Why's the probability of the unchosen doors' containing the prize reduced to the unopened door?
$begingroup$
I understand, and ask not about, how to prove that you ought switch. Let $Pr(✘)$ = probability that the unchosen doors have the prize (I chose the X mark to signify that you chose wrongly).
Please see the question in the title. Monty knows which door lacks the prize and must open it, but why exactly is $Pr(✘)$ "reduced" to the lone unopened door?
The quotes beneath don't explicitly explain this "collapsing"
of $Pr(✘)$ . This Reddit comment
Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door[.]
refers to an older version of Wikipedia:
As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player's choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.
As Keith Devlin says (Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
I first saw the idea that $Pr(✘)$ "is now concentrated on that other door", on Reddit:
It makes more sense to switch doors because Monty has changed the problem.That's the most important piece of information. Monty knows more than you do.
Imagine instead of 3 doors, there were 100 doors. You had a 1 in 100 chance of picking the door with the car behind it. Monty opens 98 doors to reveal 98 goats. So why should you switch? Well, the odds of you picking the car off the bat were 1 in 100. That means there is a 99% chance that the door you picked initially has a goat behind it. Monty has opened all of the other goat doors, meaning your odds are much better if you switch, because he eliminated all of the other goats in the problem except for one.
monty-hall
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
$endgroup$
add a comment |
$begingroup$
I understand, and ask not about, how to prove that you ought switch. Let $Pr(✘)$ = probability that the unchosen doors have the prize (I chose the X mark to signify that you chose wrongly).
Please see the question in the title. Monty knows which door lacks the prize and must open it, but why exactly is $Pr(✘)$ "reduced" to the lone unopened door?
The quotes beneath don't explicitly explain this "collapsing"
of $Pr(✘)$ . This Reddit comment
Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door[.]
refers to an older version of Wikipedia:
As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player's choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.
As Keith Devlin says (Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
I first saw the idea that $Pr(✘)$ "is now concentrated on that other door", on Reddit:
It makes more sense to switch doors because Monty has changed the problem.That's the most important piece of information. Monty knows more than you do.
Imagine instead of 3 doors, there were 100 doors. You had a 1 in 100 chance of picking the door with the car behind it. Monty opens 98 doors to reveal 98 goats. So why should you switch? Well, the odds of you picking the car off the bat were 1 in 100. That means there is a 99% chance that the door you picked initially has a goat behind it. Monty has opened all of the other goat doors, meaning your odds are much better if you switch, because he eliminated all of the other goats in the problem except for one.
monty-hall
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
$endgroup$
$begingroup$
Not sure I understand where your confusion lies. Suppose, before Monty did anything, that you could always choose between sticking with your choice or taking BOTH the other two doors. It's clear you'd switch, yes? But this is equivalent to the game at hand since you already knew one of the unchosen doors (at least) was empty.
$endgroup$
– lulu
Jan 22 at 19:18
$begingroup$
@lulu I agree with the mathematical proof that I ought switch. But it's still not clear intuitively; I have to work backwards from $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘) Longleftarrow$ Switching. $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘)$ still baffles me.
$endgroup$
– Greek - Area 51 Proposal
Jan 22 at 19:30
$begingroup$
Well, think about it. Your first guess is a winner only $frac 13$ of the time. Switching to the other two is therefore a winner $frac 23$ of the time.
$endgroup$
– lulu
Jan 22 at 19:33
$begingroup$
I think the intuition gap, so to speak, comes from thinking that the fact that Monty opened door $2$ says nothing about door $3$ (assuming I had chosen with door $1$ of course). That's a natural thought, but it is incorrect. The fact that he chose $2$ instead of $3$ is evidence (not proof) that the prize is behind $3$. That information, even though it is not decisive, is valuable and changes your estimate of the probability.
$endgroup$
– lulu
Jan 22 at 19:35
$begingroup$
For instance: let's change the rules slightly. Let's say Monty has no idea where the prize is, and that he just opens an unselected door randomly (with the game ending should he expose the prize). Now there is truly no information to be had out of the fact that he opened $2$ instead of $3$ and with these rules there is no advantage in your switching.
$endgroup$
– lulu
Jan 22 at 19:38
add a comment |
$begingroup$
I understand, and ask not about, how to prove that you ought switch. Let $Pr(✘)$ = probability that the unchosen doors have the prize (I chose the X mark to signify that you chose wrongly).
Please see the question in the title. Monty knows which door lacks the prize and must open it, but why exactly is $Pr(✘)$ "reduced" to the lone unopened door?
The quotes beneath don't explicitly explain this "collapsing"
of $Pr(✘)$ . This Reddit comment
Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door[.]
refers to an older version of Wikipedia:
As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player's choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.
As Keith Devlin says (Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
I first saw the idea that $Pr(✘)$ "is now concentrated on that other door", on Reddit:
It makes more sense to switch doors because Monty has changed the problem.That's the most important piece of information. Monty knows more than you do.
