The thread's only seven pages long...yes, it's been mentioned, briefly. It doesn't seem like a brain teaser to me, either, but I guess maybe I'm in the minority there since it keeps coming up.
The thread's only seven pages long...yes, it's been mentioned, briefly. It doesn't seem like a brain teaser to me, either, but I guess maybe I'm in the minority there since it keeps coming up.
Yeah, I didn't know how to categorize it either, but here seems to be as good a place as any.
The best way to understand the Monty Hall problem is to actually do it.
Get a deck of cards, pull out two Jokers and an Ace. The Jokers are the empty doors and the Ace is the prize door. Do it 30 times, changing doors every time. Then do it 30 times, sticking with the same door every time.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to but the odds are in reality not different. Its just the word play makes it appear to be.
The best way to understand the Monty Hall problem is to actually do it.
Get a deck of cards, pull out two Jokers and an Ace. The Jokers are the empty doors and the Ace is the prize door. Do it 30 times, changing doors every time. Then do it 30 times, sticking with the same door every time.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to but the odds are in reality not different. Its just the word play makes it appear to be.
If you change doors, the odds really are better. I'll explain:
I make one assumption: When the host/guy/whatever reveals one door and asks to switch, he never reveals the winning door.
Case 1: You do not switch doors.
If you don't switch doors, your odds of winning are the same before the reveal as after. Therefore you have a 1/3 chance of picking the correct door and 2/3 chance of picking the wrong one. Your chances of winning are thus 1 in 3.
Case 2: You DO change doors.
If you do change doors, your odds of winning are different. When you picked the door originally, you had a
1/3 chance of picking the right door and 2/3 chance of picking the wrong one.
sub-case a: Originally picked the correct door
If you originally picked the right door, and you switch you therefore lose as you have the correct door and switched it away. Your chance of this happening is 1 in 3 as you have a 1/3 chance of picking this door at the start.
sub-case b: Originally picked the incorrect door
If you originally picked the wrong door, when the host reveals one door and asks to switch He never reveals the correct prize winning door. Therefore because of this, the remaining door to switch to is the correct one to win the prize. So therefore as you have a 2/3 chance of picking the wrong door at the start, and switching with the wrong door gets you the prize, If you switch doors you have a 2 in 3 chance of winning the prize and a 1 in 3 chance of not.
So basically you always switch doors as it gives you a 2/3 chance of winning instead of a 1/3.
The best way to understand the Monty Hall problem is to actually do it.
Get a deck of cards, pull out two Jokers and an Ace. The Jokers are the empty doors and the Ace is the prize door. Do it 30 times, changing doors every time. Then do it 30 times, sticking with the same door every time.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to but the odds are in reality not different. Its just the word play makes it appear to be.
You have 3 doors. You choose Door 1. With this door you have a 33% chance of having the car. Logically then, there is a 67% chance that the car is behind one of the other 2 doors.
Critical things you know at this point: There is guaranteed to be at least 1 goat in the set of 2 doors that you did not pick. Your odd of car: 33%. Your odds of no car: 67%.
Monty Hall then reveals a goat from the set of 2 doors you didn't pick. Your information is exactly the same as it once was. You already knew that the second set contained at least one goat.
You switch to the door you didn't originally pick. Why? Because doing so effectively allowed you to choose 2 doors.
It's exactly the same as you picking door 1 and Monty then asking "would you rather have whichever prize is the best of doors 2 and 3?"
The best way to understand the Monty Hall problem is to actually do it.
Get a deck of cards, pull out two Jokers and an Ace. The Jokers are the empty doors and the Ace is the prize door. Do it 30 times, changing doors every time. Then do it 30 times, sticking with the same door every time.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to but the odds are in reality not different. Its just the word play makes it appear to be.
You have 3 doors. You choose Door 1. With this door you have a 33% chance of having the car. Logically then, there is a 67% chance that the car is behind one of the other 2 doors.
Critical things you know at this point: There is guaranteed to be at least 1 goat in the set of 2 doors that you did not pick. Your odd of car: 33%. Your odds of no car: 67%.
Monty Hall then reveals a goat from the set of 2 doors you didn't pick. Your information is exactly the same as it once was. You already knew that the second set contained at least one goat.
You switch to the door you didn't originally pick. Why? Because doing so effectively allowed you to choose 2 doors.
It's exactly the same as you picking door 1 and Monty then asking "would you rather have whichever prize is the best of doors 2 and 3?"
I assume that you were trying to refute PantsB's statement but you didn't manage to do it.
When the 3rd door is opened to reveal a goat, you are presented with a choice: Door A or Door B. The odds at this point of selecting correctly are 50/50. Selecting Door A or Door B has an equivalent chance of netting you victory regardless of whether you chose Door A or Door B the first time. If you changed it to 1000 doors, of which 998 are revealed as losers after your initial pick, you still have 50/50 odds of success by sticking with the door you picked the first time.
It's not that hard, you gain new information, let me present it to you like this:
X X X
There are your doors. Whichever one you pick, let's regroup:
P (X X)
You would agree that there is a 1/3 chance P is the car, while there is a 2/3 chance one of the two X's is the car, right?
