Mixing
Should I use conditional probability/prior probability here?
$begingroup$
I have been debating the following problem with my friends over the past few days:
“I'm going on a holiday to Mumbai. I'm curious about the weather there so I call my friend who lives there and ask him if it is raining. My friend is known for being irritating and everytime you ask him a question he has a one in third chance of lying. He assures me that it is raining in Mumbai at the moment. What is the probability of it actually raining?”
The person who posed the problem tells me the solution is two thirds by the following reasoning: The statements ‘It is raining in Mumbai’ and ‘Your friend is telling the truth’ are equivalent after you hear his response so their probabilities are the same.
But this feels weird to me. The same reasoning can be made if we replace ‘Is it raining in Mumbai’ with ’Is there a UFO parked on the town square?’ and again the answer is two thirds. Shouldn't you take the rarity of the event into account?.
If I assign a probability $p$ to the event of rain on any given day, I have calculated the probability of rain, given his answer, to be $frac{2p}{1+p}$. This was done by writing out the four possible cases and using conditional probability. If I want this expression to equal two thirds, $p$ has to be one half, which makes sense to me. If no outcome is preferred over the other, you can only make a decision based on the probability of being lied to. Otherwise the response of the friend needs to be taken with a grain of salt.
But on the other hand, the original explanation is so simple I cannot find a hole in it. My intuition needs to be sharpened but in which way? Are you really allowed to disregard the probability of it raining on a given day? Why? Maybe some analogy with coin flips or dice rolls could help... It could also be that this phenomenon has a name but I'm not aware of it. Therefore I do not know where to start looking.
probability conditional-probability
edited Jan 11 at 14:17
lioness99a
3,7492727
asked Jan 11 at 14:16
JKaertsJKaerts
506
$endgroup$
add a comment |
$begingroup$
I have been debating the following problem with my friends over the past few days:
“I'm going on a holiday to Mumbai. I'm curious about the weather there so I call my friend who lives there and ask him if it is raining. My friend is known for being irritating and everytime you ask him a question he has a one in third chance of lying. He assures me that it is raining in Mumbai at the moment. What is the probability of it actually raining?”
The person who posed the problem tells me the solution is two thirds by the following reasoning: The statements ‘It is raining in Mumbai’ and ‘Your friend is telling the truth’ are equivalent after you hear his response so their probabilities are the same.
But this feels weird to me. The same reasoning can be made if we replace ‘Is it raining in Mumbai’ with ’Is there a UFO parked on the town square?’ and again the answer is two thirds. Shouldn't you take the rarity of the event into account?.
If I assign a probability $p$ to the event of rain on any given day, I have calculated the probability of rain, given his answer, to be $frac{2p}{1+p}$. This was done by writing out the four possible cases and using conditional probability. If I want this expression to equal two thirds, $p$ has to be one half, which makes sense to me. If no outcome is preferred over the other, you can only make a decision based on the probability of being lied to. Otherwise the response of the friend needs to be taken with a grain of salt.
But on the other hand, the original explanation is so simple I cannot find a hole in it. My intuition needs to be sharpened but in which way? Are you really allowed to disregard the probability of it raining on a given day? Why? Maybe some analogy with coin flips or dice rolls could help... It could also be that this phenomenon has a name but I'm not aware of it. Therefore I do not know where to start looking.
probability conditional-probability
edited Jan 11 at 14:17
lioness99a
3,7492727
asked Jan 11 at 14:16
JKaertsJKaerts
506
$endgroup$
1
$begingroup$
Your intuition is correct. You need a prior for the probability that it is in fact raining. Suppose, say, that it is always raining in Mumbai. Then the correct answer is $1$, regardless of what your annoying friend says. For a general prior $p$, your answer, $frac {2p}{p+1}$ is correct.
$endgroup$
– lulu
Jan 11 at 14:17
add a comment |
$begingroup$
I have been debating the following problem with my friends over the past few days:
“I'm going on a holiday to Mumbai. I'm curious about the weather there so I call my friend who lives there and ask him if it is raining. My friend is known for being irritating and everytime you ask him a question he has a one in third chance of lying. He assures me that it is raining in Mumbai at the moment. What is the probability of it actually raining?”
The person who posed the problem tells me the solution is two thirds by the following reasoning: The statements ‘It is raining in Mumbai’ and ‘Your friend is telling the truth’ are equivalent after you hear his response so their probabilities are the same.
