Mixing
What is the probability of marking the same object twice or more while keeping the same ratio of objects and...
$begingroup$
I have a simple but a bit more complex probability question.
If i have 8 different kinds of hats and want to mark half of them (4), what is the probability ill mark ANY of them twice (or more)? What if there are 1000 hats and 500 picks, and what is the relationship between increasing the number of hats and picks and the probability of marking any of them twice, WHILE keeping the ration the same, half. Is there any formula? Would infinite number of hats be aproaching 100 percent probability of any of them repeating? What if i changed the ration to 1/3, would infinite number of kinds of hats still aproach 100 percent probatility?
probability
asked Feb 2 at 14:19
R ZR Z
82
$endgroup$
add a comment |
$begingroup$
I have a simple but a bit more complex probability question.
If i have 8 different kinds of hats and want to mark half of them (4), what is the probability ill mark ANY of them twice (or more)? What if there are 1000 hats and 500 picks, and what is the relationship between increasing the number of hats and picks and the probability of marking any of them twice, WHILE keeping the ration the same, half. Is there any formula? Would infinite number of hats be aproaching 100 percent probability of any of them repeating? What if i changed the ration to 1/3, would infinite number of kinds of hats still aproach 100 percent probatility?
probability
asked Feb 2 at 14:19
R ZR Z
82
$endgroup$
$begingroup$
This isn't clear. How do you go about marking the hats? Do you mean if you choose a sample of $4$ of the $8$ with replacement, what is the probability that some hat is chosen at least twice? What have you done on this problem so far?
$endgroup$
– saulspatz
Feb 2 at 14:24
$begingroup$
You have 8 hats, you will choose/mark/ 4 times- what is the probability you choose 1 of them more than once! This is the base of the question. Or you have 20 hats and choose 10 times or 1000 hats and choose 500.....1:2 ratio.
$endgroup$
– R Z
Feb 2 at 16:33
add a comment |
$begingroup$
I have a simple but a bit more complex probability question.
If i have 8 different kinds of hats and want to mark half of them (4), what is the probability ill mark ANY of them twice (or more)? What if there are 1000 hats and 500 picks, and what is the relationship between increasing the number of hats and picks and the probability of marking any of them twice, WHILE keeping the ration the same, half. Is there any formula? Would infinite number of hats be aproaching 100 percent probability of any of them repeating? What if i changed the ration to 1/3, would infinite number of kinds of hats still aproach 100 percent probatility?
probability
asked Feb 2 at 14:19
R ZR Z
82
$endgroup$
I have a simple but a bit more complex probability question.
If i have 8 different kinds of hats and want to mark half of them (4), what is the probability ill mark ANY of them twice (or more)? What if there are 1000 hats and 500 picks, and what is the relationship between increasing the number of hats and picks and the probability of marking any of them twice, WHILE keeping the ration the same, half. Is there any formula? Would infinite number of hats be aproaching 100 percent probability of any of them repeating? What if i changed the ration to 1/3, would infinite number of kinds of hats still aproach 100 percent probatility?
probability
probability
asked Feb 2 at 14:19
R ZR Z
82
asked Feb 2 at 14:19
R ZR Z
82
asked Feb 2 at 14:19
R ZR Z
82
asked Feb 2 at 14:19
R ZR Z
82
asked Feb 2 at 14:19
R ZR Z
82
82
$begingroup$
This isn't clear. How do you go about marking the hats? Do you mean if you choose a sample of $4$ of the $8$ with replacement, what is the probability that some hat is chosen at least twice? What have you done on this problem so far?
$endgroup$
– saulspatz
Feb 2 at 14:24
$begingroup$
You have 8 hats, you will choose/mark/ 4 times- what is the probability you choose 1 of them more than once! This is the base of the question. Or you have 20 hats and choose 10 times or 1000 hats and choose 500.....1:2 ratio.
$endgroup$
– R Z
Feb 2 at 16:33
add a comment |
$begingroup$
This isn't clear. How do you go about marking the hats? Do you mean if you choose a sample of $4$ of the $8$ with replacement, what is the probability that some hat is chosen at least twice? What have you done on this problem so far?
$endgroup$
– saulspatz
Feb 2 at 14:24
$begingroup$
You have 8 hats, you will choose/mark/ 4 times- what is the probability you choose 1 of them more than once! This is the base of the question. Or you have 20 hats and choose 10 times or 1000 hats and choose 500.....1:2 ratio.
$endgroup$
– R Z
Feb 2 at 16:33
$begingroup$
This isn't clear. How do you go about marking the hats? Do you mean if you choose a sample of $4$ of the $8$ with replacement, what is the probability that some hat is chosen at least twice? What have you done on this problem so far?