Imagine instead of 3 doors, there were 100 doors. You had a 1 in 100 chance of picking the door with the car behind it. Monty opens 98 doors to reveal 98 goats. So why should you switch? Well, the odds of you picking the car off the bat were 1 in 100. That means there is a 99% chance that the door you picked initially has a goat behind it. Monty has opened all of the other goat doors, meaning your odds are much better if you switch, because he eliminated all of the other goats in the problem except for one.
monty-hall
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
$endgroup$
I understand, and ask not about, how to prove that you ought switch. Let $Pr(✘)$ = probability that the unchosen doors have the prize (I chose the X mark to signify that you chose wrongly).
Please see the question in the title. Monty knows which door lacks the prize and must open it, but why exactly is $Pr(✘)$ "reduced" to the lone unopened door?
The quotes beneath don't explicitly explain this "collapsing"
of $Pr(✘)$ . This Reddit comment
Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door[.]
refers to an older version of Wikipedia:
As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player's choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.
As Keith Devlin says (Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
I first saw the idea that $Pr(✘)$ "is now concentrated on that other door", on Reddit:
It makes more sense to switch doors because Monty has changed the problem.That's the most important piece of information. Monty knows more than you do.
Imagine instead of 3 doors, there were 100 doors. You had a 1 in 100 chance of picking the door with the car behind it. Monty opens 98 doors to reveal 98 goats. So why should you switch? Well, the odds of you picking the car off the bat were 1 in 100. That means there is a 99% chance that the door you picked initially has a goat behind it. Monty has opened all of the other goat doors, meaning your odds are much better if you switch, because he eliminated all of the other goats in the problem except for one.
monty-hall
monty-hall
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
edited Jan 22 at 19:27
Greek - Area 51 Proposal
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
asked Jan 22 at 19:14
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,176769105
3,176769105
$begingroup$
Not sure I understand where your confusion lies. Suppose, before Monty did anything, that you could always choose between sticking with your choice or taking BOTH the other two doors. It's clear you'd switch, yes? But this is equivalent to the game at hand since you already knew one of the unchosen doors (at least) was empty.
$endgroup$
– lulu
Jan 22 at 19:18
$begingroup$
@lulu I agree with the mathematical proof that I ought switch. But it's still not clear intuitively; I have to work backwards from $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘) Longleftarrow$ Switching. $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘)$ still baffles me.
$endgroup$
– Greek - Area 51 Proposal
Jan 22 at 19:30
$begingroup$
Well, think about it. Your first guess is a winner only $frac 13$ of the time. Switching to the other two is therefore a winner $frac 23$ of the time.
$endgroup$
– lulu
Jan 22 at 19:33
$begingroup$
I think the intuition gap, so to speak, comes from thinking that the fact that Monty opened door $2$ says nothing about door $3$ (assuming I had chosen with door $1$ of course). That's a natural thought, but it is incorrect. The fact that he chose $2$ instead of $3$ is evidence (not proof) that the prize is behind $3$. That information, even though it is not decisive, is valuable and changes your estimate of the probability.
$endgroup$
– lulu
Jan 22 at 19:35
$begingroup$
For instance: let's change the rules slightly. Let's say Monty has no idea where the prize is, and that he just opens an unselected door randomly (with the game ending should he expose the prize). Now there is truly no information to be had out of the fact that he opened $2$ instead of $3$ and with these rules there is no advantage in your switching.
$endgroup$
– lulu
Jan 22 at 19:38
add a comment |
$begingroup$
Not sure I understand where your confusion lies. Suppose, before Monty did anything, that you could always choose between sticking with your choice or taking BOTH the other two doors. It's clear you'd switch, yes? But this is equivalent to the game at hand since you already knew one of the unchosen doors (at least) was empty.
$endgroup$
– lulu
Jan 22 at 19:18
$begingroup$
@lulu I agree with the mathematical proof that I ought switch. But it's still not clear intuitively; I have to work backwards from $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘) Longleftarrow$ Switching. $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘)$ still baffles me.
$endgroup$
– Greek - Area 51 Proposal
Jan 22 at 19:30
$begingroup$
Well, think about it. Your first guess is a winner only $frac 13$ of the time. Switching to the other two is therefore a winner $frac 23$ of the time.
$endgroup$
– lulu
Jan 22 at 19:33
$begingroup$
I think the intuition gap, so to speak, comes from thinking that the fact that Monty opened door $2$ says nothing about door $3$ (assuming I had chosen with door $1$ of course). That's a natural thought, but it is incorrect. The fact that he chose $2$ instead of $3$ is evidence (not proof) that the prize is behind $3$. That information, even though it is not decisive, is valuable and changes your estimate of the probability.
$endgroup$
– lulu
Jan 22 at 19:35
$begingroup$
For instance: let's change the rules slightly. Let's say Monty has no idea where the prize is, and that he just opens an unselected door randomly (with the game ending should he expose the prize). Now there is truly no information to be had out of the fact that he opened $2$ instead of $3$ and with these rules there is no advantage in your switching.
