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monty hall with sequentially opened door [closed]
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Let's consider the extended Monty Hall problem as follows.
There are 5 doors (A,B,C,D,E), we first choose door A, then the host opens door B.
Based on the previous discussion, we will switch because $1/5$ is smaller than $4/15$. After that, we choose door C, and the host opens D (which, of course has no car), what is the probability of getting the car for door A, C,and E based on the sequential process?
I personally get $9/29$,$8/29$,and $12/29$, could someone help with the answer?
probability
asked Jul 19 at 16:45
Hank John
61
closed as unclear what you're asking by JMoravitz, amWhy, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-2
down vote
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Let's consider the extended Monty Hall problem as follows.
There are 5 doors (A,B,C,D,E), we first choose door A, then the host opens door B.
Based on the previous discussion, we will switch because $1/5$ is smaller than $4/15$. After that, we choose door C, and the host opens D (which, of course has no car), what is the probability of getting the car for door A, C,and E based on the sequential process?
I personally get $9/29$,$8/29$,and $12/29$, could someone help with the answer?
probability
asked Jul 19 at 16:45
Hank John
61
closed as unclear what you're asking by JMoravitz, amWhy, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Do you always choose door A? Does the host always choose door B next, you always switch to door C, and the host always shows door D? And at this point, doors B and D show no car, but goats? This question is not worded in a way that can be easily answered, as it is not clear what you are asking.
– InterstellarProbe
Jul 19 at 16:49
1
One of the most crucial parts in the understanding and solution of the monty hall problem and its variants is in understanding very explicitly how the host acts. Any question which does not explicitly explain how the host acts is woefully incomplete and cannot be answered definitively.
– JMoravitz
Jul 19 at 16:53
@InterstellarProbe - perhaps better to ask whether the host only opens doors which have a goat and have not been chosen or opened before
– Henry
Jul 19 at 16:53
Well, the host always open the door which has a goat, that's the basic assumption of the Monty Hall problem. The actual doors opened should not matter.
– Hank John
Jul 19 at 16:57
Specifically: it is important to specify how Monty chooses between the worthless doors, because as the game evolves the estimated probabilities of each door will not stay the same. The easiest assumption to make is that he always chooses uniformly at random between the worthless doors, but other assumptions are possible.
– lulu
Jul 19 at 17:09
 |Â
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let's consider the extended Monty Hall problem as follows.
There are 5 doors (A,B,C,D,E), we first choose door A, then the host opens door B.
Based on the previous discussion, we will switch because $1/5$ is smaller than $4/15$. After that, we choose door C, and the host opens D (which, of course has no car), what is the probability of getting the car for door A, C,and E based on the sequential process?
I personally get $9/29$,$8/29$,and $12/29$, could someone help with the answer?
probability
asked Jul 19 at 16:45
Hank John
61
Let's consider the extended Monty Hall problem as follows.
There are 5 doors (A,B,C,D,E), we first choose door A, then the host opens door B.
Based on the previous discussion, we will switch because $1/5$ is smaller than $4/15$. After that, we choose door C, and the host opens D (which, of course has no car), what is the probability of getting the car for door A, C,and E based on the sequential process?
I personally get $9/29$,$8/29$,and $12/29$, could someone help with the answer?
probability
asked Jul 19 at 16:45
Hank John
61
asked Jul 19 at 16:45
Hank John
61
asked Jul 19 at 16:45
Hank John
61
asked Jul 19 at 16:45
Hank John
61
61
closed as unclear what you're asking by JMoravitz, amWhy, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by JMoravitz, amWhy, José Carlos Santos, Xander Henderson, Adrian Keister Jul 20 at 1:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Do you always choose door A? Does the host always choose door B next, you always switch to door C, and the host always shows door D? And at this point, doors B and D show no car, but goats? This question is not worded in a way that can be easily answered, as it is not clear what you are asking.
– InterstellarProbe
Jul 19 at 16:49
1
One of the most crucial parts in the understanding and solution of the monty hall problem and its variants is in understanding very explicitly how the host acts. Any question which does not explicitly explain how the host acts is woefully incomplete and cannot be answered definitively.