$endgroup$
– saulspatz
Feb 2 at 14:24
$begingroup$
This isn't clear. How do you go about marking the hats? Do you mean if you choose a sample of $4$ of the $8$ with replacement, what is the probability that some hat is chosen at least twice? What have you done on this problem so far?
$endgroup$
– saulspatz
Feb 2 at 14:24
$begingroup$
You have 8 hats, you will choose/mark/ 4 times- what is the probability you choose 1 of them more than once! This is the base of the question. Or you have 20 hats and choose 10 times or 1000 hats and choose 500.....1:2 ratio.
$endgroup$
– R Z
Feb 2 at 16:33
$begingroup$
You have 8 hats, you will choose/mark/ 4 times- what is the probability you choose 1 of them more than once! This is the base of the question. Or you have 20 hats and choose 10 times or 1000 hats and choose 500.....1:2 ratio.
$endgroup$
– R Z
Feb 2 at 16:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you choose $n$ of $2n$ hats without replacement then there are $2n$ ways to choose the first hat, $2n-1$ ways to choose a second hat, $2n-2$ ways to choose a third hat different from the first two, and so on. There are $${(2n)!over n!}$$ ways to choose $n$ distinct hats.
There are $(2n)^n$ way to choose $n$ hats without replacement. So the probably that no hat is chosen twice is $${(2n)!over n!(2n)^n}$$ This goes to $0$ as $n$ goes to infinity, as you guessed.
If you choose $n$ of $3n$ hats, then by similar reasoning, the probability that none is chosen twice is $$(3n)!over (2n)!(3n)^n$$ and again, this goes to $0$.
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
Wow, the formula works, great job, thx.
$endgroup$
– R Z
Feb 3 at 1:48
$begingroup$
! means factorial For $100$ hats the probability of no duplicate is about $6.666times10^{-14}$ For $1000$ hats, it's about $7.691times10^{-134}$
$endgroup$
– saulspatz
Feb 3 at 1:53
$begingroup$
How did you even manage to get those answers? Google calculator wouldnt even work with such numbers.
$endgroup$
– R Z
Feb 3 at 1:58
$begingroup$
I did it in python.
$endgroup$
– saulspatz
Feb 3 at 1:58
add a comment |
$begingroup$
This is actually the generalized birthday problem. $n$ is the number of days. The number of items you have to mark to get a $50%$ chance of a match is about $sqrt {2 ln 2} sqrt n$. For other probabilities the constant $sqrt {2 ln 2}$ changes, but the $sqrt n$ behavior does not. Roughly speaking, the chance of any pair of marks going on the same item is about $frac 1n$ so you need the number of pairs of marks to be about $n$. The number of pairs is about half the square of the number of marks.
This means that if you choose a fixed fraction of $n$ to be the number of marks, eventually the probability of a remark will approach $1$ as $n$ gets large.
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks for you reply, but i dont understand this. Imagine you have numbers 0-9 (10 number in total) whats the probability to create a 5 digit number where all the numbers are different? Like 12345, 13589 but not 11081 22222 etc
$endgroup$
– R Z
Feb 2 at 16:42
$begingroup$
There are $10$ choices for the first digit, $9$ for the second, and so on, so $10cdot 9cdot 8cdot 7 cdot 6/10^5$
$endgroup$
– Ross Millikan
Feb 2 at 17:17
$begingroup$
So the answer will be 30,24 percent of it being all different numbers? I got there manualy before by 10/10 * 9/10 * 8/10 * 7/10 * 6/10... haha thanks for this simple formula.
$endgroup$
– R Z
Feb 3 at 0:59
$begingroup$
Yes, that is correct.
$endgroup$
– Ross Millikan
Feb 3 at 1:47
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you choose $n$ of $2n$ hats without replacement then there are $2n$ ways to choose the first hat, $2n-1$ ways to choose a second hat, $2n-2$ ways to choose a third hat different from the first two, and so on. There are $${(2n)!over n!}$$ ways to choose $n$ distinct hats.
There are $(2n)^n$ way to choose $n$ hats without replacement. So the probably that no hat is chosen twice is $${(2n)!over n!(2n)^n}$$ This goes to $0$ as $n$ goes to infinity, as you guessed.
If you choose $n$ of $3n$ hats, then by similar reasoning, the probability that none is chosen twice is $$(3n)!over (2n)!(3n)^n$$ and again, this goes to $0$.
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
Wow, the formula works, great job, thx.
$endgroup$
– R Z
Feb 3 at 1:48
$begingroup$
! means factorial For $100$ hats the probability of no duplicate is about $6.666times10^{-14}$ For $1000$ hats, it's about $7.691times10^{-134}$
$endgroup$
– saulspatz
Feb 3 at 1:53
$begingroup$
How did you even manage to get those answers? Google calculator wouldnt even work with such numbers.