But this feels weird to me. The same reasoning can be made if we replace ‘Is it raining in Mumbai’ with ’Is there a UFO parked on the town square?’ and again the answer is two thirds. Shouldn't you take the rarity of the event into account?.
If I assign a probability $p$ to the event of rain on any given day, I have calculated the probability of rain, given his answer, to be $frac{2p}{1+p}$. This was done by writing out the four possible cases and using conditional probability. If I want this expression to equal two thirds, $p$ has to be one half, which makes sense to me. If no outcome is preferred over the other, you can only make a decision based on the probability of being lied to. Otherwise the response of the friend needs to be taken with a grain of salt.
But on the other hand, the original explanation is so simple I cannot find a hole in it. My intuition needs to be sharpened but in which way? Are you really allowed to disregard the probability of it raining on a given day? Why? Maybe some analogy with coin flips or dice rolls could help... It could also be that this phenomenon has a name but I'm not aware of it. Therefore I do not know where to start looking.
probability conditional-probability
edited Jan 11 at 14:17
lioness99a
3,7492727
asked Jan 11 at 14:16
JKaertsJKaerts
506
$endgroup$
I have been debating the following problem with my friends over the past few days:
“I'm going on a holiday to Mumbai. I'm curious about the weather there so I call my friend who lives there and ask him if it is raining. My friend is known for being irritating and everytime you ask him a question he has a one in third chance of lying. He assures me that it is raining in Mumbai at the moment. What is the probability of it actually raining?”
The person who posed the problem tells me the solution is two thirds by the following reasoning: The statements ‘It is raining in Mumbai’ and ‘Your friend is telling the truth’ are equivalent after you hear his response so their probabilities are the same.
But this feels weird to me. The same reasoning can be made if we replace ‘Is it raining in Mumbai’ with ’Is there a UFO parked on the town square?’ and again the answer is two thirds. Shouldn't you take the rarity of the event into account?.
If I assign a probability $p$ to the event of rain on any given day, I have calculated the probability of rain, given his answer, to be $frac{2p}{1+p}$. This was done by writing out the four possible cases and using conditional probability. If I want this expression to equal two thirds, $p$ has to be one half, which makes sense to me. If no outcome is preferred over the other, you can only make a decision based on the probability of being lied to. Otherwise the response of the friend needs to be taken with a grain of salt.
But on the other hand, the original explanation is so simple I cannot find a hole in it. My intuition needs to be sharpened but in which way? Are you really allowed to disregard the probability of it raining on a given day? Why? Maybe some analogy with coin flips or dice rolls could help... It could also be that this phenomenon has a name but I'm not aware of it. Therefore I do not know where to start looking.
probability conditional-probability
probability conditional-probability
edited Jan 11 at 14:17
lioness99a
3,7492727
asked Jan 11 at 14:16
JKaertsJKaerts
506
edited Jan 11 at 14:17
lioness99a
3,7492727
asked Jan 11 at 14:16
JKaertsJKaerts
506
edited Jan 11 at 14:17
lioness99a
3,7492727
edited Jan 11 at 14:17
lioness99a
3,7492727
edited Jan 11 at 14:17
lioness99a
3,7492727
3,7492727
asked Jan 11 at 14:16
JKaertsJKaerts
506
asked Jan 11 at 14:16
JKaertsJKaerts
506
asked Jan 11 at 14:16
JKaertsJKaerts
506
506
1
$begingroup$
Your intuition is correct. You need a prior for the probability that it is in fact raining. Suppose, say, that it is always raining in Mumbai. Then the correct answer is $1$, regardless of what your annoying friend says. For a general prior $p$, your answer, $frac {2p}{p+1}$ is correct.
$endgroup$
– lulu
Jan 11 at 14:17
add a comment |
1
$begingroup$
Your intuition is correct. You need a prior for the probability that it is in fact raining. Suppose, say, that it is always raining in Mumbai. Then the correct answer is $1$, regardless of what your annoying friend says. For a general prior $p$, your answer, $frac {2p}{p+1}$ is correct.
$endgroup$
– lulu
Jan 11 at 14:17
1
1
$begingroup$
Your intuition is correct. You need a prior for the probability that it is in fact raining. Suppose, say, that it is always raining in Mumbai. Then the correct answer is $1$, regardless of what your annoying friend says. For a general prior $p$, your answer, $frac {2p}{p+1}$ is correct.
