Probability of alien species going extinct [closed]
$begingroup$
One alien lands on a planet. Every day, the alien can perform the following actions, with their probabilities as:
Self - destruct : 1/4 chance
Do nothing - 1/4 chance
Split into three aliens - 1/4 chance
Split into two aliens - 1/4 chance
If the alien splits, each produced alien has the same probabilities.
What is the probability of the species going extinct?
probability puzzle
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
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closed as off-topic by Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak Dec 11 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
One alien lands on a planet. Every day, the alien can perform the following actions, with their probabilities as:
Self - destruct : 1/4 chance
Do nothing - 1/4 chance
Split into three aliens - 1/4 chance
Split into two aliens - 1/4 chance
If the alien splits, each produced alien has the same probabilities.
What is the probability of the species going extinct?
probability puzzle
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
$endgroup$
closed as off-topic by Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak Dec 11 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
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– Shaun
Dec 11 '18 at 10:27
add a comment |
$begingroup$
One alien lands on a planet. Every day, the alien can perform the following actions, with their probabilities as:
Self - destruct : 1/4 chance
Do nothing - 1/4 chance
Split into three aliens - 1/4 chance
Split into two aliens - 1/4 chance
If the alien splits, each produced alien has the same probabilities.
What is the probability of the species going extinct?
probability puzzle
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
$endgroup$
One alien lands on a planet. Every day, the alien can perform the following actions, with their probabilities as:
Self - destruct : 1/4 chance
Do nothing - 1/4 chance
Split into three aliens - 1/4 chance
Split into two aliens - 1/4 chance
If the alien splits, each produced alien has the same probabilities.
What is the probability of the species going extinct?
probability puzzle
probability puzzle
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
asked Dec 11 '18 at 10:24
BuluBestTapuBuluBestTapu
778
778
closed as off-topic by Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak Dec 11 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak Dec 11 '18 at 12:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Shaun, gammatester, Jon, Martin Sleziak
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 11 '18 at 10:27
add a comment |
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 11 '18 at 10:27
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 11 '18 at 10:27
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
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– Shaun
Dec 11 '18 at 10:27
add a comment |
1 Answer
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$begingroup$
Take the probability of an alien and it's entire line dying out be D. So we can write D as
$D$ = $frac14$(if it self destructs) +$frac14$(if it does nothing)$cdot$P(probability of going extinct tomorrow) + $frac14$(if it splits into $3$)$cdot$P(probability of all three going extinct) +$frac14$(if it splits into $2$)$cdot$P(probability of all two going extinct)
We can see that
P(probability of going extinct tomorrow) is same as today = $D$
P(probability of all three going extinct) = $Dcdot Dcdot D$ = $D^3$
as each of them has to go extinct which means it's an intersection of each dying case, which in case of independent events means multiplication.
Similarly, P(probability of all two going extinct) = $Dcdot D$ = $D^2$
Putting them in the equation gives us
$$D = frac{(1+D+D^2+D^3)}{4}$$
$$D^3+D^2-3D +1 =0$$
This has $1$ as an obvious solutions, which helps us determine the other two $sqrt{2}-1$ and $-sqrt{2}-1$. Since probability cannot be negative, the latter is out of the question. Now between $1$ and $sqrt{2}-1$, you can use either Expectation to see that number of aliens the next day will be $1.5>1$ which means there is non zero chance it might survive or you can use induction to get the fact that $P_k$(probability it goes extinct after $k$ days) is always less than $sqrt{2}-1$, starting from $P_1$ which is $0.25$
This means
$$D = sqrt{2}-1$$
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take the probability of an alien and it's entire line dying out be D. So we can write D as
$D$ = $frac14$(if it self destructs) +$frac14$(if it does nothing)$cdot$P(probability of going extinct tomorrow) + $frac14$(if it splits into $3$)$cdot$P(probability of all three going extinct) +$frac14$(if it splits into $2$)$cdot$P(probability of all two going extinct)
We can see that
P(probability of going extinct tomorrow) is same as today = $D$
P(probability of all three going extinct) = $Dcdot Dcdot D$ = $D^3$
as each of them has to go extinct which means it's an intersection of each dying case, which in case of independent events means multiplication.