$endgroup$
– lulu
Jan 22 at 19:38
$begingroup$
Not sure I understand where your confusion lies. Suppose, before Monty did anything, that you could always choose between sticking with your choice or taking BOTH the other two doors. It's clear you'd switch, yes? But this is equivalent to the game at hand since you already knew one of the unchosen doors (at least) was empty.
$endgroup$
– lulu
Jan 22 at 19:18
$begingroup$
Not sure I understand where your confusion lies. Suppose, before Monty did anything, that you could always choose between sticking with your choice or taking BOTH the other two doors. It's clear you'd switch, yes? But this is equivalent to the game at hand since you already knew one of the unchosen doors (at least) was empty.
$endgroup$
– lulu
Jan 22 at 19:18
$begingroup$
@lulu I agree with the mathematical proof that I ought switch. But it's still not clear intuitively; I have to work backwards from $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘) Longleftarrow$ Switching. $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘)$ still baffles me.
$endgroup$
– Greek - Area 51 Proposal
Jan 22 at 19:30
$begingroup$
@lulu I agree with the mathematical proof that I ought switch. But it's still not clear intuitively; I have to work backwards from $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘) Longleftarrow$ Switching. $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘)$ still baffles me.
$endgroup$
– Greek - Area 51 Proposal
Jan 22 at 19:30
$begingroup$
Well, think about it. Your first guess is a winner only $frac 13$ of the time. Switching to the other two is therefore a winner $frac 23$ of the time.
$endgroup$
– lulu
Jan 22 at 19:33
$begingroup$
Well, think about it. Your first guess is a winner only $frac 13$ of the time. Switching to the other two is therefore a winner $frac 23$ of the time.
$endgroup$
– lulu
Jan 22 at 19:33
$begingroup$
I think the intuition gap, so to speak, comes from thinking that the fact that Monty opened door $2$ says nothing about door $3$ (assuming I had chosen with door $1$ of course). That's a natural thought, but it is incorrect. The fact that he chose $2$ instead of $3$ is evidence (not proof) that the prize is behind $3$. That information, even though it is not decisive, is valuable and changes your estimate of the probability.
$endgroup$
– lulu
Jan 22 at 19:35
$begingroup$
I think the intuition gap, so to speak, comes from thinking that the fact that Monty opened door $2$ says nothing about door $3$ (assuming I had chosen with door $1$ of course). That's a natural thought, but it is incorrect. The fact that he chose $2$ instead of $3$ is evidence (not proof) that the prize is behind $3$. That information, even though it is not decisive, is valuable and changes your estimate of the probability.
$endgroup$
– lulu
Jan 22 at 19:35
$begingroup$
For instance: let's change the rules slightly. Let's say Monty has no idea where the prize is, and that he just opens an unselected door randomly (with the game ending should he expose the prize). Now there is truly no information to be had out of the fact that he opened $2$ instead of $3$ and with these rules there is no advantage in your switching.
$endgroup$
– lulu
Jan 22 at 19:38
$begingroup$
For instance: let's change the rules slightly. Let's say Monty has no idea where the prize is, and that he just opens an unselected door randomly (with the game ending should he expose the prize). Now there is truly no information to be had out of the fact that he opened $2$ instead of $3$ and with these rules there is no advantage in your switching.
$endgroup$
– lulu
Jan 22 at 19:38
add a comment |
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$begingroup$
Not sure I understand where your confusion lies. Suppose, before Monty did anything, that you could always choose between sticking with your choice or taking BOTH the other two doors. It's clear you'd switch, yes? But this is equivalent to the game at hand since you already knew one of the unchosen doors (at least) was empty.
$endgroup$
– lulu
Jan 22 at 19:18
$begingroup$
@lulu I agree with the mathematical proof that I ought switch. But it's still not clear intuitively; I have to work backwards from $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘) Longleftarrow$ Switching. $Pr$(lone unopened door has the prize) $color{red}{Longleftarrow} Pr(✘)$ still baffles me.
$endgroup$
– Greek - Area 51 Proposal
Jan 22 at 19:30
$begingroup$
Well, think about it. Your first guess is a winner only $frac 13$ of the time. Switching to the other two is therefore a winner $frac 23$ of the time.
$endgroup$
– lulu
Jan 22 at 19:33
$begingroup$
I think the intuition gap, so to speak, comes from thinking that the fact that Monty opened door $2$ says nothing about door $3$ (assuming I had chosen with door $1$ of course). That's a natural thought, but it is incorrect. The fact that he chose $2$ instead of $3$ is evidence (not proof) that the prize is behind $3$. That information, even though it is not decisive, is valuable and changes your estimate of the probability.
$endgroup$
– lulu
Jan 22 at 19:35
$begingroup$
For instance: let's change the rules slightly. Let's say Monty has no idea where the prize is, and that he just opens an unselected door randomly (with the game ending should he expose the prize). Now there is truly no information to be had out of the fact that he opened $2$ instead of $3$ and with these rules there is no advantage in your switching.
$endgroup$
– lulu
Jan 22 at 19:38