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What is the probability of getting any two numbers (the order is not important) in 4 rolls?
The Next CEO of Stack OverflowWhat is the probability of getting yahtzee?What is the probability two die show the same values on their second rolls?What's the probability of getting $5$ different numbers but not any $6$ when throwing $5$ dice?Basic Probability Question of Two Teams choosing numbersSelecting n real numbers between 0 and 1 at random, what is the probability they are in order, increasing?Defining a probability on the natural numbers setProbability of no two equal adjacent numbersHow do I calculate the probability of collision for four independent choices from a predetermined pool?whats the probability of a number getting picked from a pool fewer than 3 times?What is the expected number of randomly generated numbers in the range [a, b] required to reach a sum $geq X$?
$begingroup$
A random number generator will generate a number randomly from 1 to 18 and remove the number generated every time from the pool of availble numbers to generate $(1-18)$. So, for example, if it rolls 1 on the first run then it will generate numbers from $(2-18)$, if it gets a 5 on the second run it will generate numbers from $(2-18)-(5)$, etc.
What is the probability of getting any two numbers we choose (order wont matter) in 4 rolls ?
What is the probability of getting any three numbers we choose (order wont matter) in 4 rolls ?
From lurking around this forum i found this might be the probability of getting any one number i choose in 4 rolls: $1-((17/18)^4)+4*((17/18)^3)*1/18=0,3915$.
I don't know if this is correct and im a bit at a loss on how to continue for the "any two numbers", maybe multiply it by the probability of getting any other number in another roll: $(1-((17/18)^4)+4*((17/18)^3)*1/18)*(1-((16/17)^4)+4*((16/17)^3)*1/17)=0,1611$ the number feels fine.
But i just don't know.
Thanks.
PS: But really, i arrived here because i was just wondering what is the probability of getting any set of 1, 2, 3 or 4 civilizations in your team when playing age of empires 2, 4vs4 random civs, it's not a matter of life or death, i guess i must have aspergers.
probability statistics
edited Mar 28 at 3:02
Jacob Jones
14311
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A random number generator will generate a number randomly from 1 to 18 and remove the number generated every time from the pool of availble numbers to generate $(1-18)$. So, for example, if it rolls 1 on the first run then it will generate numbers from $(2-18)$, if it gets a 5 on the second run it will generate numbers from $(2-18)-(5)$, etc.
What is the probability of getting any two numbers we choose (order wont matter) in 4 rolls ?
What is the probability of getting any three numbers we choose (order wont matter) in 4 rolls ?
From lurking around this forum i found this might be the probability of getting any one number i choose in 4 rolls: $1-((17/18)^4)+4*((17/18)^3)*1/18=0,3915$.
I don't know if this is correct and im a bit at a loss on how to continue for the "any two numbers", maybe multiply it by the probability of getting any other number in another roll: $(1-((17/18)^4)+4*((17/18)^3)*1/18)*(1-((16/17)^4)+4*((16/17)^3)*1/17)=0,1611$ the number feels fine.
But i just don't know.
Thanks.
PS: But really, i arrived here because i was just wondering what is the probability of getting any set of 1, 2, 3 or 4 civilizations in your team when playing age of empires 2, 4vs4 random civs, it's not a matter of life or death, i guess i must have aspergers.
probability statistics
edited Mar 28 at 3:02
Jacob Jones
14311
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Do you have any thoughts on this yourself?
$endgroup$
– Bram28
Mar 28 at 2:21
1
$begingroup$
Welcome to Math.SE! 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.
$endgroup$
– Brian
Mar 28 at 2:22
add a comment |
$begingroup$
A random number generator will generate a number randomly from 1 to 18 and remove the number generated every time from the pool of availble numbers to generate $(1-18)$. So, for example, if it rolls 1 on the first run then it will generate numbers from $(2-18)$, if it gets a 5 on the second run it will generate numbers from $(2-18)-(5)$, etc.
What is the probability of getting any two numbers we choose (order wont matter) in 4 rolls ?
What is the probability of getting any three numbers we choose (order wont matter) in 4 rolls ?
From lurking around this forum i found this might be the probability of getting any one number i choose in 4 rolls: $1-((17/18)^4)+4*((17/18)^3)*1/18=0,3915$.
I don't know if this is correct and im a bit at a loss on how to continue for the "any two numbers", maybe multiply it by the probability of getting any other number in another roll: $(1-((17/18)^4)+4*((17/18)^3)*1/18)*(1-((16/17)^4)+4*((16/17)^3)*1/17)=0,1611$ the number feels fine.