Similarly, P(probability of all two going extinct) = $Dcdot D$ = $D^2$
Putting them in the equation gives us
$$D = frac{(1+D+D^2+D^3)}{4}$$
$$D^3+D^2-3D +1 =0$$
This has $1$ as an obvious solutions, which helps us determine the other two $sqrt{2}-1$ and $-sqrt{2}-1$. Since probability cannot be negative, the latter is out of the question. Now between $1$ and $sqrt{2}-1$, you can use either Expectation to see that number of aliens the next day will be $1.5>1$ which means there is non zero chance it might survive or you can use induction to get the fact that $P_k$(probability it goes extinct after $k$ days) is always less than $sqrt{2}-1$, starting from $P_1$ which is $0.25$
This means
$$D = sqrt{2}-1$$
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
Take the probability of an alien and it's entire line dying out be D. So we can write D as
$D$ = $frac14$(if it self destructs) +$frac14$(if it does nothing)$cdot$P(probability of going extinct tomorrow) + $frac14$(if it splits into $3$)$cdot$P(probability of all three going extinct) +$frac14$(if it splits into $2$)$cdot$P(probability of all two going extinct)
We can see that
P(probability of going extinct tomorrow) is same as today = $D$
P(probability of all three going extinct) = $Dcdot Dcdot D$ = $D^3$
as each of them has to go extinct which means it's an intersection of each dying case, which in case of independent events means multiplication.
Similarly, P(probability of all two going extinct) = $Dcdot D$ = $D^2$
Putting them in the equation gives us
$$D = frac{(1+D+D^2+D^3)}{4}$$
$$D^3+D^2-3D +1 =0$$
This has $1$ as an obvious solutions, which helps us determine the other two $sqrt{2}-1$ and $-sqrt{2}-1$. Since probability cannot be negative, the latter is out of the question. Now between $1$ and $sqrt{2}-1$, you can use either Expectation to see that number of aliens the next day will be $1.5>1$ which means there is non zero chance it might survive or you can use induction to get the fact that $P_k$(probability it goes extinct after $k$ days) is always less than $sqrt{2}-1$, starting from $P_1$ which is $0.25$
This means
$$D = sqrt{2}-1$$
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
Take the probability of an alien and it's entire line dying out be D. So we can write D as
$D$ = $frac14$(if it self destructs) +$frac14$(if it does nothing)$cdot$P(probability of going extinct tomorrow) + $frac14$(if it splits into $3$)$cdot$P(probability of all three going extinct) +$frac14$(if it splits into $2$)$cdot$P(probability of all two going extinct)
We can see that
P(probability of going extinct tomorrow) is same as today = $D$
P(probability of all three going extinct) = $Dcdot Dcdot D$ = $D^3$
as each of them has to go extinct which means it's an intersection of each dying case, which in case of independent events means multiplication.
Similarly, P(probability of all two going extinct) = $Dcdot D$ = $D^2$
Putting them in the equation gives us
$$D = frac{(1+D+D^2+D^3)}{4}$$
$$D^3+D^2-3D +1 =0$$
This has $1$ as an obvious solutions, which helps us determine the other two $sqrt{2}-1$ and $-sqrt{2}-1$. Since probability cannot be negative, the latter is out of the question. Now between $1$ and $sqrt{2}-1$, you can use either Expectation to see that number of aliens the next day will be $1.5>1$ which means there is non zero chance it might survive or you can use induction to get the fact that $P_k$(probability it goes extinct after $k$ days) is always less than $sqrt{2}-1$, starting from $P_1$ which is $0.25$
This means
$$D = sqrt{2}-1$$
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
$endgroup$
Take the probability of an alien and it's entire line dying out be D. So we can write D as
$D$ = $frac14$(if it self destructs) +$frac14$(if it does nothing)$cdot$P(probability of going extinct tomorrow) + $frac14$(if it splits into $3$)$cdot$P(probability of all three going extinct) +$frac14$(if it splits into $2$)$cdot$P(probability of all two going extinct)
We can see that
P(probability of going extinct tomorrow) is same as today = $D$
P(probability of all three going extinct) = $Dcdot Dcdot D$ = $D^3$
as each of them has to go extinct which means it's an intersection of each dying case, which in case of independent events means multiplication.
Similarly, P(probability of all two going extinct) = $Dcdot D$ = $D^2$
Putting them in the equation gives us
$$D = frac{(1+D+D^2+D^3)}{4}$$
$$D^3+D^2-3D +1 =0$$
This has $1$ as an obvious solutions, which helps us determine the other two $sqrt{2}-1$ and $-sqrt{2}-1$. Since probability cannot be negative, the latter is out of the question. Now between $1$ and $sqrt{2}-1$, you can use either Expectation to see that number of aliens the next day will be $1.5>1$ which means there is non zero chance it might survive or you can use induction to get the fact that $P_k$(probability it goes extinct after $k$ days) is always less than $sqrt{2}-1$, starting from $P_1$ which is $0.25$
This means
$$D = sqrt{2}-1$$
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
answered Dec 11 '18 at 11:26
Sauhard SharmaSauhard Sharma
953318
953318
add a comment |
add a comment |
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 11 '18 at 10:27