Mixing
Monty Hall: What's wrong with my approach
$begingroup$
Let us say we have n boxes, and one car.
The probability that one of them might have a car = p
and each box is equally likely to have a car, therefore the change that a single box has car = p/n
Now let us consider the chance that box 1 has car, given other boxes don't have car
P(Box 1 contains car | other n-1 boxes don't contain car)
$$=frac{P(Boxspace1space containsspace car cap otherspace n-1space boxesspace don'tspace containspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{P(Boxspace1space containsspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$ [1]
But if we apply this result to classical monty hall problem where we have 3 boxes. Let us say contestant chose box 1.
Other two boxes have p = 2/3 to have car, n = 2,
each box will have probability of p/n = $frac{2/3}{2} = 1/3$, of having a car
If monty opens up say box 3(which doesn't have car),
the probability of box 2 having car
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$
here p = 2/3
n = 2
$$=frac{frac{2/3}{2}}{ left( 1 - frac{2/3}{2} right)^{(2-1)} }$$
$$= 1/2$$
which is not the correct answer. I have done something wrong and [1] is not valid result.
What is wrong with it?
PS: I know how to solve monty hall problem using other approach. I want to know what is wrong with this approach?
probability probability-theory statistics monty-hall
asked Jan 13 at 10:30
q126yq126y
239212
$endgroup$
add a comment |
$begingroup$
Let us say we have n boxes, and one car.
The probability that one of them might have a car = p
and each box is equally likely to have a car, therefore the change that a single box has car = p/n
Now let us consider the chance that box 1 has car, given other boxes don't have car
P(Box 1 contains car | other n-1 boxes don't contain car)
$$=frac{P(Boxspace1space containsspace car cap otherspace n-1space boxesspace don'tspace containspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{P(Boxspace1space containsspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$ [1]
But if we apply this result to classical monty hall problem where we have 3 boxes. Let us say contestant chose box 1.
Other two boxes have p = 2/3 to have car, n = 2,
each box will have probability of p/n = $frac{2/3}{2} = 1/3$, of having a car
If monty opens up say box 3(which doesn't have car),
the probability of box 2 having car
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$
here p = 2/3
n = 2
$$=frac{frac{2/3}{2}}{ left( 1 - frac{2/3}{2} right)^{(2-1)} }$$
$$= 1/2$$
which is not the correct answer. I have done something wrong and [1] is not valid result.
What is wrong with it?
PS: I know how to solve monty hall problem using other approach. I want to know what is wrong with this approach?
probability probability-theory statistics monty-hall
asked Jan 13 at 10:30
q126yq126y
239212
$endgroup$
1
$begingroup$
The events [box 2 has no car], [box 3 has no car], etc. are not independent (at most one of them is false, since there is only one car). Hence the probability of [other n-1 boxes don't have a car] is not the product of probabilities.
$endgroup$
– D. Thomine
Jan 13 at 12:42
$begingroup$
@D.Thomine Thanks. So in that case the probability that box 2,3,...n don't have car = $Pi_{i=0}^{n-2}(1 - frac{p}{n-i}) $ Is my calculation correct?
$endgroup$
– q126y
Jan 13 at 15:01
add a comment |
$begingroup$
Let us say we have n boxes, and one car.
The probability that one of them might have a car = p
and each box is equally likely to have a car, therefore the change that a single box has car = p/n
Now let us consider the chance that box 1 has car, given other boxes don't have car
P(Box 1 contains car | other n-1 boxes don't contain car)
$$=frac{P(Boxspace1space containsspace car cap otherspace n-1space boxesspace don'tspace containspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{P(Boxspace1space containsspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$ [1]
But if we apply this result to classical monty hall problem where we have 3 boxes. Let us say contestant chose box 1.
Other two boxes have p = 2/3 to have car, n = 2,
each box will have probability of p/n = $frac{2/3}{2} = 1/3$, of having a car
If monty opens up say box 3(which doesn't have car),
the probability of box 2 having car
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$
here p = 2/3
n = 2
$$=frac{frac{2/3}{2}}{ left( 1 - frac{2/3}{2} right)^{(2-1)} }$$
$$= 1/2$$
which is not the correct answer. I have done something wrong and [1] is not valid result.
What is wrong with it?
PS: I know how to solve monty hall problem using other approach. I want to know what is wrong with this approach?
probability probability-theory statistics monty-hall
asked Jan 13 at 10:30
q126yq126y
239212
$endgroup$
Let us say we have n boxes, and one car.
