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A confusion about the Monty Hall problem [duplicate]
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What's wrong with this equal probability solution for Monty Hall Problem?
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After reading the other question about the Monty Hall problem, and seeing the solution of it, before giving my perspective to the question, I would like to point out that, to explain the possible cases, we either fix the doors that I choose and Monty Hall opens, and permutate the values of the doors, i.e what is behind them, or we fix the values of the door, and permutate the door we and Monty Hall chooses.
Given that, let WLOG I choose the first door and Monty Hall opens the 3rd door, then we have the following cases;
1-> car
2-> goat
3-> goat
or
1-> goat
2-> car
3->goat
Since, we do know that the car is not behind the door 3.
So, there is only 2 option in one of them I win, and in the other I loose, hence the there is no difference in changing the door I chose first, hence the probability of the door 1 has the case is 1/2.
So my question is what exactly wrong with this argument ? Exactly which step of the argument fails to be true ?
Edit:
In this question, I was pointing out some of the problems that I general answer to Monty Hall problems contains, so it is different than the question that is claimed that this question to be duplicate of.
probability monty-hall
asked Jul 16 at 14:53
onurcanbektas
3,1061834
marked as duplicate by Strants, Xander Henderson, max_zorn, Trần Thúc Minh TrÃ, Parcly Taxel Jul 20 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 11 more comments
up vote
0
down vote
favorite
This question already has an answer here:
What's wrong with this equal probability solution for Monty Hall Problem?
8 answers
After reading the other question about the Monty Hall problem, and seeing the solution of it, before giving my perspective to the question, I would like to point out that, to explain the possible cases, we either fix the doors that I choose and Monty Hall opens, and permutate the values of the doors, i.e what is behind them, or we fix the values of the door, and permutate the door we and Monty Hall chooses.
Given that, let WLOG I choose the first door and Monty Hall opens the 3rd door, then we have the following cases;
1-> car
2-> goat
3-> goat
or
1-> goat
2-> car
3->goat
Since, we do know that the car is not behind the door 3.
So, there is only 2 option in one of them I win, and in the other I loose, hence the there is no difference in changing the door I chose first, hence the probability of the door 1 has the case is 1/2.
So my question is what exactly wrong with this argument ? Exactly which step of the argument fails to be true ?
Edit:
In this question, I was pointing out some of the problems that I general answer to Monty Hall problems contains, so it is different than the question that is claimed that this question to be duplicate of.
probability monty-hall
asked Jul 16 at 14:53
onurcanbektas
3,1061834
marked as duplicate by Strants, Xander Henderson, max_zorn, Trần Thúc Minh TrÃ, Parcly Taxel Jul 20 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What argument? What are you trying to show? Your notation is not self-explanatory...there aren't, for instance, any probabilities attached to it.
– lulu
Jul 16 at 14:55
Are you trying to say that after they show door 3 it is $50/50$?
– Holo
Jul 16 at 14:57
7
The fact that there are only two options does not mean that they are equally likely.
– lulu
Jul 16 at 14:57
1
"I have never claimed such a thing either." Yes, you did. When you claimed that there is no difference and therefore $50%$ chance of winning either way, that's exactly what you did.
– Arthur
Jul 16 at 15:00
1
I have purchases a lottery ticket. WLOG there are two outcomes: that I have purchased the only winning ticket, or that I have purchased one of the two million losing tickets. Therefore there is a fifty-fifty chance that I will win! Yah!
– Graham Kemp
Jul 16 at 15:23
 |Â
show 11 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
What's wrong with this equal probability solution for Monty Hall Problem?
8 answers
After reading the other question about the Monty Hall problem, and seeing the solution of it, before giving my perspective to the question, I would like to point out that, to explain the possible cases, we either fix the doors that I choose and Monty Hall opens, and permutate the values of the doors, i.e what is behind them, or we fix the values of the door, and permutate the door we and Monty Hall chooses.
