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Why is this method incorrect?
Why is this Poisson distribution incorrect?Angular distribution of lines passing through two squares.Multiple Anihilating Random Walks in a Ring (cycle)Population exercise with Markov chainsFind the probability that a duel ends after the $n^th$ round of shots.Relation between: Likelihood, conditional probability and failure rateThe Three-Cornered DuelA variation of the balls-and-bins problem, with additional constraintsProbability two people will meet within 10 minutes of each other?Calculating the probability of winning a game of Rainbow Six Siege
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
A man is to be executed at random time between 00:00 and 01:00. The
firing squad's accuracy decreases linearly, so that at 00:00 they
shoot perfectly, at 00:30 miss half the time, and at 01:00 miss
always. Also, with probability 1/2, a blank round of shots will be
used.
Given that the man survived, what is the probability that he faced a
live round?
At first I drew a diagram with time on x-axis, a horizontal line at $y=1/2$ with "blind round" above it and "live round" below. Then I divided that lower region with a diagonal to represent the falling accuracy, and got the answer $frac14 / (frac14+frac12) = frac13$ which a simulation seems to confirm.
I'm not sure why the following method gives an incorrect answer: at time $theta$, $P_theta (textsurvived)=theta/2 + 1/2$ and $P_theta(textlive rounds and survived) = theta/2$. Therefore $P_theta(textlive rounds ) = fracthetatheta+1$. Now integrate to get $int_0^1 P_theta dtheta = int_0^1 fracdtheta . thetatheta+1 = 1-log 2 approx 0.3069$.
probability probability-distributions
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
$endgroup$
add a comment |
$begingroup$
A man is to be executed at random time between 00:00 and 01:00. The
firing squad's accuracy decreases linearly, so that at 00:00 they
shoot perfectly, at 00:30 miss half the time, and at 01:00 miss
always. Also, with probability 1/2, a blank round of shots will be
used.
Given that the man survived, what is the probability that he faced a
live round?
At first I drew a diagram with time on x-axis, a horizontal line at $y=1/2$ with "blind round" above it and "live round" below. Then I divided that lower region with a diagonal to represent the falling accuracy, and got the answer $frac14 / (frac14+frac12) = frac13$ which a simulation seems to confirm.
I'm not sure why the following method gives an incorrect answer: at time $theta$, $P_theta (textsurvived)=theta/2 + 1/2$ and $P_theta(textlive rounds and survived) = theta/2$. Therefore $P_theta(textlive rounds ) = fracthetatheta+1$. Now integrate to get $int_0^1 P_theta dtheta = int_0^1 fracdtheta . thetatheta+1 = 1-log 2 approx 0.3069$.
probability probability-distributions
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
$endgroup$
2
$begingroup$
Nice background story for the question. Looks taken from Games of Thrones...
$endgroup$
– DonAntonio
Jun 8 at 23:22
add a comment |
$begingroup$
A man is to be executed at random time between 00:00 and 01:00. The
firing squad's accuracy decreases linearly, so that at 00:00 they
shoot perfectly, at 00:30 miss half the time, and at 01:00 miss
always. Also, with probability 1/2, a blank round of shots will be
used.
Given that the man survived, what is the probability that he faced a
live round?
At first I drew a diagram with time on x-axis, a horizontal line at $y=1/2$ with "blind round" above it and "live round" below. Then I divided that lower region with a diagonal to represent the falling accuracy, and got the answer $frac14 / (frac14+frac12) = frac13$ which a simulation seems to confirm.
I'm not sure why the following method gives an incorrect answer: at time $theta$, $P_theta (textsurvived)=theta/2 + 1/2$ and $P_theta(textlive rounds and survived) = theta/2$. Therefore $P_theta(textlive rounds ) = fracthetatheta+1$. Now integrate to get $int_0^1 P_theta dtheta = int_0^1 fracdtheta . thetatheta+1 = 1-log 2 approx 0.3069$.
probability probability-distributions
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
$endgroup$
A man is to be executed at random time between 00:00 and 01:00. The
firing squad's accuracy decreases linearly, so that at 00:00 they
shoot perfectly, at 00:30 miss half the time, and at 01:00 miss
always. Also, with probability 1/2, a blank round of shots will be
used.