Watch what happens when we remove an X
P ([strike]X[/strike] X)
Did the value of the probability that it was in one of the two X's change? Nope.
Switching is 2/3.
Switching is 2/3, but so is sticking. You are making a second selection from a pool of choices which, at time of selection, includes your initial choice. Choosing Door A vs. Door B after Door C is eliminated ignores whether you chose Door A or Door B the first time.
The best way to understand the Monty Hall problem is to actually do it.
Get a deck of cards, pull out two Jokers and an Ace. The Jokers are the empty doors and the Ace is the prize door. Do it 30 times, changing doors every time. Then do it 30 times, sticking with the same door every time.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to but the odds are in reality not different. Its just the word play makes it appear to be.
You have 3 doors. You choose Door 1. With this door you have a 33% chance of having the car. Logically then, there is a 67% chance that the car is behind one of the other 2 doors.
Critical things you know at this point: There is guaranteed to be at least 1 goat in the set of 2 doors that you did not pick. Your odd of car: 33%. Your odds of no car: 67%.
Monty Hall then reveals a goat from the set of 2 doors you didn't pick. Your information is exactly the same as it once was. You already knew that the second set contained at least one goat.
You switch to the door you didn't originally pick. Why? Because doing so effectively allowed you to choose 2 doors.
It's exactly the same as you picking door 1 and Monty then asking "would you rather have whichever prize is the best of doors 2 and 3?"
I assume that you were trying to refute PantsB's statement but you didn't manage to do it.
When the 3rd door is opened to reveal a goat, you are presented with a choice: Door A or Door B. The odds at this point of selecting correctly are 50/50. Selecting Door A or Door B has an equivalent chance of netting you victory regardless of whether you chose Door A or Door B the first time. If you changed it to 1000 doors, of which 998 are revealed as losers after your initial pick, you still have 50/50 odds of success by sticking with the door you picked the first time.
You are mathematically wrong. It's set selection. I choose one door I have a 1/3 chance of winning. I choose the other 2 doors I have a 2/3 chance of winning. Revealing that one of the 2 doors is a goat doesn't change that.
The best way to understand the Monty Hall problem is to actually do it.
Get a deck of cards, pull out two Jokers and an Ace. The Jokers are the empty doors and the Ace is the prize door. Do it 30 times, changing doors every time. Then do it 30 times, sticking with the same door every time.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to but the odds are in reality not different. Its just the word play makes it appear to be.
You have 3 doors. You choose Door 1. With this door you have a 33% chance of having the car. Logically then, there is a 67% chance that the car is behind one of the other 2 doors.
Critical things you know at this point: There is guaranteed to be at least 1 goat in the set of 2 doors that you did not pick. Your odd of car: 33%. Your odds of no car: 67%.
Monty Hall then reveals a goat from the set of 2 doors you didn't pick. Your information is exactly the same as it once was. You already knew that the second set contained at least one goat.
You switch to the door you didn't originally pick. Why? Because doing so effectively allowed you to choose 2 doors.
It's exactly the same as you picking door 1 and Monty then asking "would you rather have whichever prize is the best of doors 2 and 3?"
I assume that you were trying to refute PantsB's statement but you didn't manage to do it.
When the 3rd door is opened to reveal a goat, you are presented with a choice: Door A or Door B. The odds at this point of selecting correctly are 50/50. Selecting Door A or Door B has an equivalent chance of netting you victory regardless of whether you chose Door A or Door B the first time. If you changed it to 1000 doors, of which 998 are revealed as losers after your initial pick, you still have 50/50 odds of success by sticking with the door you picked the first time.
You are mathematically wrong. It's set selection. I choose one door I have a 1/3 chance of winning. I choose the other 2 doors I have a 2/3 chance of winning. Revealing that one of the 2 doors is a goat doesn't change that.
If there are 3 doors and I pick one, I have 1/3 odds that I picked right.
If there are 3 doors and you tell me that Door C is a loser, I have 2/3 odds of picking correctly.
It does not matter if, before you told me not to pick Door C, I wanted to pick Door A.
Edit: The scenario makes it appear that you are getting to pick twice. This is false. You only get to pick once, after the goat is revealed. The initial pick is meaningless. It's equivalent to the setup:
Monty: Here are two doors, A and B. Which do you want?
Alice: Hmm... I think maybe A...
Monty: Well, by the way, there's a goat stage left.
Goat: Blaaa.
Alice: Huh. Okay, I'll go with B.
CptHamilton on
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MelokuAsk me about my IllusionsRegistered Userregular
edited November 2009
No, it doesn't.
You'd be right if you selected at random after all the doors were opened except for one, but that's a different problem.
I mean, this really lends itself to proof by exhaustion.
A B C
A contains the car, you pick A, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
That's it. That's all the possible cases. In two of them you win, and in one you lose. Ergo, 2/3 chance of winning if you switch, 1/3 chance of winning if you stay.
It's not that hard, you gain new information, let me present it to you like this:
X X X
There are your doors. Whichever one you pick, let's regroup:
P (X X)
You would agree that there is a 1/3 chance P is the car, while there is a 2/3 chance one of the two X's is the car, right?