Given that, let WLOG I choose the first door and Monty Hall opens the 3rd door, then we have the following cases;
1-> car
2-> goat
3-> goat
or
1-> goat
2-> car
3->goat
Since, we do know that the car is not behind the door 3.
So, there is only 2 option in one of them I win, and in the other I loose, hence the there is no difference in changing the door I chose first, hence the probability of the door 1 has the case is 1/2.
So my question is what exactly wrong with this argument ? Exactly which step of the argument fails to be true ?
Edit:
In this question, I was pointing out some of the problems that I general answer to Monty Hall problems contains, so it is different than the question that is claimed that this question to be duplicate of.
probability monty-hall
asked Jul 16 at 14:53
onurcanbektas
3,1061834
This question already has an answer here:
What's wrong with this equal probability solution for Monty Hall Problem?
8 answers
After reading the other question about the Monty Hall problem, and seeing the solution of it, before giving my perspective to the question, I would like to point out that, to explain the possible cases, we either fix the doors that I choose and Monty Hall opens, and permutate the values of the doors, i.e what is behind them, or we fix the values of the door, and permutate the door we and Monty Hall chooses.
Given that, let WLOG I choose the first door and Monty Hall opens the 3rd door, then we have the following cases;
1-> car
2-> goat
3-> goat
or
1-> goat
2-> car
3->goat
Since, we do know that the car is not behind the door 3.
So, there is only 2 option in one of them I win, and in the other I loose, hence the there is no difference in changing the door I chose first, hence the probability of the door 1 has the case is 1/2.
So my question is what exactly wrong with this argument ? Exactly which step of the argument fails to be true ?
Edit:
In this question, I was pointing out some of the problems that I general answer to Monty Hall problems contains, so it is different than the question that is claimed that this question to be duplicate of.
This question already has an answer here:
What's wrong with this equal probability solution for Monty Hall Problem?
8 answers
probability monty-hall
asked Jul 16 at 14:53
onurcanbektas
3,1061834
edited Jul 20 at 2:51
asked Jul 16 at 14:53
onurcanbektas
3,1061834
asked Jul 16 at 14:53
onurcanbektas
3,1061834
asked Jul 16 at 14:53
onurcanbektas
3,1061834
3,1061834
marked as duplicate by Strants, Xander Henderson, max_zorn, Trần Thúc Minh TrÃ, Parcly Taxel Jul 20 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Strants, Xander Henderson, max_zorn, Trần Thúc Minh TrÃ, Parcly Taxel Jul 20 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What argument? What are you trying to show? Your notation is not self-explanatory...there aren't, for instance, any probabilities attached to it.
– lulu
Jul 16 at 14:55
Are you trying to say that after they show door 3 it is $50/50$?
– Holo
Jul 16 at 14:57
7
The fact that there are only two options does not mean that they are equally likely.
– lulu
Jul 16 at 14:57
1
"I have never claimed such a thing either." Yes, you did. When you claimed that there is no difference and therefore $50%$ chance of winning either way, that's exactly what you did.
– Arthur
Jul 16 at 15:00
1
I have purchases a lottery ticket. WLOG there are two outcomes: that I have purchased the only winning ticket, or that I have purchased one of the two million losing tickets. Therefore there is a fifty-fifty chance that I will win! Yah!
– Graham Kemp
Jul 16 at 15:23
 |Â
show 11 more comments
What argument? What are you trying to show? Your notation is not self-explanatory...there aren't, for instance, any probabilities attached to it.
– lulu
Jul 16 at 14:55
Are you trying to say that after they show door 3 it is $50/50$?
– Holo
Jul 16 at 14:57
7
The fact that there are only two options does not mean that they are equally likely.
– lulu
Jul 16 at 14:57
1
"I have never claimed such a thing either." Yes, you did. When you claimed that there is no difference and therefore $50%$ chance of winning either way, that's exactly what you did.
