Probability of picking the right answer in a yes no question.
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Suppose two people have a yes / no question that the probability for each of them to answer correctly is $ p $. Each one answer is independent of the other one. What way is the best for them to pick an answer?
- Picking one of their answer randomly
- If they both think the same answer they will say it if they each choose different one they will pick randomly one of their answer and say it.
If I set $ A_i $ = probability of person $ i = 1,2 $ answer correctly. $ B $ will be the event of them picking both the same answer.
$$ P(B) = ((A_1 cap A_2)cup (A_1^ccap A_2^c))$$
if they just pick randomly is
$$ P(A_i) =p\ P(A_1 cup A_2) = P(A_1) + P(A_2) - P(A_1 cap A_2) = 2p -p^2 $$
According to the second way, if they both pick the same answer :
$$ P( right answer | B) =frac{P((A_1cap A_2)cap B)}{P(B)} = frac{p^2}{p^2+(1-p)^2} $$
and then their last option is
$$ P( right answer | B^c ) = frac{P((A_1cup A_2)cap B^c)}{P(B^c)} = frac{P((A_1cup A_2)cap((A_1cap A_2^c)cup(A_1^c cap A_2)))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = frac{P((A_1cap A_2^c)cup(A_1^ccap A_2))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = 1 $$
Which is wrong.. Yet I couldn't find what am I missing?
probability-theory conditional-probability
asked Nov 21 at 22:01
bm1125
55116
|
show 6 more comments
up vote
0
down vote
favorite
Suppose two people have a yes / no question that the probability for each of them to answer correctly is $ p $. Each one answer is independent of the other one. What way is the best for them to pick an answer?
- Picking one of their answer randomly
- If they both think the same answer they will say it if they each choose different one they will pick randomly one of their answer and say it.
If I set $ A_i $ = probability of person $ i = 1,2 $ answer correctly. $ B $ will be the event of them picking both the same answer.
$$ P(B) = ((A_1 cap A_2)cup (A_1^ccap A_2^c))$$
if they just pick randomly is
$$ P(A_i) =p\ P(A_1 cup A_2) = P(A_1) + P(A_2) - P(A_1 cap A_2) = 2p -p^2 $$
According to the second way, if they both pick the same answer :
$$ P( right answer | B) =frac{P((A_1cap A_2)cap B)}{P(B)} = frac{p^2}{p^2+(1-p)^2} $$
and then their last option is
$$ P( right answer | B^c ) = frac{P((A_1cup A_2)cap B^c)}{P(B^c)} = frac{P((A_1cup A_2)cap((A_1cap A_2^c)cup(A_1^c cap A_2)))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = frac{P((A_1cap A_2^c)cup(A_1^ccap A_2))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = 1 $$
Which is wrong.. Yet I couldn't find what am I missing?
probability-theory conditional-probability
asked Nov 21 at 22:01
bm1125
55116
Neither is good if $p<1/2$.
– Michael
Nov 21 at 22:05
How can they be independent? Presumably the answers are both dependent on the true answer.
– lulu
Nov 21 at 22:12
1
The probability $P[A_1 cup A_2]$ is the probability that either person 1 or person 2 answers correctly, and is not relevant to this problem; it does not represent a probaiblity of being correct given you randomly select one of the persons.
– Michael
Nov 21 at 22:12
@Michael so if they pick random answer, how do I define the probability of the answer to be correct?
– bm1125
Nov 21 at 22:22
1
@lulu If you know the first person answered the question correctly, you cannot improve your estimate of the probability with which the second person answers the question correctly. It is still $p$. This is what “independent” means. The probability $p$ in this question is the probability of correctly answering the question, not the probability of giving a particular answer from the set ${T,F}$. But even so, if it were the latter, knowing the first person said $T$ would not allow you to predict the second person’s answer any better, assuming you know each probability of a $T$ answer was $p$.
– Steve Kass
Nov 21 at 23:02
|
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose two people have a yes / no question that the probability for each of them to answer correctly is $ p $. Each one answer is independent of the other one. What way is the best for them to pick an answer?