Watch what happens when we remove an X
P ([strike]X[/strike] X)
Did the value of the probability that it was in one of the two X's change? Nope.
Switching is 2/3.
Switching is 2/3, but so is sticking. You are making a second selection from a pool of choices which, at time of selection, includes your initial choice. Choosing Door A vs. Door B after Door C is eliminated ignores whether you chose Door A or Door B the first time.
No, staying is 1/3. Why? Because youre set only has 1 door out of 3. Switching nets you 2 doors out of 3. It's the reveal that is confusing you. Pretend it's not there and think about the problem again.
It's not that hard, you gain new information, let me present it to you like this:
X X X
There are your doors. Whichever one you pick, let's regroup:
P (X X)
You would agree that there is a 1/3 chance P is the car, while there is a 2/3 chance one of the two X's is the car, right?
Watch what happens when we remove an X
P ([strike]X[/strike] X)
Did the value of the probability that it was in one of the two X's change? Nope.
Switching is 2/3.
Switching is 2/3, but so is sticking. You are making a second selection from a pool of choices which, at time of selection, includes your initial choice. Choosing Door A vs. Door B after Door C is eliminated ignores whether you chose Door A or Door B the first time.
No, staying is 1/3. Why? Because youre set only has 1 door out of 3. Switching nets you 2 doors out of 3. It's the reveal that is confusing you. Pretend it's not there and think about the problem again.
I understand the math perfectly well, but you don't actually get to pick twice.
You get what's behind your final pick. If the prize was in Door A and you switched to Door B, you don't get your initial pick of Prize Door A.
The best way to understand the Monty Hall problem is to actually do it.
Get a deck of cards, pull out two Jokers and an Ace. The Jokers are the empty doors and the Ace is the prize door. Do it 30 times, changing doors every time. Then do it 30 times, sticking with the same door every time.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to but the odds are in reality not different. Its just the word play makes it appear to be.
You have 3 doors. You choose Door 1. With this door you have a 33% chance of having the car. Logically then, there is a 67% chance that the car is behind one of the other 2 doors.
Critical things you know at this point: There is guaranteed to be at least 1 goat in the set of 2 doors that you did not pick. Your odd of car: 33%. Your odds of no car: 67%.
Monty Hall then reveals a goat from the set of 2 doors you didn't pick. Your information is exactly the same as it once was. You already knew that the second set contained at least one goat.
You switch to the door you didn't originally pick. Why? Because doing so effectively allowed you to choose 2 doors.
It's exactly the same as you picking door 1 and Monty then asking "would you rather have whichever prize is the best of doors 2 and 3?"
I assume that you were trying to refute PantsB's statement but you didn't manage to do it.
When the 3rd door is opened to reveal a goat, you are presented with a choice: Door A or Door B. The odds at this point of selecting correctly are 50/50. Selecting Door A or Door B has an equivalent chance of netting you victory regardless of whether you chose Door A or Door B the first time. If you changed it to 1000 doors, of which 998 are revealed as losers after your initial pick, you still have 50/50 odds of success by sticking with the door you picked the first time.
You are mathematically wrong. It's set selection. I choose one door I have a 1/3 chance of winning. I choose the other 2 doors I have a 2/3 chance of winning. Revealing that one of the 2 doors is a goat doesn't change that.
If there are 3 doors and I pick one, I have 1/3 odds that I picked right.
If there are 3 doors and you tell me that Door C is a loser, I have 2/3 odds of picking correctly.
It does not matter if, before you told me not to pick Door C, I wanted to pick Door A.
Edit: The scenario makes it appear that you are getting to pick twice. This is false. You only get to pick once, after the goat is revealed. The initial pick is meaningless. It's equivalent to the setup:
Monty: Here are two doors, A and B. Which do you want?
Alice: Hmm... I think maybe A...
Monty: Well, by the way, there's a goat stage left.
Goat: Blaaa.
Alice: Huh. Okay, I'll go with B.
No, it's the same as Monty saying "there's a car behind one of these three doors, pick one." Then you pick door 1. Then he says "you can keep door one or you can switch to doors 2 and 3 and keep the car if it is behind either door."
You'd be right if you selected at random after all the doors were opened except for one, but that's a different problem.
I mean, this really lends itself to proof by exhaustion.
A B C
A contains the car, you pick A, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
That's it. That's all the possible cases. In two of them you win, and in one you lose. Ergo, 2/3 chance of winning if you switch, 1/3 chance of winning if you stay.
Your list of options isn't exhaustive.
A B C
A contains the car, you pick A, B opens, you switch, you're wrong.
A contains the car, you pick A, C opens, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
It's not that hard, you gain new information, let me present it to you like this:
X X X
There are your doors. Whichever one you pick, let's regroup:
P (X X)
You would agree that there is a 1/3 chance P is the car, while there is a 2/3 chance one of the two X's is the car, right?
Watch what happens when we remove an X
P ([strike]X[/strike] X)
Did the value of the probability that it was in one of the two X's change? Nope.
Switching is 2/3.
Switching is 2/3, but so is sticking. You are making a second selection from a pool of choices which, at time of selection, includes your initial choice. Choosing Door A vs. Door B after Door C is eliminated ignores whether you chose Door A or Door B the first time.