– Arthur
Jul 16 at 15:00
1
I have purchases a lottery ticket. WLOG there are two outcomes: that I have purchased the only winning ticket, or that I have purchased one of the two million losing tickets. Therefore there is a fifty-fifty chance that I will win! Yah!
– Graham Kemp
Jul 16 at 15:23
What argument? What are you trying to show? Your notation is not self-explanatory...there aren't, for instance, any probabilities attached to it.
– lulu
Jul 16 at 14:55
What argument? What are you trying to show? Your notation is not self-explanatory...there aren't, for instance, any probabilities attached to it.
– lulu
Jul 16 at 14:55
Are you trying to say that after they show door 3 it is $50/50$?
– Holo
Jul 16 at 14:57
Are you trying to say that after they show door 3 it is $50/50$?
– Holo
Jul 16 at 14:57
7
7
The fact that there are only two options does not mean that they are equally likely.
– lulu
Jul 16 at 14:57
The fact that there are only two options does not mean that they are equally likely.
– lulu
Jul 16 at 14:57
1
1
"I have never claimed such a thing either." Yes, you did. When you claimed that there is no difference and therefore $50%$ chance of winning either way, that's exactly what you did.
– Arthur
Jul 16 at 15:00
"I have never claimed such a thing either." Yes, you did. When you claimed that there is no difference and therefore $50%$ chance of winning either way, that's exactly what you did.
– Arthur
Jul 16 at 15:00
1
1
I have purchases a lottery ticket. WLOG there are two outcomes: that I have purchased the only winning ticket, or that I have purchased one of the two million losing tickets. Therefore there is a fifty-fifty chance that I will win! Yah!
– Graham Kemp
Jul 16 at 15:23
I have purchases a lottery ticket. WLOG there are two outcomes: that I have purchased the only winning ticket, or that I have purchased one of the two million losing tickets. Therefore there is a fifty-fifty chance that I will win! Yah!
– Graham Kemp
Jul 16 at 15:23
 |Â
show 11 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
We have $$beginarrayc#1&mboxResult at #1&mboxresult otherwise\hline c&g&g&win&lose\hline g&c&g&lose&win\hline g&g&c&lose&winendarray$$We can't assume that the host open door $3$ because then you say "the third option(g,g,c) can't happen", what the host does is taking the column of $#2$ or $#3$ "away", but the results won't change.
Update:
You have to note that the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$.
But if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
answered Jul 16 at 15:05
Holo
4,2512629
Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen.
– onurcanbektas
Jul 16 at 15:09
1
@onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
– Holo
Jul 16 at 15:26
Well that was convincing :) I advise you to turn that comment into an answer.
– onurcanbektas
Jul 16 at 15:31
1
And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows.
– Graham Kemp
Jul 16 at 15:32
@onurcanbektas I added it to the answer
– Holo
Jul 16 at 15:32
add a comment |Â
up vote
8
down vote
You cannot do the "WLOG" and expect probabilities to stay equally distributed over your cases.
That is, you can say -- assuming that the car is behind a uniformly random door -- that you choose door 1, and then the three options are (car, goat, goat), (goat, car, goat) or (goat, goat, car) -- each with probability 1/3. That is fine.
However, when you then say "Without loss of generality, Monty opens door 3", you are excluding case 3, and implicitly treating it while you are treating case 2. If you continue you then have to add the probability of that situation occurring to the treatment of case 2. There are two options left: (car, goat, goat) and (goat, car, goat), but the first has probability 1/3, while the second has probability 2/3.
Edit: As a more general point, you seem to be using the common argument, "there are $n$ options, and therefore each occurs with probability $1/n$". This argument is dangerous: there's no reason to expect it to hold in general, but you often need it as an implicit assumption, because people forget to specify e.g. that a die or a coin is fair.