– JMoravitz
Jul 19 at 16:53
@InterstellarProbe - perhaps better to ask whether the host only opens doors which have a goat and have not been chosen or opened before
– Henry
Jul 19 at 16:53
Well, the host always open the door which has a goat, that's the basic assumption of the Monty Hall problem. The actual doors opened should not matter.
– Hank John
Jul 19 at 16:57
Specifically: it is important to specify how Monty chooses between the worthless doors, because as the game evolves the estimated probabilities of each door will not stay the same. The easiest assumption to make is that he always chooses uniformly at random between the worthless doors, but other assumptions are possible.
– lulu
Jul 19 at 17:09
 |Â
show 2 more comments
1
Do you always choose door A? Does the host always choose door B next, you always switch to door C, and the host always shows door D? And at this point, doors B and D show no car, but goats? This question is not worded in a way that can be easily answered, as it is not clear what you are asking.
– InterstellarProbe
Jul 19 at 16:49
1
One of the most crucial parts in the understanding and solution of the monty hall problem and its variants is in understanding very explicitly how the host acts. Any question which does not explicitly explain how the host acts is woefully incomplete and cannot be answered definitively.
– JMoravitz
Jul 19 at 16:53
@InterstellarProbe - perhaps better to ask whether the host only opens doors which have a goat and have not been chosen or opened before
– Henry
Jul 19 at 16:53
Well, the host always open the door which has a goat, that's the basic assumption of the Monty Hall problem. The actual doors opened should not matter.
– Hank John
Jul 19 at 16:57
Specifically: it is important to specify how Monty chooses between the worthless doors, because as the game evolves the estimated probabilities of each door will not stay the same. The easiest assumption to make is that he always chooses uniformly at random between the worthless doors, but other assumptions are possible.
– lulu
Jul 19 at 17:09
1
1
Do you always choose door A? Does the host always choose door B next, you always switch to door C, and the host always shows door D? And at this point, doors B and D show no car, but goats? This question is not worded in a way that can be easily answered, as it is not clear what you are asking.
– InterstellarProbe
Jul 19 at 16:49
Do you always choose door A? Does the host always choose door B next, you always switch to door C, and the host always shows door D? And at this point, doors B and D show no car, but goats? This question is not worded in a way that can be easily answered, as it is not clear what you are asking.
– InterstellarProbe
Jul 19 at 16:49
1
1
One of the most crucial parts in the understanding and solution of the monty hall problem and its variants is in understanding very explicitly how the host acts. Any question which does not explicitly explain how the host acts is woefully incomplete and cannot be answered definitively.
– JMoravitz
Jul 19 at 16:53
One of the most crucial parts in the understanding and solution of the monty hall problem and its variants is in understanding very explicitly how the host acts. Any question which does not explicitly explain how the host acts is woefully incomplete and cannot be answered definitively.
– JMoravitz
Jul 19 at 16:53
@InterstellarProbe - perhaps better to ask whether the host only opens doors which have a goat and have not been chosen or opened before
– Henry
Jul 19 at 16:53
@InterstellarProbe - perhaps better to ask whether the host only opens doors which have a goat and have not been chosen or opened before
– Henry
Jul 19 at 16:53
Well, the host always open the door which has a goat, that's the basic assumption of the Monty Hall problem. The actual doors opened should not matter.
– Hank John
Jul 19 at 16:57
Well, the host always open the door which has a goat, that's the basic assumption of the Monty Hall problem. The actual doors opened should not matter.
– Hank John
Jul 19 at 16:57
Specifically: it is important to specify how Monty chooses between the worthless doors, because as the game evolves the estimated probabilities of each door will not stay the same. The easiest assumption to make is that he always chooses uniformly at random between the worthless doors, but other assumptions are possible.
– lulu
Jul 19 at 17:09
Specifically: it is important to specify how Monty chooses between the worthless doors, because as the game evolves the estimated probabilities of each door will not stay the same. The easiest assumption to make is that he always chooses uniformly at random between the worthless doors, but other assumptions are possible.
