05 Aug 08
Originally posted by CeSingeThe only assumption is that the presenter will never open the door with the car in it when you haven't made your final selection. That's not an invalid assumption since that's exactly the way the show "Let's make a deal" worked - the host was Monty Hall and hence why it's called the Monty Hall problem.
In all replies I read here, assumptions are made that the presenter will this or would that. In the initial question, there is no hint towards such assumptions.
In such a case, your adds are 50%, and they are independent of the presenter and of the door you then choose (out of 2).
If this is incorrect, I'd be glad to hear it and understand why.
Beyo ...[text shortened]... presenter when opening a door, and doing so, you can modify the outcome of the problem at will.
The game worked by the contestant choosing one door. Then Monty would open a door showing the non-car item. Then he'd ask you whether you wanted to keep your choice or switch. There isn't any real assumption other than that he wouldn't open the door with the car first - which would ruin the game anyways.
The odds do work out to be better if you switch as people who have posted have explained and it is unintuitive to most. That's why so many people got goats on the show 🙂
05 Aug 08
Originally posted by Wheelyhttp://www.theproblemsite.com/games/monty_hall_game.asp
No, the original post mentions that the presenter will show you a door with the salt. This is crucial. The presenter knows which door is which (otherwise he couldn´t guarantee showing you a door with the salt) and therefore all this nice math we see around boils down to the following.
You are most likely to have picked a door with salt. The presenter wi ...[text shortened]... then change your selection, you are more likely to achieve the goal. It is not a 50/50 chance.
05 Aug 08
Originally posted by PsychoPawnActualy it isn´t even an assumption. Shav said in the original post that the presenter opens a door containing salt. To guarantee that, the presenter must know what is behind each door.
The only assumption is that the presenter will never open the door with the car in it when you haven't made your final selection. That's not an invalid assumption since that's exactly the way the show "Let's make a deal" worked - the host was Monty Hall and hence why it's called the Monty Hall problem.
The game worked by the contestant choosing one doo ...[text shortened]... ned and it is unintuitive to most. That's why so many people got goats on the show 🙂
05 Aug 08
1 edit
Originally posted by WheelyShav described a problem so incredibly similar to the famous Monty Hall problem that this is what's being discussed. Shav says that one's odds improve if the door is changed, which also suggests it's the Monty Hall problem.
Actualy it isn´t even an assumption. Shav said in the original post that the presenter opens a door containing salt. To guarantee that, the presenter must know what is behind each door.
Yeah, Shav worded the problem badly, I'll give you that.
Without the assumption that the presenter intends to show salt, Shav's claim that odds improve with changing is false. Anyone want to prove that one?
05 Aug 08
Originally posted by AThousandYoungAs far as I can tell the odds are 2/3 that you got it wrong to start with and then, given that the presenter shows you another wrong one (not an assumption from shav´s text, he tells you that you got shown the other one) That leaves you with a 2/3 chance of getting the right one if you switch. As the original reply says, you now have more information. a) You now can eliminate one door and b) you know you are likely to have selected the wrong door too.
Shav described a problem so incredibly similar to the famous Monty Hall problem that this is what's being discussed. Shav says that one's odds improve if the door is changed, which also suggests it's the Monty Hall problem.
Yeah, Shav worded the problem badly, I'll give you that.
Without the assumption that the presenter intends to show salt, Shav's claim that odds improve with changing is false. Anyone want to prove that one?
05 Aug 08
Originally posted by WheelyI suppose you're right.
As far as I can tell the odds are 2/3 that you got it wrong to start with and then, given that the presenter shows you another wrong one (not an assumption from shav´s text, he tells you that you got shown the other one) That leaves you with a 2/3 chance of getting the right one if you switch. As the original reply says, you now have more information. a) You now can eliminate one door and b) you know you are likely to have selected the wrong door too.
05 Aug 08
Originally posted by WheelyThis is because you make the assumption that the presenter *knows* what's behind the doors. I do not read that in the original text.
The presenter knows which door is which (otherwise he couldn´t guarantee showing you a door with the salt)
"Before it's opened, the presenter opens a different door showing a sack of salt and asks"...
I read: the presenter opens a door, and there is salt. Nothing else. There is no hint that he knows what's behind the doors.
Making assumptions you could think that the presenter will only show you a door because he knows your first try is a good one, to mislead you to the third door... or just the opposite. So what do you prove? Then it is obvious that the odds change, because the presenter is actually telling or hinting towards something. So bvious that it wouldn't even be worth a discussion...
05 Aug 08
Originally posted by CeSingeI think Wheely has the best understanding of this problem and this thread of anyone here. As far as I can tell he's complety right.
This is because you make the assumption that the presenter *knows* what's behind the doors. I do not read that in the original text.
"Before it's opened, the presenter opens a different door showing a sack of salt and asks"...
I read: the presenter opens a door, and there is salt. Nothing else. There is no hint that he knows what's behind the doors.