Consider the following similar argument: you throw two fair 6-sided dice, and your roll is the sum of the two numbers. The possible outcomes are $2,3,4,5,6,7,8,9,10,11,12$. You might be tempted to conclude that, since there are 11 possible outcomes, the probability of rolling 2 is $1/11$ -- but if you've played enough Settlers of Catan you know that to be false!
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
Nicely explained (+1).
– Mark Viola
Jul 16 at 15:04
But, the set $2,3,4,5,6,7,8,9,10,11,12$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors.
– onurcanbektas
Jul 16 at 15:13
@onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you.
– Mees de Vries
Jul 16 at 15:15
Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it.
– Graham Kemp
Jul 16 at 15:17
@GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ?
– onurcanbektas
Jul 16 at 15:25
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We have $$beginarrayc#1&mboxResult at #1&mboxresult otherwise\hline c&g&g&win&lose\hline g&c&g&lose&win\hline g&g&c&lose&winendarray$$We can't assume that the host open door $3$ because then you say "the third option(g,g,c) can't happen", what the host does is taking the column of $#2$ or $#3$ "away", but the results won't change.
Update:
You have to note that the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$.
But if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
answered Jul 16 at 15:05
Holo
4,2512629
Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen.
– onurcanbektas
Jul 16 at 15:09
1
@onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
– Holo
Jul 16 at 15:26
Well that was convincing :) I advise you to turn that comment into an answer.
– onurcanbektas
Jul 16 at 15:31
1
And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows.
– Graham Kemp
Jul 16 at 15:32
@onurcanbektas I added it to the answer
– Holo
Jul 16 at 15:32
add a comment |Â
up vote
2
down vote
accepted
We have $$beginarrayc#1&mboxResult at #1&mboxresult otherwise\hline c&g&g&win&lose\hline g&c&g&lose&win\hline g&g&c&lose&winendarray$$We can't assume that the host open door $3$ because then you say "the third option(g,g,c) can't happen", what the host does is taking the column of $#2$ or $#3$ "away", but the results won't change.
Update:
You have to note that the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$.
But if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
answered Jul 16 at 15:05
Holo
4,2512629
Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen.
– onurcanbektas
Jul 16 at 15:09
1
@onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
– Holo
Jul 16 at 15:26
Well that was convincing :) I advise you to turn that comment into an answer.
– onurcanbektas
Jul 16 at 15:31
1
And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows.
– Graham Kemp
Jul 16 at 15:32
@onurcanbektas I added it to the answer
– Holo
Jul 16 at 15:32
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We have $$beginarrayc#1&mboxResult at #1&mboxresult otherwise\hline c&g&g&win&lose\hline g&c&g&lose&win\hline g&g&c&lose&winendarray$$We can't assume that the host open door $3$ because then you say "the third option(g,g,c) can't happen", what the host does is taking the column of $#2$ or $#3$ "away", but the results won't change.
Update:
You have to note that the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$.
But if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
answered Jul 16 at 15:05
Holo
4,2512629
We have $$beginarrayc#1&mboxResult at #1&mboxresult otherwise\hline c&g&g&win&lose\hline g&c&g&lose&win\hline g&g&c&lose&winendarray$$We can't assume that the host open door $3$ because then you say "the third option(g,g,c) can't happen", what the host does is taking the column of $#2$ or $#3$ "away", but the results won't change.
Update:
You have to note that the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$.
But if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
answered Jul 16 at 15:05
Holo
4,2512629
edited Jul 16 at 15:32
answered Jul 16 at 15:05
Holo
4,2512629
answered Jul 16 at 15:05
Holo
4,2512629
answered Jul 16 at 15:05
Holo
4,2512629
4,2512629
Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen.
– onurcanbektas
Jul 16 at 15:09
1
@onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
– Holo
Jul 16 at 15:26
Well that was convincing :) I advise you to turn that comment into an answer.
– onurcanbektas
Jul 16 at 15:31
1
And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows.