$endgroup$
– lulu
Jan 11 at 14:17
$begingroup$
Your intuition is correct. You need a prior for the probability that it is in fact raining. Suppose, say, that it is always raining in Mumbai. Then the correct answer is $1$, regardless of what your annoying friend says. For a general prior $p$, your answer, $frac {2p}{p+1}$ is correct.
$endgroup$
– lulu
Jan 11 at 14:17
add a comment |
2 Answers
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oldest
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$begingroup$
A practical version of this problem is the old standard: "Suppose you have a test for a disease which occurs in $.001%$ of the population. Your test is accurate in $99%$ of cases (that is, a person with the disease tests negative for it $1%$ of the time and a health person tests positive for it $1%$ of the time. Your patient tests positive for the disease....what is the probability that the patient actually has it?"
Here, as in your case, the prior probability that the patient has the disease has to be taken into account. Sure the test result is evidence for the disease, but that must be weighed against the rarity of the disease. In your case, your annoying friend's claim that it is raining is certainly evidence that it is raining, but that evidence must be wighed against the underlying probability of rain.
For the disease problem: there are two ways to get a positive result; either you have the disease and the test is correct, or you are healthy and the test is in error. Thus $Psi$, the probability that the test result is positive is given by
$$Psi=.00001times .99+ (1-.00001)*.01=.0100098$$
Of that, $.00001times .99 = .0000099$ is explained by actually having the disease so the probability that you have the disease, given a positive test result, is $$frac {.0000099}{.0100098}=.000989$$
Thus, your estimate of the probability has gone way up (from $.00001$ to slightly less than $.001$), but it is still very unlikely that you actually have the disease.
In your case, there are two reasons your friend might say it is raining. Either it is raining and he tells the truth or it isn't raining and he lies. If $p$ is the probability of rain then the probability that he says it is raining is therefore $$ptimes frac 23+(1-p)times frac 13=frac {p+1}3$$
Of that, the portion which is explained by actual rain is $ptimes frac 23$ so the probability that it is raining is $$frac {2p/3}{(p+1)/3}=frac {2p}{p+1}$$ just as you say.
Sanity checks: Note that if $p=0$ this is $0$, confirming your thought about the UFO. And if $p=1$ then this is $1$ as it should be. If $p=frac 12$ then you do get $frac 23$ which is, I think, intuitively correct...if you had no idea at all whether or not it was raining then your friend's claim is all the evidence there is, so you are just stuck with the hope that he is telling the truth. If, say, someone tossed a coin and you (unwisely) asked your friend to say if it came up Heads, his claim that it did would imply a $frac 23$ chance that it actually did.
More subtle sanity check: Your friend's claim ought to be evidence, even if he is unreliable. That is, it would violate intuition if his claim that it is raining lowered your estimate of the probability that it is raining. Here you can check that $frac {2p}{p+1}≥p$ for $0≤p≤1$ and that equality holds only in the extreme cases $p=0,1$.
answered Jan 11 at 14:59
lulululu
41.1k24979
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add a comment |
$begingroup$
You are correct if the original event does need to be taken into account (as would usually be the case) - if an event has no chance of happening and your friend says it does, then obviously your friend is lying and the chance is $0$.
However, if your friend just lies $frac{1}{3}$ of the time regardless of the event, then the chance would always be $frac{2}{3}$ of the event happening. Of course, the problem with this argument is that your friend shouldn't be able to dictate the result of an event, but that would be the case if your friend just lies a third of the time no matter what.
tl;dr I think you're right realistically, but as a math problem the problem writer would technically be right, though if I were the writer I'd probably pose a different scenario to get the idea to work. With what you did, this might be a nice conditional probability question, though.
answered Jan 11 at 14:21
pie314271pie314271
2,114819
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
A practical version of this problem is the old standard: "Suppose you have a test for a disease which occurs in $.001%$ of the population. Your test is accurate in $99%$ of cases (that is, a person with the disease tests negative for it $1%$ of the time and a health person tests positive for it $1%$ of the time. Your patient tests positive for the disease....what is the probability that the patient actually has it?"
Here, as in your case, the prior probability that the patient has the disease has to be taken into account. Sure the test result is evidence for the disease, but that must be weighed against the rarity of the disease. In your case, your annoying friend's claim that it is raining is certainly evidence that it is raining, but that evidence must be wighed against the underlying probability of rain.