The probability that one of them might have a car = p
and each box is equally likely to have a car, therefore the change that a single box has car = p/n
Now let us consider the chance that box 1 has car, given other boxes don't have car
P(Box 1 contains car | other n-1 boxes don't contain car)
$$=frac{P(Boxspace1space containsspace car cap otherspace n-1space boxesspace don'tspace containspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{P(Boxspace1space containsspace car)}{P(otherspace n-1space boxesspace don'tspace containspace car)}$$
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$ [1]
But if we apply this result to classical monty hall problem where we have 3 boxes. Let us say contestant chose box 1.
Other two boxes have p = 2/3 to have car, n = 2,
each box will have probability of p/n = $frac{2/3}{2} = 1/3$, of having a car
If monty opens up say box 3(which doesn't have car),
the probability of box 2 having car
$$=frac{frac{p}{n}}{ left( 1 - frac{p}{n} right)^{(n-1)} }$$
here p = 2/3
n = 2
$$=frac{frac{2/3}{2}}{ left( 1 - frac{2/3}{2} right)^{(2-1)} }$$
$$= 1/2$$
which is not the correct answer. I have done something wrong and [1] is not valid result.
What is wrong with it?
PS: I know how to solve monty hall problem using other approach. I want to know what is wrong with this approach?
probability probability-theory statistics monty-hall
probability probability-theory statistics monty-hall
asked Jan 13 at 10:30
q126yq126y
239212
asked Jan 13 at 10:30
q126yq126y
239212
asked Jan 13 at 10:30
q126yq126y
239212
asked Jan 13 at 10:30
q126yq126y
239212
asked Jan 13 at 10:30
q126yq126y
239212
239212
1
$begingroup$
The events [box 2 has no car], [box 3 has no car], etc. are not independent (at most one of them is false, since there is only one car). Hence the probability of [other n-1 boxes don't have a car] is not the product of probabilities.
$endgroup$
– D. Thomine
Jan 13 at 12:42
$begingroup$
@D.Thomine Thanks. So in that case the probability that box 2,3,...n don't have car = $Pi_{i=0}^{n-2}(1 - frac{p}{n-i}) $ Is my calculation correct?
$endgroup$
– q126y
Jan 13 at 15:01
add a comment |
1
$begingroup$
The events [box 2 has no car], [box 3 has no car], etc. are not independent (at most one of them is false, since there is only one car). Hence the probability of [other n-1 boxes don't have a car] is not the product of probabilities.
$endgroup$
– D. Thomine
Jan 13 at 12:42
$begingroup$
@D.Thomine Thanks. So in that case the probability that box 2,3,...n don't have car = $Pi_{i=0}^{n-2}(1 - frac{p}{n-i}) $ Is my calculation correct?
$endgroup$
– q126y
Jan 13 at 15:01
1
1
$begingroup$
The events [box 2 has no car], [box 3 has no car], etc. are not independent (at most one of them is false, since there is only one car). Hence the probability of [other n-1 boxes don't have a car] is not the product of probabilities.
$endgroup$
– D. Thomine
Jan 13 at 12:42
$begingroup$
The events [box 2 has no car], [box 3 has no car], etc. are not independent (at most one of them is false, since there is only one car). Hence the probability of [other n-1 boxes don't have a car] is not the product of probabilities.
$endgroup$
– D. Thomine
Jan 13 at 12:42
$begingroup$
@D.Thomine Thanks. So in that case the probability that box 2,3,...n don't have car = $Pi_{i=0}^{n-2}(1 - frac{p}{n-i}) $ Is my calculation correct?
$endgroup$
– q126y
Jan 13 at 15:01
$begingroup$
@D.Thomine Thanks. So in that case the probability that box 2,3,...n don't have car = $Pi_{i=0}^{n-2}(1 - frac{p}{n-i}) $ Is my calculation correct?
$endgroup$
– q126y
Jan 13 at 15:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First off, your are not actually computing the Monty Hall Problem. You are computing the probability:
$$
P(text{box 2}midtext{not box 3})=frac12
$$
Whereas the Monty Hall Problem is concerned with the probability:
$$
P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23
$$
To comment on your derivation, by $n$ you do not mean the total number of boxes since otherwise specifying that $n-1$ boxes do not contain a car would be redundant. This was unclear when I read it at first.
Instead suppose we have $N>n$ boxes in total. Then you are asking:
What is the probability that the car is found in the first of the $n<N$ boxes given that it is NOT found in the other $(n-1)$ boxes.