Given that the man survived, what is the probability that he faced a
live round?
At first I drew a diagram with time on x-axis, a horizontal line at $y=1/2$ with "blind round" above it and "live round" below. Then I divided that lower region with a diagonal to represent the falling accuracy, and got the answer $frac14 / (frac14+frac12) = frac13$ which a simulation seems to confirm.
I'm not sure why the following method gives an incorrect answer: at time $theta$, $P_theta (textsurvived)=theta/2 + 1/2$ and $P_theta(textlive rounds and survived) = theta/2$. Therefore $P_theta(textlive rounds ) = fracthetatheta+1$. Now integrate to get $int_0^1 P_theta dtheta = int_0^1 fracdtheta . thetatheta+1 = 1-log 2 approx 0.3069$.
probability probability-distributions
probability probability-distributions
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
edited Jun 9 at 10:18
Klangen
2,1051 gold badge13 silver badges36 bronze badges
2,1051 gold badge13 silver badges36 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
asked Jun 8 at 22:57
MoronicMoronic
523 bronze badges
523 bronze badges
2
$begingroup$
Nice background story for the question. Looks taken from Games of Thrones...
$endgroup$
– DonAntonio
Jun 8 at 23:22
add a comment |
2
$begingroup$
Nice background story for the question. Looks taken from Games of Thrones...
$endgroup$
– DonAntonio
Jun 8 at 23:22
2
2
$begingroup$
Nice background story for the question. Looks taken from Games of Thrones...
$endgroup$
– DonAntonio
Jun 8 at 23:22
$begingroup$
Nice background story for the question. Looks taken from Games of Thrones...
$endgroup$
– DonAntonio
Jun 8 at 23:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer by @paulinho is excellent, but here's another way to explain why your second answer is wrong.
Let $L=$ live rounds, and $S =$ survive. I will use $P()$ for probability and $p()$ for pdf.
Your $theta over theta + 1 = P(L | theta, S)$. So if you integrate $int_0^1 theta over theta + 1; dtheta = int_0^1 P(L | theta, S); d theta$, since $theta$ is time, you are assuming that every moment is equally likely. However, when conditioned on survival, every moment is not equally likely!
You needed something akin to the pdf of $theta$, conditioned on $S$, shown in red below:
- $P(L|S) = int_0^1 p(L, theta | S); d theta = int_0^1 P(L | theta, S) colorredp(theta ; d theta$
Frankly, finding $p(theta | S)$ requires basically solving the entire problem, so this is not a good way to proceed. Instead:
$p(L, theta | S) = p(L, theta, S) / P(S)$
$p(L, theta, S) = theta / 2$
So, $P(L|S) = int_0^1 theta / 2 over P(S) ; dtheta = int_0^1 (theta /2) dtheta over P(S)$ and you are back to the answer by @paulinho
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
This is similar to the question, "does $displaystyle int fracf(x)g(x) dx= fracint f(x) dxint g(x) dx?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here is $$p = fracint_0^1 (theta/2)dthetaint_0^1 (theta/2 + 1/2)d theta=frac1/21 + 1/2 = boxed1/3.$$
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
$endgroup$
$begingroup$
Yes (+1), that pinpoints the error.
$endgroup$
– quasi
Jun 8 at 23:45
$begingroup$
I understand why these integrals aren't equal, but I don't understand why "$int_0^1 fracx.dxx+1$" does not define the correct answer. What is this value?
$endgroup$
– Moronic
Jun 9 at 0:02
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer by @paulinho is excellent, but here's another way to explain why your second answer is wrong.
Let $L=$ live rounds, and $S =$ survive. I will use $P()$ for probability and $p()$ for pdf.
Your $theta over theta + 1 = P(L | theta, S)$. So if you integrate $int_0^1 theta over theta + 1; dtheta = int_0^1 P(L | theta, S); d theta$, since $theta$ is time, you are assuming that every moment is equally likely. However, when conditioned on survival, every moment is not equally likely!