For the disease problem: there are two ways to get a positive result; either you have the disease and the test is correct, or you are healthy and the test is in error. Thus $Psi$, the probability that the test result is positive is given by
$$Psi=.00001times .99+ (1-.00001)*.01=.0100098$$
Of that, $.00001times .99 = .0000099$ is explained by actually having the disease so the probability that you have the disease, given a positive test result, is $$frac {.0000099}{.0100098}=.000989$$
Thus, your estimate of the probability has gone way up (from $.00001$ to slightly less than $.001$), but it is still very unlikely that you actually have the disease.
In your case, there are two reasons your friend might say it is raining. Either it is raining and he tells the truth or it isn't raining and he lies. If $p$ is the probability of rain then the probability that he says it is raining is therefore $$ptimes frac 23+(1-p)times frac 13=frac {p+1}3$$
Of that, the portion which is explained by actual rain is $ptimes frac 23$ so the probability that it is raining is $$frac {2p/3}{(p+1)/3}=frac {2p}{p+1}$$ just as you say.
Sanity checks: Note that if $p=0$ this is $0$, confirming your thought about the UFO. And if $p=1$ then this is $1$ as it should be. If $p=frac 12$ then you do get $frac 23$ which is, I think, intuitively correct...if you had no idea at all whether or not it was raining then your friend's claim is all the evidence there is, so you are just stuck with the hope that he is telling the truth. If, say, someone tossed a coin and you (unwisely) asked your friend to say if it came up Heads, his claim that it did would imply a $frac 23$ chance that it actually did.
More subtle sanity check: Your friend's claim ought to be evidence, even if he is unreliable. That is, it would violate intuition if his claim that it is raining lowered your estimate of the probability that it is raining. Here you can check that $frac {2p}{p+1}≥p$ for $0≤p≤1$ and that equality holds only in the extreme cases $p=0,1$.
answered Jan 11 at 14:59
lulululu
41.1k24979
$endgroup$
add a comment |
$begingroup$
A practical version of this problem is the old standard: "Suppose you have a test for a disease which occurs in $.001%$ of the population. Your test is accurate in $99%$ of cases (that is, a person with the disease tests negative for it $1%$ of the time and a health person tests positive for it $1%$ of the time. Your patient tests positive for the disease....what is the probability that the patient actually has it?"
Here, as in your case, the prior probability that the patient has the disease has to be taken into account. Sure the test result is evidence for the disease, but that must be weighed against the rarity of the disease. In your case, your annoying friend's claim that it is raining is certainly evidence that it is raining, but that evidence must be wighed against the underlying probability of rain.
For the disease problem: there are two ways to get a positive result; either you have the disease and the test is correct, or you are healthy and the test is in error. Thus $Psi$, the probability that the test result is positive is given by
$$Psi=.00001times .99+ (1-.00001)*.01=.0100098$$
Of that, $.00001times .99 = .0000099$ is explained by actually having the disease so the probability that you have the disease, given a positive test result, is $$frac {.0000099}{.0100098}=.000989$$
Thus, your estimate of the probability has gone way up (from $.00001$ to slightly less than $.001$), but it is still very unlikely that you actually have the disease.
In your case, there are two reasons your friend might say it is raining. Either it is raining and he tells the truth or it isn't raining and he lies. If $p$ is the probability of rain then the probability that he says it is raining is therefore $$ptimes frac 23+(1-p)times frac 13=frac {p+1}3$$
Of that, the portion which is explained by actual rain is $ptimes frac 23$ so the probability that it is raining is $$frac {2p/3}{(p+1)/3}=frac {2p}{p+1}$$ just as you say.
Sanity checks: Note that if $p=0$ this is $0$, confirming your thought about the UFO. And if $p=1$ then this is $1$ as it should be. If $p=frac 12$ then you do get $frac 23$ which is, I think, intuitively correct...if you had no idea at all whether or not it was raining then your friend's claim is all the evidence there is, so you are just stuck with the hope that he is telling the truth. If, say, someone tossed a coin and you (unwisely) asked your friend to say if it came up Heads, his claim that it did would imply a $frac 23$ chance that it actually did.
More subtle sanity check: Your friend's claim ought to be evidence, even if he is unreliable. That is, it would violate intuition if his claim that it is raining lowered your estimate of the probability that it is raining. Here you can check that $frac {2p}{p+1}≥p$ for $0≤p≤1$ and that equality holds only in the extreme cases $p=0,1$.
answered Jan 11 at 14:59
lulululu
41.1k24979
$endgroup$
add a comment |
$begingroup$
A practical version of this problem is the old standard: "Suppose you have a test for a disease which occurs in $.001%$ of the population. Your test is accurate in $99%$ of cases (that is, a person with the disease tests negative for it $1%$ of the time and a health person tests positive for it $1%$ of the time. Your patient tests positive for the disease....what is the probability that the patient actually has it?"