– lulu
Jul 19 at 17:09
 |Â
show 2 more comments
1 Answer
1
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Let's say you chose Door A. There is a $1/5$ chance that you chose the correct door. At this moment, there is a $4/5$ chance that the correct door is among B,C,D,E.
Once Door B is shown to be a sham, there is still a $4/5$ chance the door is among C,D,E. You choose to switch to Door C because, on average, these doors have a higher probability of being correct. Assuming the correct door is among C,D,E, there is a $1/3$ chance door C is the correct door. That gives it a probability of $1/3$ * $4/5$ = $4/15$. Now, there is a $2/3$ * $4/5$ = $8/15$ chance that the correct door is between D and E.
Now Door D is shown to be a sham. This means that there is a $8/15$ chance that Door E is the correct Door. If you keep switching doors, there is a $8/15$ chance you will have the correct door.
($1/5$, $4/15$, $8/15$)
As a side node, the general pattern for $N$ doors for any odd $N$ where $i$ represents the $(i+1)$th door opened:
$$
f(0) = 1/n \
f(i) = f(i-1) * n(2 - 2*i + 1) * frac12i
$$
answered Jul 19 at 17:01
Sidd Singal
2,53731630
I can follow your thoughts, thanks. Will your answer change if the second opened door is A? (and thus the probability is for C,D,E)
– Hank John
Jul 19 at 17:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let's say you chose Door A. There is a $1/5$ chance that you chose the correct door. At this moment, there is a $4/5$ chance that the correct door is among B,C,D,E.
Once Door B is shown to be a sham, there is still a $4/5$ chance the door is among C,D,E. You choose to switch to Door C because, on average, these doors have a higher probability of being correct. Assuming the correct door is among C,D,E, there is a $1/3$ chance door C is the correct door. That gives it a probability of $1/3$ * $4/5$ = $4/15$. Now, there is a $2/3$ * $4/5$ = $8/15$ chance that the correct door is between D and E.
Now Door D is shown to be a sham. This means that there is a $8/15$ chance that Door E is the correct Door. If you keep switching doors, there is a $8/15$ chance you will have the correct door.
($1/5$, $4/15$, $8/15$)
As a side node, the general pattern for $N$ doors for any odd $N$ where $i$ represents the $(i+1)$th door opened:
$$
f(0) = 1/n \
f(i) = f(i-1) * n(2 - 2*i + 1) * frac12i
$$
answered Jul 19 at 17:01
Sidd Singal
2,53731630
I can follow your thoughts, thanks. Will your answer change if the second opened door is A? (and thus the probability is for C,D,E)
– Hank John
Jul 19 at 17:17
add a comment |Â
up vote
1
down vote
Let's say you chose Door A. There is a $1/5$ chance that you chose the correct door. At this moment, there is a $4/5$ chance that the correct door is among B,C,D,E.
Once Door B is shown to be a sham, there is still a $4/5$ chance the door is among C,D,E. You choose to switch to Door C because, on average, these doors have a higher probability of being correct. Assuming the correct door is among C,D,E, there is a $1/3$ chance door C is the correct door. That gives it a probability of $1/3$ * $4/5$ = $4/15$. Now, there is a $2/3$ * $4/5$ = $8/15$ chance that the correct door is between D and E.
Now Door D is shown to be a sham. This means that there is a $8/15$ chance that Door E is the correct Door. If you keep switching doors, there is a $8/15$ chance you will have the correct door.
($1/5$, $4/15$, $8/15$)
As a side node, the general pattern for $N$ doors for any odd $N$ where $i$ represents the $(i+1)$th door opened:
$$
f(0) = 1/n \
f(i) = f(i-1) * n(2 - 2*i + 1) * frac12i
$$
answered Jul 19 at 17:01
Sidd Singal
2,53731630
I can follow your thoughts, thanks. Will your answer change if the second opened door is A? (and thus the probability is for C,D,E)
– Hank John
Jul 19 at 17:17
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let's say you chose Door A. There is a $1/5$ chance that you chose the correct door. At this moment, there is a $4/5$ chance that the correct door is among B,C,D,E.