...[text shortened]... ing or hinting towards something. So bvious that it wouldn't even be worth a discussion...
06 Aug 08
Originally posted by AThousandYoungIf the presenter opens a random door, and you switch
Shav described a problem so incredibly similar to the famous Monty Hall problem that this is what's being discussed. Shav says that one's odds improve if the door is changed, which also suggests it's the Monty Hall problem.
Yeah, Shav worded the problem badly, I'll give you that.
Without the assumption that the presenter intends to show salt, Shav's claim that odds improve with changing is false. Anyone want to prove that one?
1/3 : you choose correctly, switch, and lose
2/3 * 1/2 = 1/3: you choose incorrectly, the other salt door is opened, you win
2/3 * 1/2 = 1/3: you choose incorrectly, the porshe door is opened, you lose
therefore: 1/3 chance of winning by switching
if you don't switch:
1/3 : you choose correctly and win
2/3 * 1/2 = 1/3 : you choose incorrectly, the salt door is opened, you lose
2/3 * 1/2 = 1/3 : you choose incorrectly, the porshe door is opened, you lose
The odds are the same without that assumption.
06 Aug 08
Originally posted by forkedknightGiven that this is a well known question, we know (ATY has the link) that it isn't, in fact, a random choice the presenter makes.
If the presenter opens a random door, and you switch
1/3 : you choose correctly, switch, and lose
2/3 * 1/2 = 1/3: you choose incorrectly, the other salt door is opened, you win
2/3 * 1/2 = 1/3: you choose incorrectly, the porshe door is opened, you lose
therefore: 1/3 chance of winning by switching
if you don't switch:
1/3 : you choose co ...[text shortened]... orrectly, the porshe door is opened, you lose
The odds are the same without that assumption.
Even if this wasn't the case, Shav tells us that the presenter opened a door that had the salt behind it. At that point, you have a 2/3 chance of winning if you switch and a 1/3 chance of winning if you don't. This is simply because you probably chose the wrong door to start with and you know there is only one other door that could contain the porche.
06 Aug 08
Originally posted by WheelyNope.
Given that this is a well known question, we know (ATY has the link) that it isn't, in fact, a random choice the presenter makes.
Even if this wasn't the case, Shav tells us that the presenter opened a door that had the salt behind it. At that point, you have a 2/3 chance of winning if you switch and a 1/3 chance of winning if you don't. This is simply be ...[text shortened]... g door to start with and you know there is only one other door that could contain the porche.
If the door was opened randomly, then it's 50/50. In the original problem, the only reason why the odds increase for switching is that you know the presenter would not open the door with the car.
06 Aug 08
3 edits
Originally posted by AThousandYoungIf the door is opened randomly then the possibilities are:
Without the assumption that the presenter intends to show salt, Shav's claim that odds improve with changing is false. Anyone want to prove that one?
A: door with car
B and C: doors with salt
P(choose A) = 1/3
P(open B|A) = 1/2
P(open C|A) = 1/2
P(choose B) = 1/3
P(open A|B) = 1/2 *
P(open C|B) = 1/2
P(choose C) = 1/3
P(open B|C) = 1/2
P(open A|C) = 1/2 *
We know the door opened was not A, so we rule out the cases with *. The unconditional odds of these cases sum up to 1/3, so the conditional odds on not switching are the unconditional ones divided by 2/3.
P(choose A|salt) = (1/3) / (2/3) = 1/2
------------------
In the original case you have:
P(choose A) = 1/3
P(open B|A) = 1/2
P(open C|A) = 1/2
P(choose B) = 1/3
P(open A|B) = 0 *
P(open C|B) = 1
P(choose C) = 1/3
P(open B|C) = 1
P(open A|C) = 0 *
So the odds of those cases sum up to 0, so the opening of the door does not change the conditional odds of winning with your initial choice (1/3).
06 Aug 08
Originally posted by PalynkaYou guys are being far to technical with all this when actually the reason why you need to switch your choice is obvious when you think about it.. We're not talking about winning with your initial choice which is, as you say 1/3. We are talking about winning in the end.
If the door is opened randomly then the possibilities are:
A: door with car
B and C: doors with salt
P(choose A) = 1/3
P(open B|A) = 1/2
P(open C|A) = 1/2
P(choose B) = 1/3
P(open A|B) = 1/2 *
P(open C|B) = 1/2
P(choose C) = 1/3
P(open B|C) = 1/2
P(open A|C) = 1/2 *
We know the door opened was not A, so we rule out the cases w ...[text shortened]... ning of the door does not change the conditional odds of winning with your initial choice (1/3).
If we take the specific example in this thread the presenter shows us the salt door. There is therefore no point in discussion whether this was arrived at randomly or not. In the story as told by Shav, the situation is exactly the same as if the presenter intentionally shows us the salt door and knows what's behind each door.