No, staying is 1/3. Why? Because youre set only has 1 door out of 3. Switching nets you 2 doors out of 3. It's the reveal that is confusing you. Pretend it's not there and think about the problem again.
I understand the math perfectly well, but you don't actually get to pick twice.
You get what's behind your final pick. If the prize was in Door A and you switched to Door B, you don't get your initial pick of Prize Door A.
And what you don't get is that your initial pick is important because it determines the sets. The reveal is not important. You're choosing door 1 creates these sets
(Door 1)
(Door 2, Door 3)
After the (meaningless) reveal your real choice is Set containing only door 1 or set containing doors 2 and 3. The odds of winning are better on the larger set. Every time.
The trick to the Monty Hall problem is that the host isn't actually providing any new information when he opens the door with nothing. Try looking at it this way.
At the beginning you choose door A. There's a 1/3 chance it's the right one, and a 2/3 chance it's not.
Of the remaining two doors, either one has nothing and one has the prize, or both have nothing. In other words, at least one of the other doors has nothing in it. So when the host reveals that one of the other doors has nothing it, it doesn't mean anything.
It doesn't make it any less likely that the prize is one of other doors. There's still a 1/3 chance of it being in door A, and 2/3 chance it's not.
You'd be right if you selected at random after all the doors were opened except for one, but that's a different problem.
I mean, this really lends itself to proof by exhaustion.
A B C
A contains the car, you pick A, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
That's it. That's all the possible cases. In two of them you win, and in one you lose. Ergo, 2/3 chance of winning if you switch, 1/3 chance of winning if you stay.
Your list of options isn't exhaustive.
A B C
A contains the car, you pick A, B opens, you switch, you're wrong.
A contains the car, you pick A, C opens, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
Choice 1 and 2 are the same!
You're ignoring that you've been given additional information and somehow are making this bizarre leap that your odds of picking the car change from 1/3 to 2/3 by magic. They don't.
You will be wrong initially 2/3 of the time.
Extend the problem. 100 doors.
You are 1/100 of getting it right initial try. There is a 99% chance that the car is in the other doors. 98 of those wrong doors are opened. Did most of that chance magically change to the door you picked? No!
enlightenedbum on
Self-righteousness is incompatible with coalition building.
You'd be right if you selected at random after all the doors were opened except for one, but that's a different problem.
I mean, this really lends itself to proof by exhaustion.
A B C
A contains the car, you pick A, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
That's it. That's all the possible cases. In two of them you win, and in one you lose. Ergo, 2/3 chance of winning if you switch, 1/3 chance of winning if you stay.
Your list of options isn't exhaustive.
A B C
A contains the car, you pick A, B opens, you switch, you're wrong.
A contains the car, you pick A, C opens, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
Those aren't the real cases. These are:
A contains the car, you pick A creating Set 1 (A), half of Set 2 (B, C) that contains a goat opens, you switch, you lose.
A contains the car, you pick B creating Set 1 (B), half of Set 2 (A, C) that contains a goat opens, you switch, you win.
A contains the car, you pick C creating Set 1 (C), half of Set 2 (A, that contains a goat opens, you switch, you win.
A contains the car, you pick A creating Set 1 (A), half of Set 2 (B, C) that contains a goat opens, you stay, you win.
A contains the car, you pick B creating Set 1 (B), half of Set 2 (A, C) that contains a goat opens, you stay, you lose.
A contains the car, you pick C creating Set 1 (C), half of Set 2 (A, that contains a goat opens, you stay, you lose.
That's 2 wins/1 loss when you switch and 1 win/2 losses when you stay. It has now been proven to you that switching is better than staying.
The crux of the monty hall problem is that it's 'Do you change?' not 'Which remaining door do you pick?'. If it were the latter, and they were randomly assigned, then it'd be 50/50. But as it's the former, 'do you change?', then you have to consider where you already are. And you're much more likely to already be on a losing one, thus changing is more likely to be successful.
A contains the car, you pick A, B opens, you switch, you're wrong.
A contains the car, you pick A, C opens, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
The first two cases do not have the same chance of occurring as the last two cases.
Feral on
every person who doesn't like an acquired taste always seems to think everyone who likes it is faking it. it should be an official fallacy.
If you've selected a losing door and you switch it will be a win, yes?
If you've selected a winning door and you stay it will be a win, yes?
The odds that you selected a losing door are greater than the odds that you selected a winning door. There is a 33% chance that you picked the winning door and a 66% chance that you picked a losing door. Switching wins in more cases than staying.
Hrm. I am uncertain why I didn't get that a page and a half ago. It's fairly obvious that for n doors there are n-1 possible courses of action if you switch and 1 if you don't. I even remember doing it in freshman math, now.
I must be having a really stupid day or something. Blah. Sorry for wasting the thread's time.
My problem with Monty Hall is not the math but the presentation. I get the math, as counter-intuitive as it is at first. The problem I have with it is treating "switching" and "staying" as the choices. If we approach the problem the other way:
There are two doors. One has a car, one has a goat. Choose one. Clearly it is a 1/2 probability.