– Graham Kemp
Jul 16 at 15:32
@onurcanbektas I added it to the answer
– Holo
Jul 16 at 15:32
add a comment |Â
Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen.
– onurcanbektas
Jul 16 at 15:09
1
@onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
– Holo
Jul 16 at 15:26
Well that was convincing :) I advise you to turn that comment into an answer.
– onurcanbektas
Jul 16 at 15:31
1
And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows.
– Graham Kemp
Jul 16 at 15:32
@onurcanbektas I added it to the answer
– Holo
Jul 16 at 15:32
Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen.
– onurcanbektas
Jul 16 at 15:09
Ok, but we do know that Monty will never open the door with the car, so yes that option will never happen.
– onurcanbektas
Jul 16 at 15:09
1
1
@onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
– Holo
Jul 16 at 15:26
@onurcanbektas but the host choice depends on your choice: if you choose the correct door then the host will open a door with probability $50%$ for each door. you have $#1$ your choice and host open $#2$ to be $1/3cdot1/2=1/6$($1/3$ is the probability you are correct at the start) and you have $#1$ your choice and host open $#3$ to be $1/3cdot1/2=1/6$, together you gave $1/3$. if you chose incorrect there is $100%$ that the host will open a particular door: $#1$ your choice and host open $#2$ to be $1/3cdot1=1/3$ and and host open $#3$ to be $1/3cdot1=1/3$, together it is $=2/3$.
– Holo
Jul 16 at 15:26
Well that was convincing :) I advise you to turn that comment into an answer.
– onurcanbektas
Jul 16 at 15:31
Well that was convincing :) I advise you to turn that comment into an answer.
– onurcanbektas
Jul 16 at 15:31
1
1
And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows.
– Graham Kemp
Jul 16 at 15:32
And that is where generality is lost. We can only say WLOG that you select door 1 - and the car will be behind a door with equal probability for each (1/3). If the car is behind door 3, Monty cannot open it; he must open door 2 instead - and switching will win. Just as if the car is behind door 2, Monty must open door 3 - and switching will win. Yet if the car is behind door 1, Monty may freely select either door 2 or 3 - and whichever the case switching will lose. So switching will win in two of the three rows.
– Graham Kemp
Jul 16 at 15:32
@onurcanbektas I added it to the answer
– Holo
Jul 16 at 15:32
@onurcanbektas I added it to the answer
– Holo
Jul 16 at 15:32
add a comment |Â
up vote
8
down vote
You cannot do the "WLOG" and expect probabilities to stay equally distributed over your cases.
That is, you can say -- assuming that the car is behind a uniformly random door -- that you choose door 1, and then the three options are (car, goat, goat), (goat, car, goat) or (goat, goat, car) -- each with probability 1/3. That is fine.
However, when you then say "Without loss of generality, Monty opens door 3", you are excluding case 3, and implicitly treating it while you are treating case 2. If you continue you then have to add the probability of that situation occurring to the treatment of case 2. There are two options left: (car, goat, goat) and (goat, car, goat), but the first has probability 1/3, while the second has probability 2/3.
Edit: As a more general point, you seem to be using the common argument, "there are $n$ options, and therefore each occurs with probability $1/n$". This argument is dangerous: there's no reason to expect it to hold in general, but you often need it as an implicit assumption, because people forget to specify e.g. that a die or a coin is fair.
Consider the following similar argument: you throw two fair 6-sided dice, and your roll is the sum of the two numbers. The possible outcomes are $2,3,4,5,6,7,8,9,10,11,12$. You might be tempted to conclude that, since there are 11 possible outcomes, the probability of rolling 2 is $1/11$ -- but if you've played enough Settlers of Catan you know that to be false!
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
Nicely explained (+1).
– Mark Viola
Jul 16 at 15:04
But, the set $2,3,4,5,6,7,8,9,10,11,12$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors.
– onurcanbektas
Jul 16 at 15:13
@onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you.