But i just don't know.
Thanks.
PS: But really, i arrived here because i was just wondering what is the probability of getting any set of 1, 2, 3 or 4 civilizations in your team when playing age of empires 2, 4vs4 random civs, it's not a matter of life or death, i guess i must have aspergers.
probability statistics
edited Mar 28 at 3:02
Jacob Jones
14311
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A random number generator will generate a number randomly from 1 to 18 and remove the number generated every time from the pool of availble numbers to generate $(1-18)$. So, for example, if it rolls 1 on the first run then it will generate numbers from $(2-18)$, if it gets a 5 on the second run it will generate numbers from $(2-18)-(5)$, etc.
What is the probability of getting any two numbers we choose (order wont matter) in 4 rolls ?
What is the probability of getting any three numbers we choose (order wont matter) in 4 rolls ?
From lurking around this forum i found this might be the probability of getting any one number i choose in 4 rolls: $1-((17/18)^4)+4*((17/18)^3)*1/18=0,3915$.
I don't know if this is correct and im a bit at a loss on how to continue for the "any two numbers", maybe multiply it by the probability of getting any other number in another roll: $(1-((17/18)^4)+4*((17/18)^3)*1/18)*(1-((16/17)^4)+4*((16/17)^3)*1/17)=0,1611$ the number feels fine.
But i just don't know.
Thanks.
PS: But really, i arrived here because i was just wondering what is the probability of getting any set of 1, 2, 3 or 4 civilizations in your team when playing age of empires 2, 4vs4 random civs, it's not a matter of life or death, i guess i must have aspergers.
probability statistics
probability statistics
edited Mar 28 at 3:02
Jacob Jones
14311
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 3:02
Jacob Jones
14311
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 3:02
Jacob Jones
14311
edited Mar 28 at 3:02
Jacob Jones
14311
edited Mar 28 at 3:02
Jacob Jones
14311
14311
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
asked Mar 28 at 2:18
heiner weinerheiner weiner
11
11
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
heiner weiner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Do you have any thoughts on this yourself?
$endgroup$
– Bram28
Mar 28 at 2:21
1
$begingroup$
Welcome to Math.SE! 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.
$endgroup$
– Brian
Mar 28 at 2:22
add a comment |
2
$begingroup$
Do you have any thoughts on this yourself?
$endgroup$
– Bram28
Mar 28 at 2:21
1
$begingroup$
Welcome to Math.SE! 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.
$endgroup$
– Brian
Mar 28 at 2:22
2
2
$begingroup$
Do you have any thoughts on this yourself?
$endgroup$
– Bram28
Mar 28 at 2:21
$begingroup$
Do you have any thoughts on this yourself?
$endgroup$
– Bram28
Mar 28 at 2:21
1
1
$begingroup$
Welcome to Math.SE! 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.
$endgroup$
– Brian
Mar 28 at 2:22
$begingroup$
Welcome to Math.SE! 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.
$endgroup$
– Brian
Mar 28 at 2:22
add a comment |
1 Answer
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$begingroup$
There are $binom182$ pairs we could choose. Each set of four contains $binom42$ pairs. The probability we get a particular pair is $dfracbinom42binom182$.
The same would apply to whatever set sizes and pool sizes you're using.
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
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$begingroup$
There are $binom182$ pairs we could choose. Each set of four contains $binom42$ pairs. The probability we get a particular pair is $dfracbinom42binom182$.
The same would apply to whatever set sizes and pool sizes you're using.
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
There are $binom182$ pairs we could choose. Each set of four contains $binom42$ pairs. The probability we get a particular pair is $dfracbinom42binom182$.
The same would apply to whatever set sizes and pool sizes you're using.
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
There are $binom182$ pairs we could choose. Each set of four contains $binom42$ pairs. The probability we get a particular pair is $dfracbinom42binom182$.
The same would apply to whatever set sizes and pool sizes you're using.
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
$endgroup$
There are $binom182$ pairs we could choose. Each set of four contains $binom42$ pairs. The probability we get a particular pair is $dfracbinom42binom182$.
The same would apply to whatever set sizes and pool sizes you're using.
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
answered Mar 28 at 3:21
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
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$begingroup$
Do you have any thoughts on this yourself?
$endgroup$
– Bram28
Mar 28 at 2:21
1
$begingroup$
Welcome to Math.SE! 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.
$endgroup$
– Brian
Mar 28 at 2:22