To be clear, this entails the possibility that the car may in fact be in one of the remaining $N-n$ boxes.
Then most of your derivation is correct, except for your denominator. It considers the $(n-1)$ boxes as $(n-1)$ independent trials, which it is not. We are only considering a single trial here. Rather the denominator should be:
$$
P(text{not the other $n-1$ boxes})=1-(n-1)cdottfrac pn
$$
answered Jan 13 at 15:49
StringString
13.7k32756
$endgroup$
$begingroup$
"Whereas the Monty Hall Problem is concerned with the probability: $$ P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23 $$" Could your please elaborate on this a bit. I am not able to wrap my head around this formulation.
$endgroup$
– q126y
Jan 14 at 8:34
$begingroup$
Also, Monty knows car is in which box, how is that knowledge factored in this formulation?
$endgroup$
– q126y
Jan 14 at 8:46
$begingroup$
@q126y: In the classical problem, the contestant has (WLOG) chosen box 1. Then one of the other boxes is opened revealing either "not box 2" OR "not box 3". Now we are asked to evaluate the probability that the contestant should change their mind (away from box 1). This gives us two scenarios: $$P(text{box 2}midtext{not box 3})$$ OR $$P(text{box 3}midtext{not box 2})$$, which is what I tried to combine into the compressed expression: $$P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23$$
$endgroup$
– String
Jan 14 at 9:20
$begingroup$
@q126y: The knowledge Monty holds about the car is factored in exactly by stating the two possibilities "not box 2" OR "not box 3". He would never reveal a box containing the car, so if box 2 contains the car he would be forced to opt for the "not box 3"-condition.
$endgroup$
– String
Jan 14 at 9:30
$begingroup$
this is some new insight into the problem. Thanks.
$endgroup$
– q126y
Jan 14 at 12:43
|
show 1 more comment
$begingroup$
Firstly, your definition of $p$ is rather vague.
Try something in the lines of: Assuming there are $n$ boxes, each box has the probability of containing the car equal to $p = frac{1}{n}$.
Now, the probability you propose $$P(text{Box 1 contains car} | text{other n-1 boxes don't contain car}) = 1$$
since the statement "Box 1 contains car" is equivalent to "other n-1 boxes don't contain car" because there is only 1 car.
The probability you are looking for is
$$P(text{Box 1 contains car} | text{1 other box don't contain car}).$$
Other than that, I think you're on the right track.
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First off, your are not actually computing the Monty Hall Problem. You are computing the probability:
$$
P(text{box 2}midtext{not box 3})=frac12
$$
Whereas the Monty Hall Problem is concerned with the probability:
$$
P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23
$$
To comment on your derivation, by $n$ you do not mean the total number of boxes since otherwise specifying that $n-1$ boxes do not contain a car would be redundant. This was unclear when I read it at first.
Instead suppose we have $N>n$ boxes in total. Then you are asking:
What is the probability that the car is found in the first of the $n<N$ boxes given that it is NOT found in the other $(n-1)$ boxes.
To be clear, this entails the possibility that the car may in fact be in one of the remaining $N-n$ boxes.
Then most of your derivation is correct, except for your denominator. It considers the $(n-1)$ boxes as $(n-1)$ independent trials, which it is not. We are only considering a single trial here. Rather the denominator should be:
$$
P(text{not the other $n-1$ boxes})=1-(n-1)cdottfrac pn
$$
answered Jan 13 at 15:49
StringString
13.7k32756
$endgroup$
$begingroup$
"Whereas the Monty Hall Problem is concerned with the probability: $$ P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23 $$" Could your please elaborate on this a bit. I am not able to wrap my head around this formulation.
$endgroup$
– q126y
Jan 14 at 8:34
$begingroup$
Also, Monty knows car is in which box, how is that knowledge factored in this formulation?
$endgroup$
– q126y
Jan 14 at 8:46
$begingroup$
@q126y: In the classical problem, the contestant has (WLOG) chosen box 1. Then one of the other boxes is opened revealing either "not box 2" OR "not box 3". Now we are asked to evaluate the probability that the contestant should change their mind (away from box 1). This gives us two scenarios: $$P(text{box 2}midtext{not box 3})$$ OR $$P(text{box 3}midtext{not box 2})$$, which is what I tried to combine into the compressed expression: $$P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23$$
$endgroup$
– String
Jan 14 at 9:20
$begingroup$
@q126y: The knowledge Monty holds about the car is factored in exactly by stating the two possibilities "not box 2" OR "not box 3". He would never reveal a box containing the car, so if box 2 contains the car he would be forced to opt for the "not box 3"-condition.