– Mees de Vries
Jul 16 at 15:15
Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it.
– Graham Kemp
Jul 16 at 15:17
@GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ?
– onurcanbektas
Jul 16 at 15:25
 |Â
show 1 more comment
up vote
8
down vote
You cannot do the "WLOG" and expect probabilities to stay equally distributed over your cases.
That is, you can say -- assuming that the car is behind a uniformly random door -- that you choose door 1, and then the three options are (car, goat, goat), (goat, car, goat) or (goat, goat, car) -- each with probability 1/3. That is fine.
However, when you then say "Without loss of generality, Monty opens door 3", you are excluding case 3, and implicitly treating it while you are treating case 2. If you continue you then have to add the probability of that situation occurring to the treatment of case 2. There are two options left: (car, goat, goat) and (goat, car, goat), but the first has probability 1/3, while the second has probability 2/3.
Edit: As a more general point, you seem to be using the common argument, "there are $n$ options, and therefore each occurs with probability $1/n$". This argument is dangerous: there's no reason to expect it to hold in general, but you often need it as an implicit assumption, because people forget to specify e.g. that a die or a coin is fair.
Consider the following similar argument: you throw two fair 6-sided dice, and your roll is the sum of the two numbers. The possible outcomes are $2,3,4,5,6,7,8,9,10,11,12$. You might be tempted to conclude that, since there are 11 possible outcomes, the probability of rolling 2 is $1/11$ -- but if you've played enough Settlers of Catan you know that to be false!
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
Nicely explained (+1).
– Mark Viola
Jul 16 at 15:04
But, the set $2,3,4,5,6,7,8,9,10,11,12$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors.
– onurcanbektas
Jul 16 at 15:13
@onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you.
– Mees de Vries
Jul 16 at 15:15
Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it.
– Graham Kemp
Jul 16 at 15:17
@GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ?
– onurcanbektas
Jul 16 at 15:25
 |Â
show 1 more comment
up vote
8
down vote
up vote
8
down vote
You cannot do the "WLOG" and expect probabilities to stay equally distributed over your cases.
That is, you can say -- assuming that the car is behind a uniformly random door -- that you choose door 1, and then the three options are (car, goat, goat), (goat, car, goat) or (goat, goat, car) -- each with probability 1/3. That is fine.
However, when you then say "Without loss of generality, Monty opens door 3", you are excluding case 3, and implicitly treating it while you are treating case 2. If you continue you then have to add the probability of that situation occurring to the treatment of case 2. There are two options left: (car, goat, goat) and (goat, car, goat), but the first has probability 1/3, while the second has probability 2/3.
Edit: As a more general point, you seem to be using the common argument, "there are $n$ options, and therefore each occurs with probability $1/n$". This argument is dangerous: there's no reason to expect it to hold in general, but you often need it as an implicit assumption, because people forget to specify e.g. that a die or a coin is fair.
Consider the following similar argument: you throw two fair 6-sided dice, and your roll is the sum of the two numbers. The possible outcomes are $2,3,4,5,6,7,8,9,10,11,12$. You might be tempted to conclude that, since there are 11 possible outcomes, the probability of rolling 2 is $1/11$ -- but if you've played enough Settlers of Catan you know that to be false!
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
You cannot do the "WLOG" and expect probabilities to stay equally distributed over your cases.
That is, you can say -- assuming that the car is behind a uniformly random door -- that you choose door 1, and then the three options are (car, goat, goat), (goat, car, goat) or (goat, goat, car) -- each with probability 1/3. That is fine.
However, when you then say "Without loss of generality, Monty opens door 3", you are excluding case 3, and implicitly treating it while you are treating case 2. If you continue you then have to add the probability of that situation occurring to the treatment of case 2. There are two options left: (car, goat, goat) and (goat, car, goat), but the first has probability 1/3, while the second has probability 2/3.