Next element is "There was an additional door that had a goat before you were given this choice." OK. Doesn't change the final probability.
Now if we say you hit your head and don't remember which door you chose, its a 50-50 choice to pick the car or the goat. That's the only way the second decision becomes important.
Only if we remember which door we picked initially does it matter. And if we look at it that way it becomes "did you pick a goat?" not "do you switch choices?" Its not the switching doors that is important, its that you are switching binary states so that picking a goat initially becomes a good thing, not "changing your decision." Everyone does that problem and comes away with "you should switch" which isn't right. Its that you probably got the goat the first time, so go with that.
The slightly misleading presentation (which mimics how it is presented on the show) is half the point, to me. The Monty Hall Problem is my go-to retort when someone cites "common sense," so presenting the situation in a very natural way serves to prove my point.
A contains the car, you pick A, B opens, you switch, you're wrong.
A contains the car, you pick A, C opens, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
The first two cases do not have the same chance of occurring as the last two cases.
To elaborate: there is a one in three chance that you picked A and it is the winning door. Assuming the reveal is at random in that case, there is a one in six chance that you picked A, it is the winning door, and C opens. There is a one in six chance that you picked A, it is the winning door, and B opens. There is a one in three chance you picked A and B is the winning door. There is a one in three chance you picked A and C is the winning door.
If it helps you to understand, let's use a deck of cards:
I randomly write down a card on a piece of paper. I have you randomly draw a card out of the deck. I then go through the rest of the deck other than the card you drew, remove all of the cards except one, and tell you that the card I wrote down is either the one in your hand, or the one remaining from the deck, and you can pick which one you'd like to use. Would you say that the odds of winning are 50/50 regardless of whether you switched or not?
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JuliusCaptain of Serenityon my shipRegistered Userregular
My problem with Monty Hall is not the math but the presentation. I get the math, as counter-intuitive as it is at first. The problem I have with it is treating "switching" and "staying" as the choices. If we approach the problem the other way:
There are two doors. One has a car, one has a goat. Choose one. Clearly it is a 1/2 probability.
Next element is "There was an additional door that had a goat before you were given this choice." OK. Doesn't change the final probability.
Now if we say you hit your head and don't remember which door you chose, its a 50-50 choice to pick the car or the goat. That's the only way the second decision becomes important.
Only if we remember which door we picked initially does it matter. And if we look at it that way it becomes "did you pick a goat?" not "do you switch choices?" Its not the switching doors that is important, its that you are switching binary states so that picking a goat initially becomes a good thing, not "changing your decision." Everyone does that problem and comes away with "you should switch" which isn't right. Its that you probably got the goat the first time, so go with that.
The point of the Monty Hall problem isn't really the math, it's how bad at it we actually are. "Changing your decision" is probably the reason why so many people get it wrong the first time. The math is easy to grasp, it's just that people don't see it in these cases.
So the spin on the lazy susan isn't random? It's just rotating one position clockwise/ccw?
The spin on the lazy Susan is random. I framed the problem so that a shot glasses always ends up directly in front of you, and the website I linked framed the problem so that the shot glasses always end up [two in front, two in back], but that detail isn't really important. There are four discrete rotations the lazy Susan can take, and the specific rotation chosen each turn is random.
So surely you can't guarantee succeeding in any number of turns. I mean there's always a (small) chance that you'll just see the same two glasses in front of you for the first 2000 turns.
You can grab diagonally, so as long as you switch up between grabbing glasses next to each other and grabbing diagonally, no, there's not.
It's the reveal that screws people up because then, in their minds, they are choosing between 2 doors instead of choosing between 2 sets with different odds attached to them. Yeah, you saw a goat. Everybody saw a goat. And you know what else? Somewhere or another, you already knew that one of the 2 doors that you didn't pick was a goat - it is guaranteed.
So, I understand that island/logician problem now, but my issue with it and every other 'group of people on an island' scenario is not the logic involved, but more the assumptions of behavior. It's always implicit in those that the people behave in a certain way to solve the problem, which is never intuitive to me.
edit:: also yes, let's not do the .999 = 1 thing again, because it's not even a puzzle, it's a difference of definitions.
So the spin on the lazy susan isn't random? It's just rotating one position clockwise/ccw?
The spin on the lazy Susan is random. I framed the problem so that a shot glasses always ends up directly in front of you, and the website I linked framed the problem so that the shot glasses always end up [two in front, two in back], but that detail isn't really important. There are four discrete rotations the lazy Susan can take, and the specific rotation chosen each turn is random.
So surely you can't guarantee succeeding in any number of turns. I mean there's always a (small) chance that you'll just see the same two glasses in front of you for the first 2000 turns.
You can grab diagonally, so as long as you switch up between grabbing glasses next to each other and grabbing diagonally, no, there's not.
There's still a chance that you'll never lay your hands on one of the glasses, but since it's only one you can simply flip all the others to match it.