Here, as in your case, the prior probability that the patient has the disease has to be taken into account. Sure the test result is evidence for the disease, but that must be weighed against the rarity of the disease. In your case, your annoying friend's claim that it is raining is certainly evidence that it is raining, but that evidence must be wighed against the underlying probability of rain.
For the disease problem: there are two ways to get a positive result; either you have the disease and the test is correct, or you are healthy and the test is in error. Thus $Psi$, the probability that the test result is positive is given by
$$Psi=.00001times .99+ (1-.00001)*.01=.0100098$$
Of that, $.00001times .99 = .0000099$ is explained by actually having the disease so the probability that you have the disease, given a positive test result, is $$frac {.0000099}{.0100098}=.000989$$
Thus, your estimate of the probability has gone way up (from $.00001$ to slightly less than $.001$), but it is still very unlikely that you actually have the disease.
In your case, there are two reasons your friend might say it is raining. Either it is raining and he tells the truth or it isn't raining and he lies. If $p$ is the probability of rain then the probability that he says it is raining is therefore $$ptimes frac 23+(1-p)times frac 13=frac {p+1}3$$
Of that, the portion which is explained by actual rain is $ptimes frac 23$ so the probability that it is raining is $$frac {2p/3}{(p+1)/3}=frac {2p}{p+1}$$ just as you say.
Sanity checks: Note that if $p=0$ this is $0$, confirming your thought about the UFO. And if $p=1$ then this is $1$ as it should be. If $p=frac 12$ then you do get $frac 23$ which is, I think, intuitively correct...if you had no idea at all whether or not it was raining then your friend's claim is all the evidence there is, so you are just stuck with the hope that he is telling the truth. If, say, someone tossed a coin and you (unwisely) asked your friend to say if it came up Heads, his claim that it did would imply a $frac 23$ chance that it actually did.
More subtle sanity check: Your friend's claim ought to be evidence, even if he is unreliable. That is, it would violate intuition if his claim that it is raining lowered your estimate of the probability that it is raining. Here you can check that $frac {2p}{p+1}≥p$ for $0≤p≤1$ and that equality holds only in the extreme cases $p=0,1$.
answered Jan 11 at 14:59
lulululu
41.1k24979
$endgroup$
A practical version of this problem is the old standard: "Suppose you have a test for a disease which occurs in $.001%$ of the population. Your test is accurate in $99%$ of cases (that is, a person with the disease tests negative for it $1%$ of the time and a health person tests positive for it $1%$ of the time. Your patient tests positive for the disease....what is the probability that the patient actually has it?"
Here, as in your case, the prior probability that the patient has the disease has to be taken into account. Sure the test result is evidence for the disease, but that must be weighed against the rarity of the disease. In your case, your annoying friend's claim that it is raining is certainly evidence that it is raining, but that evidence must be wighed against the underlying probability of rain.
For the disease problem: there are two ways to get a positive result; either you have the disease and the test is correct, or you are healthy and the test is in error. Thus $Psi$, the probability that the test result is positive is given by
$$Psi=.00001times .99+ (1-.00001)*.01=.0100098$$
Of that, $.00001times .99 = .0000099$ is explained by actually having the disease so the probability that you have the disease, given a positive test result, is $$frac {.0000099}{.0100098}=.000989$$
Thus, your estimate of the probability has gone way up (from $.00001$ to slightly less than $.001$), but it is still very unlikely that you actually have the disease.
In your case, there are two reasons your friend might say it is raining. Either it is raining and he tells the truth or it isn't raining and he lies. If $p$ is the probability of rain then the probability that he says it is raining is therefore $$ptimes frac 23+(1-p)times frac 13=frac {p+1}3$$
Of that, the portion which is explained by actual rain is $ptimes frac 23$ so the probability that it is raining is $$frac {2p/3}{(p+1)/3}=frac {2p}{p+1}$$ just as you say.
Sanity checks: Note that if $p=0$ this is $0$, confirming your thought about the UFO. And if $p=1$ then this is $1$ as it should be. If $p=frac 12$ then you do get $frac 23$ which is, I think, intuitively correct...if you had no idea at all whether or not it was raining then your friend's claim is all the evidence there is, so you are just stuck with the hope that he is telling the truth. If, say, someone tossed a coin and you (unwisely) asked your friend to say if it came up Heads, his claim that it did would imply a $frac 23$ chance that it actually did.