You needed something akin to the pdf of $theta$, conditioned on $S$, shown in red below:
- $P(L|S) = int_0^1 p(L, theta | S); d theta = int_0^1 P(L | theta, S) colorredp(theta ; d theta$
Frankly, finding $p(theta | S)$ requires basically solving the entire problem, so this is not a good way to proceed. Instead:
$p(L, theta | S) = p(L, theta, S) / P(S)$
$p(L, theta, S) = theta / 2$
So, $P(L|S) = int_0^1 theta / 2 over P(S) ; dtheta = int_0^1 (theta /2) dtheta over P(S)$ and you are back to the answer by @paulinho
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
The answer by @paulinho is excellent, but here's another way to explain why your second answer is wrong.
Let $L=$ live rounds, and $S =$ survive. I will use $P()$ for probability and $p()$ for pdf.
Your $theta over theta + 1 = P(L | theta, S)$. So if you integrate $int_0^1 theta over theta + 1; dtheta = int_0^1 P(L | theta, S); d theta$, since $theta$ is time, you are assuming that every moment is equally likely. However, when conditioned on survival, every moment is not equally likely!
You needed something akin to the pdf of $theta$, conditioned on $S$, shown in red below:
- $P(L|S) = int_0^1 p(L, theta | S); d theta = int_0^1 P(L | theta, S) colorredp(theta ; d theta$
Frankly, finding $p(theta | S)$ requires basically solving the entire problem, so this is not a good way to proceed. Instead:
$p(L, theta | S) = p(L, theta, S) / P(S)$
$p(L, theta, S) = theta / 2$
So, $P(L|S) = int_0^1 theta / 2 over P(S) ; dtheta = int_0^1 (theta /2) dtheta over P(S)$ and you are back to the answer by @paulinho
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
The answer by @paulinho is excellent, but here's another way to explain why your second answer is wrong.
Let $L=$ live rounds, and $S =$ survive. I will use $P()$ for probability and $p()$ for pdf.
Your $theta over theta + 1 = P(L | theta, S)$. So if you integrate $int_0^1 theta over theta + 1; dtheta = int_0^1 P(L | theta, S); d theta$, since $theta$ is time, you are assuming that every moment is equally likely. However, when conditioned on survival, every moment is not equally likely!
You needed something akin to the pdf of $theta$, conditioned on $S$, shown in red below:
- $P(L|S) = int_0^1 p(L, theta | S); d theta = int_0^1 P(L | theta, S) colorredp(theta ; d theta$
Frankly, finding $p(theta | S)$ requires basically solving the entire problem, so this is not a good way to proceed. Instead:
$p(L, theta | S) = p(L, theta, S) / P(S)$
$p(L, theta, S) = theta / 2$
So, $P(L|S) = int_0^1 theta / 2 over P(S) ; dtheta = int_0^1 (theta /2) dtheta over P(S)$ and you are back to the answer by @paulinho
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
$endgroup$
The answer by @paulinho is excellent, but here's another way to explain why your second answer is wrong.
Let $L=$ live rounds, and $S =$ survive. I will use $P()$ for probability and $p()$ for pdf.
Your $theta over theta + 1 = P(L | theta, S)$. So if you integrate $int_0^1 theta over theta + 1; dtheta = int_0^1 P(L | theta, S); d theta$, since $theta$ is time, you are assuming that every moment is equally likely. However, when conditioned on survival, every moment is not equally likely!
You needed something akin to the pdf of $theta$, conditioned on $S$, shown in red below:
- $P(L|S) = int_0^1 p(L, theta | S); d theta = int_0^1 P(L | theta, S) colorredp(theta ; d theta$
Frankly, finding $p(theta | S)$ requires basically solving the entire problem, so this is not a good way to proceed. Instead:
$p(L, theta | S) = p(L, theta, S) / P(S)$
$p(L, theta, S) = theta / 2$
So, $P(L|S) = int_0^1 theta / 2 over P(S) ; dtheta = int_0^1 (theta /2) dtheta over P(S)$ and you are back to the answer by @paulinho
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
edited Jun 9 at 7:27
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
answered Jun 9 at 6:58
antkamantkam
5,3604 silver badges15 bronze badges
5,3604 silver badges15 bronze badges
add a comment |
add a comment |
$begingroup$
This is similar to the question, "does $displaystyle int fracf(x)g(x) dx= fracint f(x) dxint g(x) dx?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here is $$p = fracint_0^1 (theta/2)dthetaint_0^1 (theta/2 + 1/2)d theta=frac1/21 + 1/2 = boxed1/3.$$
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
$endgroup$
$begingroup$
Yes (+1), that pinpoints the error.