I was thinking about the glass swapping problem and I came to the same conclusion as Bama. Realistically there is a solution that is close to what was posted earlier (MrB... I lost the name) however there is no guarantee that you will ever touch every glass. The spinning of the lazy-susan is random, and grabbing diagonally and adjacently cannot solve the random spin (for example the time you grab diagonally the spin could have put those as two glasses you touched before, then next spin the adjacent two you have touched before etc.) - There could be that one glass that just doesn't get grabbed....
The probability of this happening is low, but it is likely enough that you could not consistently solve the puzzle in 10 moves or less
And switching the others to match wouldn't work either as you don't know if you have touched all the glasses or not.
Well if you know that you've touched three unique glasses and turned them all the same way, then you know that the fourth glass is turned the opposite way if the game doesn't end. The algorithm that was posted earlier should work.
I was thinking about the glass swapping problem and I came to the same conclusion as Bama. Realistically there is a solution that is close to what was posted earlier (MrB... I lost the name) however there is no guarantee that you will ever touch every glass. The spinning of the lazy-susan is random, and grabbing diagonally and adjacently cannot solve the random spin (for example the time you grab diagonally the spin could have put those as two glasses you touched before, then next spin the adjacent two you have touched before etc.) - There could be that one glass that just doesn't get grabbed....
The probability of this happening is low, but it is likely enough that you could not consistently solve the puzzle in 10 moves or less
And switching the others to match wouldn't work either as you don't know if you have touched all the glasses or not.
On the first turn, place two glasses on the diagonal face-down.
On the second turn, place two adjacent glasses face-down.
You now know which way all the glasses are facing, as either they're all down (and you've won) or you have three face-down (and you haven't won).
On the third turn, grab diagonal glasses. If one is up, turn it down and you win. If both are down, flip one over, and now you have two adjacent pairs of glasses.
On the fourth turn, grab adjacent glasses and flip them. Either they were the same and you win, or they were different and you have two diagonal pairs of glasses
On the fifth turn, grab diagonal glasses and flip them. As the two pairs are oriented diagonally, you will always be grabbing a pair and therefore will win either way.
I'm not sure if this is the same as was posted before as I only skimmed it, all I know is that it's more straightforward.
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SmasherStarting to get dizzyRegistered Userregular
I was thinking about the glass swapping problem and I came to the same conclusion as Bama. Realistically there is a solution that is close to what was posted earlier (MrB... I lost the name) however there is no guarantee that you will ever touch every glass. The spinning of the lazy-susan is random, and grabbing diagonally and adjacently cannot solve the random spin (for example the time you grab diagonally the spin could have put those as two glasses you touched before, then next spin the adjacent two you have touched before etc.) - There could be that one glass that just doesn't get grabbed....
The probability of this happening is low, but it is likely enough that you could not consistently solve the puzzle in 10 moves or less
And switching the others to match wouldn't work either as you don't know if you have touched all the glasses or not.
You're right that you're not guaranteed to ever touch all four glasses, but wrong about not knowing if you've touched the other three.
Partial solution to demonstrate why
Move 1: Grab two adjacent glasses and turn them both up.
Move 2: Grab two opposite glasses and turn them both up (though at least one will already be this way).
You will have grabbed three glasses by this point and they will now all be up. Either the fourth is up as well and you're done (which you'll know because the guy will tell you), or it is down and you know the state of all four glasses.
The keys to the puzzle are
1) Getting the puzzle into a state where it can be solved no matter which pair of glasses you get (given that you can choose whether to pick adjacent or opposite ones).
2) Being able to get the puzzle into that state even if you never touch one of the glasses (and yes, it's possible).
3) Knowing that you can guarantee you'll be able to control three glasses by grabbing both adjacent and opposite ones.
So, I understand that island/logician problem now, but my issue with it and every other 'group of people on an island' scenario is not the logic involved, but more the assumptions of behavior. It's always implicit in those that the people behave in a certain way to solve the problem, which is never intuitive to me.
That's why the problem specifies that everyone on the island is a perfect Logician, and that everyone knows everyone else is too.
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This solution is apocryphally attributed to Gauss as a child.
IOS Game Center ID: Isotope-X
The thread's only seven pages long...yes, it's been mentioned, briefly. It doesn't seem like a brain teaser to me, either, but I guess maybe I'm in the minority there since it keeps coming up.
Yeah, I didn't know how to categorize it either, but here seems to be as good a place as any.
A Battle of Wits
EDIT: Bonus! The clip is cut before the spoiler is revealed.
The Monty Hall problem is bullshit because its presenting things in a such a way as to make its not actually descriptive.
You pick 1 of 3 doors hoping to get a car (1/3 chance of being correct).
They reveal one incorrect door. At this point, you are given the option to change doors. Its presented that you now have a 1/2 chance to get it correct if you change doors. The way its presented, changing doors increases your odds to 1/2.
This is not accurate. Its the presentation of a incorrect door that changes the odds.
To use the card analogy, say you followed the directions until the first Joker is revealed. You then mixed the cards and picked a card at random from the two remaining. If you select the same card as before or the other one, each has an identical chance of being correct.
So in reality, you have the same odds of being correct by not changing doors as if you did change doors. You can choose A(remain on door A) or choose B(switch to
QEDMF xbl: PantsB G+
If you change doors, the odds really are better. I'll explain:
I make one assumption: When the host/guy/whatever reveals one door and asks to switch, he never reveals the winning door.