$endgroup$
– String
Jan 14 at 9:30
$begingroup$
this is some new insight into the problem. Thanks.
$endgroup$
– q126y
Jan 14 at 12:43
|
show 1 more comment
$begingroup$
First off, your are not actually computing the Monty Hall Problem. You are computing the probability:
$$
P(text{box 2}midtext{not box 3})=frac12
$$
Whereas the Monty Hall Problem is concerned with the probability:
$$
P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23
$$
To comment on your derivation, by $n$ you do not mean the total number of boxes since otherwise specifying that $n-1$ boxes do not contain a car would be redundant. This was unclear when I read it at first.
Instead suppose we have $N>n$ boxes in total. Then you are asking:
What is the probability that the car is found in the first of the $n<N$ boxes given that it is NOT found in the other $(n-1)$ boxes.
To be clear, this entails the possibility that the car may in fact be in one of the remaining $N-n$ boxes.
Then most of your derivation is correct, except for your denominator. It considers the $(n-1)$ boxes as $(n-1)$ independent trials, which it is not. We are only considering a single trial here. Rather the denominator should be:
$$
P(text{not the other $n-1$ boxes})=1-(n-1)cdottfrac pn
$$
answered Jan 13 at 15:49
StringString
13.7k32756
$endgroup$
$begingroup$
"Whereas the Monty Hall Problem is concerned with the probability: $$ P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23 $$" Could your please elaborate on this a bit. I am not able to wrap my head around this formulation.
$endgroup$
– q126y
Jan 14 at 8:34
$begingroup$
Also, Monty knows car is in which box, how is that knowledge factored in this formulation?
$endgroup$
– q126y
Jan 14 at 8:46
$begingroup$
@q126y: In the classical problem, the contestant has (WLOG) chosen box 1. Then one of the other boxes is opened revealing either "not box 2" OR "not box 3". Now we are asked to evaluate the probability that the contestant should change their mind (away from box 1). This gives us two scenarios: $$P(text{box 2}midtext{not box 3})$$ OR $$P(text{box 3}midtext{not box 2})$$, which is what I tried to combine into the compressed expression: $$P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23$$
$endgroup$
– String
Jan 14 at 9:20
$begingroup$
@q126y: The knowledge Monty holds about the car is factored in exactly by stating the two possibilities "not box 2" OR "not box 3". He would never reveal a box containing the car, so if box 2 contains the car he would be forced to opt for the "not box 3"-condition.
$endgroup$
– String
Jan 14 at 9:30
$begingroup$
this is some new insight into the problem. Thanks.
$endgroup$
– q126y
Jan 14 at 12:43
|
show 1 more comment
$begingroup$
First off, your are not actually computing the Monty Hall Problem. You are computing the probability:
$$
P(text{box 2}midtext{not box 3})=frac12
$$
Whereas the Monty Hall Problem is concerned with the probability:
$$
P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23
$$
To comment on your derivation, by $n$ you do not mean the total number of boxes since otherwise specifying that $n-1$ boxes do not contain a car would be redundant. This was unclear when I read it at first.
Instead suppose we have $N>n$ boxes in total. Then you are asking:
What is the probability that the car is found in the first of the $n<N$ boxes given that it is NOT found in the other $(n-1)$ boxes.
To be clear, this entails the possibility that the car may in fact be in one of the remaining $N-n$ boxes.
Then most of your derivation is correct, except for your denominator. It considers the $(n-1)$ boxes as $(n-1)$ independent trials, which it is not. We are only considering a single trial here. Rather the denominator should be:
$$
P(text{not the other $n-1$ boxes})=1-(n-1)cdottfrac pn
$$
answered Jan 13 at 15:49
StringString
13.7k32756
$endgroup$
First off, your are not actually computing the Monty Hall Problem. You are computing the probability:
$$
P(text{box 2}midtext{not box 3})=frac12
$$
Whereas the Monty Hall Problem is concerned with the probability:
$$
P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23
$$
To comment on your derivation, by $n$ you do not mean the total number of boxes since otherwise specifying that $n-1$ boxes do not contain a car would be redundant. This was unclear when I read it at first.