- Picking one of their answer randomly
- If they both think the same answer they will say it if they each choose different one they will pick randomly one of their answer and say it.
If I set $ A_i $ = probability of person $ i = 1,2 $ answer correctly. $ B $ will be the event of them picking both the same answer.
$$ P(B) = ((A_1 cap A_2)cup (A_1^ccap A_2^c))$$
if they just pick randomly is
$$ P(A_i) =p\ P(A_1 cup A_2) = P(A_1) + P(A_2) - P(A_1 cap A_2) = 2p -p^2 $$
According to the second way, if they both pick the same answer :
$$ P( right answer | B) =frac{P((A_1cap A_2)cap B)}{P(B)} = frac{p^2}{p^2+(1-p)^2} $$
and then their last option is
$$ P( right answer | B^c ) = frac{P((A_1cup A_2)cap B^c)}{P(B^c)} = frac{P((A_1cup A_2)cap((A_1cap A_2^c)cup(A_1^c cap A_2)))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = frac{P((A_1cap A_2^c)cup(A_1^ccap A_2))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = 1 $$
Which is wrong.. Yet I couldn't find what am I missing?
probability-theory conditional-probability
asked Nov 21 at 22:01
bm1125
55116
Suppose two people have a yes / no question that the probability for each of them to answer correctly is $ p $. Each one answer is independent of the other one. What way is the best for them to pick an answer?
- Picking one of their answer randomly
- If they both think the same answer they will say it if they each choose different one they will pick randomly one of their answer and say it.
If I set $ A_i $ = probability of person $ i = 1,2 $ answer correctly. $ B $ will be the event of them picking both the same answer.
$$ P(B) = ((A_1 cap A_2)cup (A_1^ccap A_2^c))$$
if they just pick randomly is
$$ P(A_i) =p\ P(A_1 cup A_2) = P(A_1) + P(A_2) - P(A_1 cap A_2) = 2p -p^2 $$
According to the second way, if they both pick the same answer :
$$ P( right answer | B) =frac{P((A_1cap A_2)cap B)}{P(B)} = frac{p^2}{p^2+(1-p)^2} $$
and then their last option is
$$ P( right answer | B^c ) = frac{P((A_1cup A_2)cap B^c)}{P(B^c)} = frac{P((A_1cup A_2)cap((A_1cap A_2^c)cup(A_1^c cap A_2)))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = frac{P((A_1cap A_2^c)cup(A_1^ccap A_2))}{P((A_1cap A_2^c)cup(A_1^ccap A_2))} = 1 $$
Which is wrong.. Yet I couldn't find what am I missing?
probability-theory conditional-probability
probability-theory conditional-probability
asked Nov 21 at 22:01
bm1125
55116
asked Nov 21 at 22:01
bm1125
55116
asked Nov 21 at 22:01
bm1125
55116
asked Nov 21 at 22:01
bm1125
55116
asked Nov 21 at 22:01
bm1125
55116
55116
Neither is good if $p<1/2$.
– Michael
Nov 21 at 22:05
How can they be independent? Presumably the answers are both dependent on the true answer.
– lulu
Nov 21 at 22:12
1
The probability $P[A_1 cup A_2]$ is the probability that either person 1 or person 2 answers correctly, and is not relevant to this problem; it does not represent a probaiblity of being correct given you randomly select one of the persons.
– Michael
Nov 21 at 22:12
@Michael so if they pick random answer, how do I define the probability of the answer to be correct?
– bm1125
Nov 21 at 22:22
1
@lulu If you know the first person answered the question correctly, you cannot improve your estimate of the probability with which the second person answers the question correctly. It is still $p$. This is what “independent” means. The probability $p$ in this question is the probability of correctly answering the question, not the probability of giving a particular answer from the set ${T,F}$. But even so, if it were the latter, knowing the first person said $T$ would not allow you to predict the second person’s answer any better, assuming you know each probability of a $T$ answer was $p$.
– Steve Kass
Nov 21 at 23:02
|
show 6 more comments
Neither is good if $p<1/2$.
– Michael
Nov 21 at 22:05
How can they be independent? Presumably the answers are both dependent on the true answer.