Case 1: You do not switch doors.
If you don't switch doors, your odds of winning are the same before the reveal as after. Therefore you have a 1/3 chance of picking the correct door and 2/3 chance of picking the wrong one. Your chances of winning are thus 1 in 3.
Case 2: You DO change doors.
If you do change doors, your odds of winning are different. When you picked the door originally, you had a
1/3 chance of picking the right door and 2/3 chance of picking the wrong one.
sub-case a: Originally picked the correct door
If you originally picked the right door, and you switch you therefore lose as you have the correct door and switched it away. Your chance of this happening is 1 in 3 as you have a 1/3 chance of picking this door at the start.
sub-case b: Originally picked the incorrect door
If you originally picked the wrong door, when the host reveals one door and asks to switch He never reveals the correct prize winning door. Therefore because of this, the remaining door to switch to is the correct one to win the prize. So therefore as you have a 2/3 chance of picking the wrong door at the start, and switching with the wrong door gets you the prize, If you switch doors you have a 2 in 3 chance of winning the prize and a 1 in 3 chance of not.
So basically you always switch doors as it gives you a 2/3 chance of winning instead of a 1/3.
You have 3 doors. You choose Door 1. With this door you have a 33% chance of having the car. Logically then, there is a 67% chance that the car is behind one of the other 2 doors.
Critical things you know at this point: There is guaranteed to be at least 1 goat in the set of 2 doors that you did not pick. Your odd of car: 33%. Your odds of no car: 67%.
Monty Hall then reveals a goat from the set of 2 doors you didn't pick. Your information is exactly the same as it once was. You already knew that the second set contained at least one goat.
You switch to the door you didn't originally pick. Why? Because doing so effectively allowed you to choose 2 doors.
It's exactly the same as you picking door 1 and Monty then asking "would you rather have whichever prize is the best of doors 2 and 3?"
X X X
There are your doors. Whichever one you pick, let's regroup:
P (X X)
You would agree that there is a 1/3 chance P is the car, while there is a 2/3 chance one of the two X's is the car, right?
Watch what happens when we remove an X
P ([strike]X[/strike] X)
Did the value of the probability that it was in one of the two X's change? Nope.
Switching is 2/3.
I assume that you were trying to refute PantsB's statement but you didn't manage to do it.
When the 3rd door is opened to reveal a goat, you are presented with a choice: Door A or Door B. The odds at this point of selecting correctly are 50/50. Selecting Door A or Door B has an equivalent chance of netting you victory regardless of whether you chose Door A or Door B the first time. If you changed it to 1000 doors, of which 998 are revealed as losers after your initial pick, you still have 50/50 odds of success by sticking with the door you picked the first time.
Switching is 2/3, but so is sticking. You are making a second selection from a pool of choices which, at time of selection, includes your initial choice. Choosing Door A vs. Door B after Door C is eliminated ignores whether you chose Door A or Door B the first time.
You are mathematically wrong. It's set selection. I choose one door I have a 1/3 chance of winning. I choose the other 2 doors I have a 2/3 chance of winning. Revealing that one of the 2 doors is a goat doesn't change that.
If there are 3 doors and I pick one, I have 1/3 odds that I picked right.
If there are 3 doors and you tell me that Door C is a loser, I have 2/3 odds of picking correctly.
It does not matter if, before you told me not to pick Door C, I wanted to pick Door A.
Edit: The scenario makes it appear that you are getting to pick twice. This is false. You only get to pick once, after the goat is revealed. The initial pick is meaningless. It's equivalent to the setup:
Monty: Here are two doors, A and B. Which do you want?
Alice: Hmm... I think maybe A...
Monty: Well, by the way, there's a goat stage left.
Goat: Blaaa.
Alice: Huh. Okay, I'll go with B.
You'd be right if you selected at random after all the doors were opened except for one, but that's a different problem.
I mean, this really lends itself to proof by exhaustion.
A B C
A contains the car, you pick A, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
That's it. That's all the possible cases. In two of them you win, and in one you lose. Ergo, 2/3 chance of winning if you switch, 1/3 chance of winning if you stay.
I understand the math perfectly well, but you don't actually get to pick twice.
You get what's behind your final pick. If the prize was in Door A and you switched to Door B, you don't get your initial pick of Prize Door A.
No, it's the same as Monty saying "there's a car behind one of these three doors, pick one." Then you pick door 1. Then he says "you can keep door one or you can switch to doors 2 and 3 and keep the car if it is behind either door."
Your list of options isn't exhaustive.
A B C
A contains the car, you pick A, B opens, you switch, you're wrong.
A contains the car, you pick A, C opens, you switch, you're wrong.
A contains the car, you pick B, C opens, you switch, you win.
A contains the car, you pick C, B opens, you switch, you win.
And what you don't get is that your initial pick is important because it determines the sets. The reveal is not important. You're choosing door 1 creates these sets
(Door 1)
(Door 2, Door 3)
After the (meaningless) reveal your real choice is Set containing only door 1 or set containing doors 2 and 3. The odds of winning are better on the larger set. Every time.
