your original guess (you'll lose 100% of the time by switching).

I hope it's clear that the host's motivation DOES matter, and that simply finding yourself in the situation where he has offered a switch is not enough information without knowing more about how Monty plays the game.
 
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CurtC said:
Let's take three assumptions that we can make:

A: He knows where the prize is, will always reveal a goat door, and offer you the choice to switch. We've already determined that you'll win 2/3 of the time by switching.

B. Monty doesn't know where the prize is, he just happened to pick one door which was a goat, and now he's giving you the choice. In this case, if you simulate it yourself with pen and paper, you'll quickly see that your chances are not improved by switching: it's a 50/50 proposition.

C. Monty knows where the prize is, but wants you to lose the game, therefore he offers the choice to switch only for contestants who picked right the first time. In this case, obviously, you need to stick with your original guess (you'll lose 100% of the time by switching).

I hope it's clear that the host's motivation DOES matter, and that simply finding yourself in the situation where he has offered a switch is not enough information without knowing more about how Monty plays the game.
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Agreed on all counts.

I might add one:

D) Monty doesn't know where the car is, but will always open one of the two doors you didn't pick and will always offer you the chance to switch. In that case you want to switch to the door he opens if it's the car (obviously!), and it doesn't matter whether you stay or switch (to the door he didn't open) if he opens a door to a goat.

Of the 2/3 times he opens a door to a goat, you win half no matter whether you stay or switch (that's CurtC's option B). Of the 1/3 times he opens the door to the car, you win all. So you win 2/3rds of the time in total.
 
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