[Brain Teaser] Three doors

[hedgy73]

Does the game master have to give you the chance to change your mind? Does he want the contestant to win or lose?

Yes he does. His intention is unknown and should be irrelevant.

Ok thanks, then I'll say there should be no obvious explanation that there would be an advantage switching as it would be 50/50.

[matt4054]

Ok thanks, then I'll say there should be no obvious explanation that there would be an advantage switching as it would be 50/50.

It's one of the most counterintuitive problem I've ever seen.

Some have already given elements of the answer. I will add the following hints:

The game master will always have to open 1 of the 2 doors that you didn't pick. Now you may want to know if he will always open a losing door or not (supposing you didn't pick the winning one in the first place). I will answer that opening the winning door would make the game over instantly, and the switch-or-not question irrelevant. As far as I'm concerned, the key to the explanation was grouping the 2 unpicked doors together when considering probabilities.

[(oYo)]

1) Is there an advantage to choosing "No" or "It doesn't matter" in the poll?
2) Is there a 50/50 chance those answers both mean the same thing?
3) Is there no advantage, since there's 100% absolutely no difference in their meaning?

Having posted 3 questions above, is there a 0%, 33%, 50%, 66%, or 100% chance that one of them is a logical question?

[matt4054]

1) Is there an advantage to choosing "No" or "It doesn't matter" in the poll?
2) Is there a 50/50 chance those answers both mean the same thing?
3) Is there no advantage, since there's 100% absolutely no difference in their meaning?

Having posted 3 questions above, is there a 0%, 33%, 50%, 66%, or 100% chance that one of them is a logical question?

I find them very logical indeed, it's all about the law of excluded middle

I admit that there's an ambiguity in the formulation between something that is not to your advantage and something that is to your disadvantage.

[hedgy73]

I'm sorry but I have no idea to the question but will stay tuned for an answer.

I'll go to bed thinking about this tonight....

[hedgy73]

1) Is there an advantage to choosing "No" or "It doesn't matter" in the poll?
2) Is there a 50/50 chance those answers both mean the same thing?
3) Is there no advantage, since there's 100% absolutely no difference in their meaning?

Having posted 3 questions above, is there a 0%, 33%, 50%, 66%, or 100% chance that one of them is a logical question?

I find them very logical indeed, it's all about the law of excluded middle

I admit that there's an ambiguity in the formulation between something that is not to your advantage and something that is to your disadvantage.

Christ that's complex looking forward to the answer

[ndnh]

1) Is there an advantage to choosing "No" or "It doesn't matter" in the poll?
2) Is there a 50/50 chance those answers both mean the same thing?
3) Is there no advantage, since there's 100% absolutely no difference in their meaning?

1) No advantage. Both are apparently wrong.

2) They don't mean the same thing. No means you would NOT switch. It doesn't matter means, whether you switch or not it has got the same odds.
In other words: No applies whenever, odd of winning after not switching are more or equal to the other.
It doesn't matter, applies only when they are equal, i.e. 50/50

3) What advantage are you talking about? Both options are not mutually exclusive, but they certainly does not mean the same.

[patt0]

My initial choice has a 33% chance of being right. (You could also say there's a 66% chance of choosing the wrong door.) That's a gimme, since there are initially 3 doors to choose from.

If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one.  Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.

This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.

You should switch to the other door. xD
I don't want to spoil but you can't ignore the 33% chance you had in the beginning. That didn't change to 50%. Those are different things.

[Astargath]

My initial choice has a 33% chance of being right. (You could also say there's a 66% chance of choosing the wrong door.) That's a gimme, since there are initially 3 doors to choose from.

If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one.  Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.

This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.

You should switch to the other door. xD
I don't want to spoil but you can't ignore the 33% chance you had in the beginning. That didn't change to 50%. Those are different things.

OyO is wrong, the chances of winning are 66% if you switch, it has been explained already with this example:

If the red door is the good door you have 3 possibilities:
1 pick red door and lose if you switch
2 pick yellow door and win if you switch
3 pick green door and win if you switch

Win 2/3 and lose 1/3 so 66%

[patt0]

My initial choice has a 33% chance of being right. (You could also say there's a 66% chance of choosing the wrong door.) That's a gimme, since there are initially 3 doors to choose from.

If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one.  Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.

This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.

You should switch to the other door. xD
I don't want to spoil but you can't ignore the 33% chance you had in the beginning. That didn't change to 50%. Those are different things.

OyO is wrong, the chances of winning are 66% if you switch, it has been explained already with this example:

If the red door is the good door you have 3 possibilities:
1 pick red door and lose if you switch
2 pick yellow door and win if you switch
3 pick green door and win if you switch

Win 2/3 and lose 1/3 so 66%

Yes and that explanation was posted before his answer and he didn't read it or didn't care lol. So I was using his answer to explain where he went wrong.
By the way, is Bardman your alt account then? xD

[D4C]

Fuck btc
I want my goat

[jmintuck]

I read the explanation on wikipedia a while back. Still doesn't feel right, and I will never have an intuitive understanding of it I suppose.
I think of it like dice at times , and its like saying that select a range between 1 and 100, between 1-33,34-67,68-100(dont worry about the differnce of 1, just consider it 1/3) . Now you select 1-33. Dice site owner checks seed to tell you its not 68-100. Then how does switching to 34-67 increase odds ?