At the beginning you choose door A. There's a 1/3 chance it's the right one, and a 2/3 chance it's not.
Of the remaining two doors, either one has nothing and one has the prize, or both have nothing. In other words, at least one of the other doors has nothing in it. So when the host reveals that one of the other doors has nothing it, it doesn't mean anything.
It doesn't make it any less likely that the prize is one of other doors. There's still a 1/3 chance of it being in door A, and 2/3 chance it's not.
Choice 1 and 2 are the same!
You're ignoring that you've been given additional information and somehow are making this bizarre leap that your odds of picking the car change from 1/3 to 2/3 by magic. They don't.
You will be wrong initially 2/3 of the time.
Extend the problem. 100 doors.
You are 1/100 of getting it right initial try. There is a 99% chance that the car is in the other doors. 98 of those wrong doors are opened. Did most of that chance magically change to the door you picked? No!
Those aren't the real cases. These are:
A contains the car, you pick A creating Set 1 (A), half of Set 2 (B, C) that contains a goat opens, you switch, you lose.
A contains the car, you pick B creating Set 1 (B), half of Set 2 (A, C) that contains a goat opens, you switch, you win.
A contains the car, you pick C creating Set 1 (C), half of Set 2 (A,
A contains the car, you pick A creating Set 1 (A), half of Set 2 (B, C) that contains a goat opens, you stay, you win.
A contains the car, you pick B creating Set 1 (B), half of Set 2 (A, C) that contains a goat opens, you stay, you lose.
A contains the car, you pick C creating Set 1 (C), half of Set 2 (A,
That's 2 wins/1 loss when you switch and 1 win/2 losses when you stay. It has now been proven to you that switching is better than staying.
The first two cases do not have the same chance of occurring as the last two cases.
the "no true scotch man" fallacy.
If you've selected a winning door and you stay it will be a win, yes?
The odds that you selected a losing door are greater than the odds that you selected a winning door. There is a 33% chance that you picked the winning door and a 66% chance that you picked a losing door. Switching wins in more cases than staying.
I must be having a really stupid day or something. Blah. Sorry for wasting the thread's time.
There are two doors. One has a car, one has a goat. Choose one. Clearly it is a 1/2 probability.
Next element is "There was an additional door that had a goat before you were given this choice." OK. Doesn't change the final probability.
Now if we say you hit your head and don't remember which door you chose, its a 50-50 choice to pick the car or the goat. That's the only way the second decision becomes important.
Only if we remember which door we picked initially does it matter. And if we look at it that way it becomes "did you pick a goat?" not "do you switch choices?" Its not the switching doors that is important, its that you are switching binary states so that picking a goat initially becomes a good thing, not "changing your decision." Everyone does that problem and comes away with "you should switch" which isn't right. Its that you probably got the goat the first time, so go with that.
QEDMF xbl: PantsB G+
If it helps you to understand, let's use a deck of cards:
I randomly write down a card on a piece of paper. I have you randomly draw a card out of the deck. I then go through the rest of the deck other than the card you drew, remove all of the cards except one, and tell you that the card I wrote down is either the one in your hand, or the one remaining from the deck, and you can pick which one you'd like to use. Would you say that the odds of winning are 50/50 regardless of whether you switched or not?
The point of the Monty Hall problem isn't really the math, it's how bad at it we actually are. "Changing your decision" is probably the reason why so many people get it wrong the first time. The math is easy to grasp, it's just that people don't see it in these cases.
edit:: also yes, let's not do the .999 = 1 thing again, because it's not even a puzzle, it's a difference of definitions.
The probability of this happening is low, but it is likely enough that you could not consistently solve the puzzle in 10 moves or less
And switching the others to match wouldn't work either as you don't know if you have touched all the glasses or not.
On the first turn, place two glasses on the diagonal face-down.
On the second turn, place two adjacent glasses face-down.
You now know which way all the glasses are facing, as either they're all down (and you've won) or you have three face-down (and you haven't won).
On the third turn, grab diagonal glasses. If one is up, turn it down and you win. If both are down, flip one over, and now you have two adjacent pairs of glasses.
On the fourth turn, grab adjacent glasses and flip them. Either they were the same and you win, or they were different and you have two diagonal pairs of glasses
On the fifth turn, grab diagonal glasses and flip them. As the two pairs are oriented diagonally, you will always be grabbing a pair and therefore will win either way.
I'm not sure if this is the same as was posted before as I only skimmed it, all I know is that it's more straightforward.
You're right that you're not guaranteed to ever touch all four glasses, but wrong about not knowing if you've touched the other three.
Partial solution to demonstrate why
Move 2: Grab two opposite glasses and turn them both up (though at least one will already be this way).
You will have grabbed three glasses by this point and they will now all be up. Either the fourth is up as well and you're done (which you'll know because the guy will tell you), or it is down and you know the state of all four glasses.
2) Being able to get the puzzle into that state even if you never touch one of the glasses (and yes, it's possible).
3) Knowing that you can guarantee you'll be able to control three glasses by grabbing both adjacent and opposite ones.
That's why the problem specifies that everyone on the island is a perfect Logician, and that everyone knows everyone else is too.