Instead suppose we have $N>n$ boxes in total. Then you are asking:
What is the probability that the car is found in the first of the $n<N$ boxes given that it is NOT found in the other $(n-1)$ boxes.
To be clear, this entails the possibility that the car may in fact be in one of the remaining $N-n$ boxes.
Then most of your derivation is correct, except for your denominator. It considers the $(n-1)$ boxes as $(n-1)$ independent trials, which it is not. We are only considering a single trial here. Rather the denominator should be:
$$
P(text{not the other $n-1$ boxes})=1-(n-1)cdottfrac pn
$$
answered Jan 13 at 15:49
StringString
13.7k32756
edited Jan 13 at 15:54
answered Jan 13 at 15:49
StringString
13.7k32756
answered Jan 13 at 15:49
StringString
13.7k32756
answered Jan 13 at 15:49
StringString
13.7k32756
13.7k32756
$begingroup$
"Whereas the Monty Hall Problem is concerned with the probability: $$ P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23 $$" Could your please elaborate on this a bit. I am not able to wrap my head around this formulation.
$endgroup$
– q126y
Jan 14 at 8:34
$begingroup$
Also, Monty knows car is in which box, how is that knowledge factored in this formulation?
$endgroup$
– q126y
Jan 14 at 8:46
$begingroup$
@q126y: In the classical problem, the contestant has (WLOG) chosen box 1. Then one of the other boxes is opened revealing either "not box 2" OR "not box 3". Now we are asked to evaluate the probability that the contestant should change their mind (away from box 1). This gives us two scenarios: $$P(text{box 2}midtext{not box 3})$$ OR $$P(text{box 3}midtext{not box 2})$$, which is what I tried to combine into the compressed expression: $$P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23$$
$endgroup$
– String
Jan 14 at 9:20
$begingroup$
@q126y: The knowledge Monty holds about the car is factored in exactly by stating the two possibilities "not box 2" OR "not box 3". He would never reveal a box containing the car, so if box 2 contains the car he would be forced to opt for the "not box 3"-condition.
$endgroup$
– String
Jan 14 at 9:30
$begingroup$
this is some new insight into the problem. Thanks.
$endgroup$
– q126y
Jan 14 at 12:43
|
show 1 more comment
$begingroup$
"Whereas the Monty Hall Problem is concerned with the probability: $$ P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23 $$" Could your please elaborate on this a bit. I am not able to wrap my head around this formulation.
$endgroup$
– q126y
Jan 14 at 8:34
$begingroup$
Also, Monty knows car is in which box, how is that knowledge factored in this formulation?
$endgroup$
– q126y
Jan 14 at 8:46
$begingroup$
@q126y: In the classical problem, the contestant has (WLOG) chosen box 1. Then one of the other boxes is opened revealing either "not box 2" OR "not box 3". Now we are asked to evaluate the probability that the contestant should change their mind (away from box 1). This gives us two scenarios: $$P(text{box 2}midtext{not box 3})$$ OR $$P(text{box 3}midtext{not box 2})$$, which is what I tried to combine into the compressed expression: $$P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23$$
$endgroup$
– String
Jan 14 at 9:20
$begingroup$
@q126y: The knowledge Monty holds about the car is factored in exactly by stating the two possibilities "not box 2" OR "not box 3". He would never reveal a box containing the car, so if box 2 contains the car he would be forced to opt for the "not box 3"-condition.
$endgroup$
– String
Jan 14 at 9:30
$begingroup$
this is some new insight into the problem. Thanks.
$endgroup$
– q126y
Jan 14 at 12:43
$begingroup$
"Whereas the Monty Hall Problem is concerned with the probability: $$ P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23 $$" Could your please elaborate on this a bit. I am not able to wrap my head around this formulation.
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– q126y
Jan 14 at 8:34
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"Whereas the Monty Hall Problem is concerned with the probability: $$ P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23 $$" Could your please elaborate on this a bit. I am not able to wrap my head around this formulation.
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– q126y
Jan 14 at 8:34
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Also, Monty knows car is in which box, how is that knowledge factored in this formulation?
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– q126y
Jan 14 at 8:46
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Also, Monty knows car is in which box, how is that knowledge factored in this formulation?