There is an example to make it easier: You have 3 doors, Red Yellow and Green.

In the case that the Bitcoin us behind the red door:
If you pick the red door switching would make you lose
If you pick the yellow door switching would make you win
If you pick the green door switching would make you win

Lose 1/3
Win 2/3

Thanks that does clear it up. But with the knowledge of this, the game show host might trick you into a trap , and you may actually lose, if you had selected the red from the start.

[sdp]

I would agree that oYo.  I have to agree with this analysis.  Once you know that one of the other doors is worthless you have a 50% chance of having picked the right door.  Just like if or when you find out you lost, you can say you have a 0% chance of having picked the right door, regardless of how good your chances were when you didn't have that knowledge.

In other words, it doesn't matter whether you switch or not.

sdp

[patt0]

I would agree that oYo.  I have to agree with this analysis.  Once you know that one of the other doors is worthless you have a 50% chance of having picked the right door.  Just like if or when you find out you lost, you can say you have a 0% chance of having picked the right door, regardless of how good your chances were when you didn't have that knowledge.

In other words, it doesn't matter whether you switch or not.

sdp

Lol. Maybe this is better to help see why that is wrong: imagine you have 1000 doors instead of 3.
You pick 1 out of 1000, and the host opens 998 doors that don't have the prize. Would you change to the door he left then or keep the same? xD

[criptix]

I would agree that oYo.  I have to agree with this analysis.  Once you know that one of the other doors is worthless you have a 50% chance of having picked the right door.  Just like if or when you find out you lost, you can say you have a 0% chance of having picked the right door, regardless of how good your chances were when you didn't have that knowledge.

In other words, it doesn't matter whether you switch or not.

sdp

Lol. Maybe this is better to help see why that is wrong: imagine you have 1000 doors instead of 3.
You pick 1 out of 1000, and the host opens 998 doors that don't have the prize. Would you change to the door he left then or keep the same? xD

I think right now its not about mathematical probalities but about the game show.
if the moderator opens 1 out of the three (998 out of 1000) and we know for sure it is not the prize then only two doors remain.
1 out of 2 is 50%

Like you would not need to pick at all in the first place

[patt0]

^ I thought with more doors would be more intuitive. : /
With 1000 doors, when you choose a door in the beginning you only have 0.1% probability of choosing the right one. The game host knows where the prize is and opens every door except the one you choose and another one.
But your choice was still made with a 0.1% chance. It doesn't change to 50% now, because that would be losing information. You could only say it was 50% chance if you started from 2 doors, not 1000.

[criptix]

^ I thought with more doors would be more intuitive. : /
With 1000 doors, when you choose a door in the beginning you only have 0.1% probability of choosing the right one. The game host knows where the prize is and opens every door except the one you choose and another one.
But your choice was still made with a 0.1% chance. It doesn't change to 50% now, because that would be losing information. You could only say it was 50% chance if you started from 2 doors, not 1000.

yes, that was exactly what i was trying to point out.

people think of it as there would be only 2 doors to choose from and then logically the probality would be 50%

[Xprim777]

I heard the solution in the film "Las Vegas 21"

[(oYo)]

There are two separate instances (equations) here, of which you are trying to represent as one.

The first round consisted of a 1/3 chance, on account there were 3 doors. That round ended with no winners, but one of the doors was shown to be a loser.

The next round essentially only had two doors to choose from, ergo there was a 1/2 chance to choose the right door.

Stop trying to insist that the losing door in the first round is even a legitimate consideration anymore. There is a 0% percent chance that that door (or another 997 just like it) is the right choice now, as it has been proven to be the wrong one. It is no longer an option, now. In this new round (equation), consideration is only regarded for two brand new choices essentially. Implying there are any other choices or odds is simply misleading.

So, the correct answer to the poll's question should be "No, it doesn't matter." "No" there is no advantage to switching, nor is there any disadvantage, as "it doesn't matter" which door you pick, since they both have a 50% chance of winning, now.

 TL;DR - Ok, here's a great graphic to help make my point. Pick the "not a goat" door. Here's 3 doors to choose from. There's a goat behind 2 of them. Door number 3 absolutely has a goat and one of the other 2 doesn't. Picking door 3 is an automatic loss, so if you want a (50%) chance to win, you will only pick door 1 or door 2.

[matt4054]

Thanks for your contributions!

While most of you already knew about the Monty Hall problem, it's always interesting to discuss, especially how to turn it into something more intuitive. The meta problem of the difference between 'No' and 'It doesn't matter' was an interesting development too