Edit: As a more general point, you seem to be using the common argument, "there are $n$ options, and therefore each occurs with probability $1/n$". This argument is dangerous: there's no reason to expect it to hold in general, but you often need it as an implicit assumption, because people forget to specify e.g. that a die or a coin is fair.
Consider the following similar argument: you throw two fair 6-sided dice, and your roll is the sum of the two numbers. The possible outcomes are $2,3,4,5,6,7,8,9,10,11,12$. You might be tempted to conclude that, since there are 11 possible outcomes, the probability of rolling 2 is $1/11$ -- but if you've played enough Settlers of Catan you know that to be false!
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
edited Jul 16 at 15:09
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
answered Jul 16 at 14:59
Mees de Vries
13.7k12345
13.7k12345
Nicely explained (+1).
– Mark Viola
Jul 16 at 15:04
But, the set $2,3,4,5,6,7,8,9,10,11,12$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors.
– onurcanbektas
Jul 16 at 15:13
@onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you.
– Mees de Vries
Jul 16 at 15:15
Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it.
– Graham Kemp
Jul 16 at 15:17
@GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ?
– onurcanbektas
Jul 16 at 15:25
 |Â
show 1 more comment
Nicely explained (+1).
– Mark Viola
Jul 16 at 15:04
But, the set $2,3,4,5,6,7,8,9,10,11,12$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors.
– onurcanbektas
Jul 16 at 15:13
@onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you.
– Mees de Vries
Jul 16 at 15:15
Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it.
– Graham Kemp
Jul 16 at 15:17
@GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ?
– onurcanbektas
Jul 16 at 15:25
Nicely explained (+1).
– Mark Viola
Jul 16 at 15:04
Nicely explained (+1).
– Mark Viola
Jul 16 at 15:04
But, the set $2,3,4,5,6,7,8,9,10,11,12$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors.
– onurcanbektas
Jul 16 at 15:13
But, the set $2,3,4,5,6,7,8,9,10,11,12$ is not the sample space of the dice, whereas in this case, I'm directly looking into the case where we considering the each possible cases for the doors.
– onurcanbektas
Jul 16 at 15:13
@onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you.
– Mees de Vries
Jul 16 at 15:15
@onurcanbektas, it could be the sample space for the dice, if all you are interested in is the outcome of their sum. And you are explicitly not looking into each of the possible cases for the doors; that is what the "WLOG" is doing for you.
– Mees de Vries
Jul 16 at 15:15
Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it.
– Graham Kemp
Jul 16 at 15:17
Exactly. Saying that you open door 1 and Monty opens door 3 loses generality, because that can only happen in the specific case there is not a car behind it.
– Graham Kemp
Jul 16 at 15:17
@GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ?
– onurcanbektas
Jul 16 at 15:25
@GrahamKemp Then think like this; I'm in the contest right now, and I have chosen the door 1, and Monty opened the door 3, what about now ? I mean this situation is occurring right now, so I'm not saying "WLOG", I'm just saying what I'm seeing. What is the chance of my winning if I stay with 1 ?
– onurcanbektas
Jul 16 at 15:25
 |Â
show 1 more comment
What argument? What are you trying to show? Your notation is not self-explanatory...there aren't, for instance, any probabilities attached to it.
– lulu
Jul 16 at 14:55
Are you trying to say that after they show door 3 it is $50/50$?
– Holo
Jul 16 at 14:57
7
The fact that there are only two options does not mean that they are equally likely.
– lulu
Jul 16 at 14:57
1
"I have never claimed such a thing either." Yes, you did. When you claimed that there is no difference and therefore $50%$ chance of winning either way, that's exactly what you did.
– Arthur
Jul 16 at 15:00
1
I have purchases a lottery ticket. WLOG there are two outcomes: that I have purchased the only winning ticket, or that I have purchased one of the two million losing tickets. Therefore there is a fifty-fifty chance that I will win! Yah!
– Graham Kemp
Jul 16 at 15:23