More subtle sanity check: Your friend's claim ought to be evidence, even if he is unreliable. That is, it would violate intuition if his claim that it is raining lowered your estimate of the probability that it is raining. Here you can check that $frac {2p}{p+1}≥p$ for $0≤p≤1$ and that equality holds only in the extreme cases $p=0,1$.
answered Jan 11 at 14:59
lulululu
41.1k24979
edited Jan 11 at 15:05
answered Jan 11 at 14:59
lulululu
41.1k24979
answered Jan 11 at 14:59
lulululu
41.1k24979
answered Jan 11 at 14:59
lulululu
41.1k24979
41.1k24979
add a comment |
add a comment |
$begingroup$
You are correct if the original event does need to be taken into account (as would usually be the case) - if an event has no chance of happening and your friend says it does, then obviously your friend is lying and the chance is $0$.
However, if your friend just lies $frac{1}{3}$ of the time regardless of the event, then the chance would always be $frac{2}{3}$ of the event happening. Of course, the problem with this argument is that your friend shouldn't be able to dictate the result of an event, but that would be the case if your friend just lies a third of the time no matter what.
tl;dr I think you're right realistically, but as a math problem the problem writer would technically be right, though if I were the writer I'd probably pose a different scenario to get the idea to work. With what you did, this might be a nice conditional probability question, though.
answered Jan 11 at 14:21
pie314271pie314271
2,114819
$endgroup$
add a comment |
$begingroup$
You are correct if the original event does need to be taken into account (as would usually be the case) - if an event has no chance of happening and your friend says it does, then obviously your friend is lying and the chance is $0$.
However, if your friend just lies $frac{1}{3}$ of the time regardless of the event, then the chance would always be $frac{2}{3}$ of the event happening. Of course, the problem with this argument is that your friend shouldn't be able to dictate the result of an event, but that would be the case if your friend just lies a third of the time no matter what.
tl;dr I think you're right realistically, but as a math problem the problem writer would technically be right, though if I were the writer I'd probably pose a different scenario to get the idea to work. With what you did, this might be a nice conditional probability question, though.
answered Jan 11 at 14:21
pie314271pie314271
2,114819
$endgroup$
add a comment |
$begingroup$
You are correct if the original event does need to be taken into account (as would usually be the case) - if an event has no chance of happening and your friend says it does, then obviously your friend is lying and the chance is $0$.
However, if your friend just lies $frac{1}{3}$ of the time regardless of the event, then the chance would always be $frac{2}{3}$ of the event happening. Of course, the problem with this argument is that your friend shouldn't be able to dictate the result of an event, but that would be the case if your friend just lies a third of the time no matter what.
tl;dr I think you're right realistically, but as a math problem the problem writer would technically be right, though if I were the writer I'd probably pose a different scenario to get the idea to work. With what you did, this might be a nice conditional probability question, though.
answered Jan 11 at 14:21
pie314271pie314271
2,114819
$endgroup$
You are correct if the original event does need to be taken into account (as would usually be the case) - if an event has no chance of happening and your friend says it does, then obviously your friend is lying and the chance is $0$.
However, if your friend just lies $frac{1}{3}$ of the time regardless of the event, then the chance would always be $frac{2}{3}$ of the event happening. Of course, the problem with this argument is that your friend shouldn't be able to dictate the result of an event, but that would be the case if your friend just lies a third of the time no matter what.
tl;dr I think you're right realistically, but as a math problem the problem writer would technically be right, though if I were the writer I'd probably pose a different scenario to get the idea to work. With what you did, this might be a nice conditional probability question, though.
answered Jan 11 at 14:21
pie314271pie314271
2,114819
answered Jan 11 at 14:21
pie314271pie314271
2,114819
answered Jan 11 at 14:21
pie314271pie314271
2,114819
answered Jan 11 at 14:21
pie314271pie314271
2,114819
2,114819
add a comment |
add a comment |
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$begingroup$
Your intuition is correct. You need a prior for the probability that it is in fact raining. Suppose, say, that it is always raining in Mumbai. Then the correct answer is $1$, regardless of what your annoying friend says. For a general prior $p$, your answer, $frac {2p}{p+1}$ is correct.
$endgroup$
– lulu
Jan 11 at 14:17