$endgroup$
– quasi
Jun 8 at 23:45
$begingroup$
I understand why these integrals aren't equal, but I don't understand why "$int_0^1 fracx.dxx+1$" does not define the correct answer. What is this value?
$endgroup$
– Moronic
Jun 9 at 0:02
add a comment |
$begingroup$
This is similar to the question, "does $displaystyle int fracf(x)g(x) dx= fracint f(x) dxint g(x) dx?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here is $$p = fracint_0^1 (theta/2)dthetaint_0^1 (theta/2 + 1/2)d theta=frac1/21 + 1/2 = boxed1/3.$$
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
$endgroup$
$begingroup$
Yes (+1), that pinpoints the error.
$endgroup$
– quasi
Jun 8 at 23:45
$begingroup$
I understand why these integrals aren't equal, but I don't understand why "$int_0^1 fracx.dxx+1$" does not define the correct answer. What is this value?
$endgroup$
– Moronic
Jun 9 at 0:02
add a comment |
$begingroup$
This is similar to the question, "does $displaystyle int fracf(x)g(x) dx= fracint f(x) dxint g(x) dx?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here is $$p = fracint_0^1 (theta/2)dthetaint_0^1 (theta/2 + 1/2)d theta=frac1/21 + 1/2 = boxed1/3.$$
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
$endgroup$
This is similar to the question, "does $displaystyle int fracf(x)g(x) dx= fracint f(x) dxint g(x) dx?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here is $$p = fracint_0^1 (theta/2)dthetaint_0^1 (theta/2 + 1/2)d theta=frac1/21 + 1/2 = boxed1/3.$$
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
answered Jun 8 at 23:38
paulinhopaulinho
9451 silver badge11 bronze badges
9451 silver badge11 bronze badges
$begingroup$
Yes (+1), that pinpoints the error.
$endgroup$
– quasi
Jun 8 at 23:45
$begingroup$
I understand why these integrals aren't equal, but I don't understand why "$int_0^1 fracx.dxx+1$" does not define the correct answer. What is this value?
$endgroup$
– Moronic
Jun 9 at 0:02
add a comment |
$begingroup$
Yes (+1), that pinpoints the error.
$endgroup$
– quasi
Jun 8 at 23:45
$begingroup$
I understand why these integrals aren't equal, but I don't understand why "$int_0^1 fracx.dxx+1$" does not define the correct answer. What is this value?
$endgroup$
– Moronic
Jun 9 at 0:02
$begingroup$
Yes (+1), that pinpoints the error.
$endgroup$
– quasi
Jun 8 at 23:45
$begingroup$
Yes (+1), that pinpoints the error.
$endgroup$
– quasi
Jun 8 at 23:45
$begingroup$
I understand why these integrals aren't equal, but I don't understand why "$int_0^1 fracx.dxx+1$" does not define the correct answer. What is this value?
$endgroup$
– Moronic
Jun 9 at 0:02
$begingroup$
I understand why these integrals aren't equal, but I don't understand why "$int_0^1 fracx.dxx+1$" does not define the correct answer. What is this value?
$endgroup$
– Moronic
Jun 9 at 0:02
add a comment |
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T7,6domztZyaT,RuLEyYudnbcOFXZKG,iM1uF78 bWS,uFA6v LABRmJJXa4dHY,mCwP9MOic8,xPpNb 0
2
$begingroup$
Nice background story for the question. Looks taken from Games of Thrones...
$endgroup$
– DonAntonio
Jun 8 at 23:22