Once Door B is shown to be a sham, there is still a $4/5$ chance the door is among C,D,E. You choose to switch to Door C because, on average, these doors have a higher probability of being correct. Assuming the correct door is among C,D,E, there is a $1/3$ chance door C is the correct door. That gives it a probability of $1/3$ * $4/5$ = $4/15$. Now, there is a $2/3$ * $4/5$ = $8/15$ chance that the correct door is between D and E.
Now Door D is shown to be a sham. This means that there is a $8/15$ chance that Door E is the correct Door. If you keep switching doors, there is a $8/15$ chance you will have the correct door.
($1/5$, $4/15$, $8/15$)
As a side node, the general pattern for $N$ doors for any odd $N$ where $i$ represents the $(i+1)$th door opened:
$$
f(0) = 1/n \
f(i) = f(i-1) * n(2 - 2*i + 1) * frac12i
$$
answered Jul 19 at 17:01
Sidd Singal
2,53731630
Let's say you chose Door A. There is a $1/5$ chance that you chose the correct door. At this moment, there is a $4/5$ chance that the correct door is among B,C,D,E.
Once Door B is shown to be a sham, there is still a $4/5$ chance the door is among C,D,E. You choose to switch to Door C because, on average, these doors have a higher probability of being correct. Assuming the correct door is among C,D,E, there is a $1/3$ chance door C is the correct door. That gives it a probability of $1/3$ * $4/5$ = $4/15$. Now, there is a $2/3$ * $4/5$ = $8/15$ chance that the correct door is between D and E.
Now Door D is shown to be a sham. This means that there is a $8/15$ chance that Door E is the correct Door. If you keep switching doors, there is a $8/15$ chance you will have the correct door.
($1/5$, $4/15$, $8/15$)
As a side node, the general pattern for $N$ doors for any odd $N$ where $i$ represents the $(i+1)$th door opened:
$$
f(0) = 1/n \
f(i) = f(i-1) * n(2 - 2*i + 1) * frac12i
$$
answered Jul 19 at 17:01
Sidd Singal
2,53731630
edited Jul 19 at 17:17
answered Jul 19 at 17:01
Sidd Singal
2,53731630
answered Jul 19 at 17:01
Sidd Singal
2,53731630
answered Jul 19 at 17:01
Sidd Singal
2,53731630
2,53731630
I can follow your thoughts, thanks. Will your answer change if the second opened door is A? (and thus the probability is for C,D,E)
– Hank John
Jul 19 at 17:17
add a comment |Â
I can follow your thoughts, thanks. Will your answer change if the second opened door is A? (and thus the probability is for C,D,E)
– Hank John
Jul 19 at 17:17
I can follow your thoughts, thanks. Will your answer change if the second opened door is A? (and thus the probability is for C,D,E)
– Hank John
Jul 19 at 17:17
I can follow your thoughts, thanks. Will your answer change if the second opened door is A? (and thus the probability is for C,D,E)
– Hank John
Jul 19 at 17:17
add a comment |Â
1
Do you always choose door A? Does the host always choose door B next, you always switch to door C, and the host always shows door D? And at this point, doors B and D show no car, but goats? This question is not worded in a way that can be easily answered, as it is not clear what you are asking.
– InterstellarProbe
Jul 19 at 16:49
1
One of the most crucial parts in the understanding and solution of the monty hall problem and its variants is in understanding very explicitly how the host acts. Any question which does not explicitly explain how the host acts is woefully incomplete and cannot be answered definitively.
– JMoravitz
Jul 19 at 16:53
@InterstellarProbe - perhaps better to ask whether the host only opens doors which have a goat and have not been chosen or opened before
– Henry
Jul 19 at 16:53
Well, the host always open the door which has a goat, that's the basic assumption of the Monty Hall problem. The actual doors opened should not matter.
– Hank John
Jul 19 at 16:57
Specifically: it is important to specify how Monty chooses between the worthless doors, because as the game evolves the estimated probabilities of each door will not stay the same. The easiest assumption to make is that he always chooses uniformly at random between the worthless doors, but other assumptions are possible.
– lulu
Jul 19 at 17:09