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– q126y
Jan 14 at 8:46
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@q126y: In the classical problem, the contestant has (WLOG) chosen box 1. Then one of the other boxes is opened revealing either "not box 2" OR "not box 3". Now we are asked to evaluate the probability that the contestant should change their mind (away from box 1). This gives us two scenarios: $$P(text{box 2}midtext{not box 3})$$ OR $$P(text{box 3}midtext{not box 2})$$, which is what I tried to combine into the compressed expression: $$P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23$$
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– String
Jan 14 at 9:20
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@q126y: In the classical problem, the contestant has (WLOG) chosen box 1. Then one of the other boxes is opened revealing either "not box 2" OR "not box 3". Now we are asked to evaluate the probability that the contestant should change their mind (away from box 1). This gives us two scenarios: $$P(text{box 2}midtext{not box 3})$$ OR $$P(text{box 3}midtext{not box 2})$$, which is what I tried to combine into the compressed expression: $$P(text{box 2 OR box 3}midtext{(not box 3) OR (not box 2)})=frac23$$
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– String
Jan 14 at 9:20
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@q126y: The knowledge Monty holds about the car is factored in exactly by stating the two possibilities "not box 2" OR "not box 3". He would never reveal a box containing the car, so if box 2 contains the car he would be forced to opt for the "not box 3"-condition.
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– String
Jan 14 at 9:30
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@q126y: The knowledge Monty holds about the car is factored in exactly by stating the two possibilities "not box 2" OR "not box 3". He would never reveal a box containing the car, so if box 2 contains the car he would be forced to opt for the "not box 3"-condition.
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– String
Jan 14 at 9:30
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this is some new insight into the problem. Thanks.
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– q126y
Jan 14 at 12:43
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this is some new insight into the problem. Thanks.
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– q126y
Jan 14 at 12:43
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show 1 more comment
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Firstly, your definition of $p$ is rather vague.
Try something in the lines of: Assuming there are $n$ boxes, each box has the probability of containing the car equal to $p = frac{1}{n}$.
Now, the probability you propose $$P(text{Box 1 contains car} | text{other n-1 boxes don't contain car}) = 1$$
since the statement "Box 1 contains car" is equivalent to "other n-1 boxes don't contain car" because there is only 1 car.
The probability you are looking for is
$$P(text{Box 1 contains car} | text{1 other box don't contain car}).$$
Other than that, I think you're on the right track.
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
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add a comment |
$begingroup$
Firstly, your definition of $p$ is rather vague.
Try something in the lines of: Assuming there are $n$ boxes, each box has the probability of containing the car equal to $p = frac{1}{n}$.
Now, the probability you propose $$P(text{Box 1 contains car} | text{other n-1 boxes don't contain car}) = 1$$
since the statement "Box 1 contains car" is equivalent to "other n-1 boxes don't contain car" because there is only 1 car.
The probability you are looking for is
$$P(text{Box 1 contains car} | text{1 other box don't contain car}).$$
Other than that, I think you're on the right track.
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
$endgroup$
add a comment |
$begingroup$
Firstly, your definition of $p$ is rather vague.
Try something in the lines of: Assuming there are $n$ boxes, each box has the probability of containing the car equal to $p = frac{1}{n}$.
Now, the probability you propose $$P(text{Box 1 contains car} | text{other n-1 boxes don't contain car}) = 1$$
since the statement "Box 1 contains car" is equivalent to "other n-1 boxes don't contain car" because there is only 1 car.
The probability you are looking for is
$$P(text{Box 1 contains car} | text{1 other box don't contain car}).$$
Other than that, I think you're on the right track.
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
$endgroup$
Firstly, your definition of $p$ is rather vague.
Try something in the lines of: Assuming there are $n$ boxes, each box has the probability of containing the car equal to $p = frac{1}{n}$.
Now, the probability you propose $$P(text{Box 1 contains car} | text{other n-1 boxes don't contain car}) = 1$$
since the statement "Box 1 contains car" is equivalent to "other n-1 boxes don't contain car" because there is only 1 car.
The probability you are looking for is
$$P(text{Box 1 contains car} | text{1 other box don't contain car}).$$
Other than that, I think you're on the right track.
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
answered Jan 13 at 10:53
Mads HulgaardMads Hulgaard
748
748
add a comment |
add a comment |
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The events [box 2 has no car], [box 3 has no car], etc. are not independent (at most one of them is false, since there is only one car). Hence the probability of [other n-1 boxes don't have a car] is not the product of probabilities.
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– D. Thomine
Jan 13 at 12:42
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@D.Thomine Thanks. So in that case the probability that box 2,3,...n don't have car = $Pi_{i=0}^{n-2}(1 - frac{p}{n-i}) $ Is my calculation correct?
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– q126y
Jan 13 at 15:01