– lulu
Nov 21 at 22:12
1
The probability $P[A_1 cup A_2]$ is the probability that either person 1 or person 2 answers correctly, and is not relevant to this problem; it does not represent a probaiblity of being correct given you randomly select one of the persons.
– Michael
Nov 21 at 22:12
@Michael so if they pick random answer, how do I define the probability of the answer to be correct?
– bm1125
Nov 21 at 22:22
1
@lulu If you know the first person answered the question correctly, you cannot improve your estimate of the probability with which the second person answers the question correctly. It is still $p$. This is what “independent” means. The probability $p$ in this question is the probability of correctly answering the question, not the probability of giving a particular answer from the set ${T,F}$. But even so, if it were the latter, knowing the first person said $T$ would not allow you to predict the second person’s answer any better, assuming you know each probability of a $T$ answer was $p$.
– Steve Kass
Nov 21 at 23:02
Neither is good if $p<1/2$.
– Michael
Nov 21 at 22:05
Neither is good if $p<1/2$.
– Michael
Nov 21 at 22:05
How can they be independent? Presumably the answers are both dependent on the true answer.
– lulu
Nov 21 at 22:12
How can they be independent? Presumably the answers are both dependent on the true answer.
– lulu
Nov 21 at 22:12
1
1
The probability $P[A_1 cup A_2]$ is the probability that either person 1 or person 2 answers correctly, and is not relevant to this problem; it does not represent a probaiblity of being correct given you randomly select one of the persons.
– Michael
Nov 21 at 22:12
The probability $P[A_1 cup A_2]$ is the probability that either person 1 or person 2 answers correctly, and is not relevant to this problem; it does not represent a probaiblity of being correct given you randomly select one of the persons.
– Michael
Nov 21 at 22:12
@Michael so if they pick random answer, how do I define the probability of the answer to be correct?
– bm1125
Nov 21 at 22:22
@Michael so if they pick random answer, how do I define the probability of the answer to be correct?
– bm1125
Nov 21 at 22:22
1
1
@lulu If you know the first person answered the question correctly, you cannot improve your estimate of the probability with which the second person answers the question correctly. It is still $p$. This is what “independent” means. The probability $p$ in this question is the probability of correctly answering the question, not the probability of giving a particular answer from the set ${T,F}$. But even so, if it were the latter, knowing the first person said $T$ would not allow you to predict the second person’s answer any better, assuming you know each probability of a $T$ answer was $p$.
– Steve Kass
Nov 21 at 23:02
@lulu If you know the first person answered the question correctly, you cannot improve your estimate of the probability with which the second person answers the question correctly. It is still $p$. This is what “independent” means. The probability $p$ in this question is the probability of correctly answering the question, not the probability of giving a particular answer from the set ${T,F}$. But even so, if it were the latter, knowing the first person said $T$ would not allow you to predict the second person’s answer any better, assuming you know each probability of a $T$ answer was $p$.
– Steve Kass
Nov 21 at 23:02
|
show 6 more comments
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Neither is good if $p<1/2$.
– Michael
Nov 21 at 22:05
How can they be independent? Presumably the answers are both dependent on the true answer.
– lulu
Nov 21 at 22:12
1
The probability $P[A_1 cup A_2]$ is the probability that either person 1 or person 2 answers correctly, and is not relevant to this problem; it does not represent a probaiblity of being correct given you randomly select one of the persons.
– Michael
Nov 21 at 22:12
@Michael so if they pick random answer, how do I define the probability of the answer to be correct?
– bm1125
Nov 21 at 22:22
1
@lulu If you know the first person answered the question correctly, you cannot improve your estimate of the probability with which the second person answers the question correctly. It is still $p$. This is what “independent” means. The probability $p$ in this question is the probability of correctly answering the question, not the probability of giving a particular answer from the set ${T,F}$. But even so, if it were the latter, knowing the first person said $T$ would not allow you to predict the second person’s answer any better, assuming you know each probability of a $T$ answer was $p$.
– Steve Kass
Nov 21 at 23:02