$endgroup$
– R Z
Feb 3 at 1:58
$begingroup$
I did it in python.
$endgroup$
– saulspatz
Feb 3 at 1:58
add a comment |
$begingroup$
If you choose $n$ of $2n$ hats without replacement then there are $2n$ ways to choose the first hat, $2n-1$ ways to choose a second hat, $2n-2$ ways to choose a third hat different from the first two, and so on. There are $${(2n)!over n!}$$ ways to choose $n$ distinct hats.
There are $(2n)^n$ way to choose $n$ hats without replacement. So the probably that no hat is chosen twice is $${(2n)!over n!(2n)^n}$$ This goes to $0$ as $n$ goes to infinity, as you guessed.
If you choose $n$ of $3n$ hats, then by similar reasoning, the probability that none is chosen twice is $$(3n)!over (2n)!(3n)^n$$ and again, this goes to $0$.
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
Wow, the formula works, great job, thx.
$endgroup$
– R Z
Feb 3 at 1:48
$begingroup$
! means factorial For $100$ hats the probability of no duplicate is about $6.666times10^{-14}$ For $1000$ hats, it's about $7.691times10^{-134}$
$endgroup$
– saulspatz
Feb 3 at 1:53
$begingroup$
How did you even manage to get those answers? Google calculator wouldnt even work with such numbers.
$endgroup$
– R Z
Feb 3 at 1:58
$begingroup$
I did it in python.
$endgroup$
– saulspatz
Feb 3 at 1:58
add a comment |
$begingroup$
If you choose $n$ of $2n$ hats without replacement then there are $2n$ ways to choose the first hat, $2n-1$ ways to choose a second hat, $2n-2$ ways to choose a third hat different from the first two, and so on. There are $${(2n)!over n!}$$ ways to choose $n$ distinct hats.
There are $(2n)^n$ way to choose $n$ hats without replacement. So the probably that no hat is chosen twice is $${(2n)!over n!(2n)^n}$$ This goes to $0$ as $n$ goes to infinity, as you guessed.
If you choose $n$ of $3n$ hats, then by similar reasoning, the probability that none is chosen twice is $$(3n)!over (2n)!(3n)^n$$ and again, this goes to $0$.
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
$endgroup$
If you choose $n$ of $2n$ hats without replacement then there are $2n$ ways to choose the first hat, $2n-1$ ways to choose a second hat, $2n-2$ ways to choose a third hat different from the first two, and so on. There are $${(2n)!over n!}$$ ways to choose $n$ distinct hats.
There are $(2n)^n$ way to choose $n$ hats without replacement. So the probably that no hat is chosen twice is $${(2n)!over n!(2n)^n}$$ This goes to $0$ as $n$ goes to infinity, as you guessed.
If you choose $n$ of $3n$ hats, then by similar reasoning, the probability that none is chosen twice is $$(3n)!over (2n)!(3n)^n$$ and again, this goes to $0$.
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
answered Feb 2 at 16:51
saulspatzsaulspatz
17.3k31435
17.3k31435
$begingroup$
Wow, the formula works, great job, thx.
$endgroup$
– R Z
Feb 3 at 1:48
$begingroup$
! means factorial For $100$ hats the probability of no duplicate is about $6.666times10^{-14}$ For $1000$ hats, it's about $7.691times10^{-134}$
$endgroup$
– saulspatz
Feb 3 at 1:53
$begingroup$
How did you even manage to get those answers? Google calculator wouldnt even work with such numbers.
$endgroup$
– R Z
Feb 3 at 1:58
$begingroup$
I did it in python.
$endgroup$
– saulspatz
Feb 3 at 1:58
add a comment |
$begingroup$
Wow, the formula works, great job, thx.
$endgroup$
– R Z
Feb 3 at 1:48
$begingroup$
! means factorial For $100$ hats the probability of no duplicate is about $6.666times10^{-14}$ For $1000$ hats, it's about $7.691times10^{-134}$
$endgroup$
– saulspatz
Feb 3 at 1:53
$begingroup$
How did you even manage to get those answers? Google calculator wouldnt even work with such numbers.
$endgroup$
– R Z
Feb 3 at 1:58
$begingroup$
I did it in python.
$endgroup$
– saulspatz
Feb 3 at 1:58
$begingroup$
Wow, the formula works, great job, thx.
$endgroup$
– R Z
Feb 3 at 1:48
$begingroup$
Wow, the formula works, great job, thx.
$endgroup$
– R Z
Feb 3 at 1:48
$begingroup$
! means factorial For $100$ hats the probability of no duplicate is about $6.666times10^{-14}$ For $1000$ hats, it's about $7.691times10^{-134}$
$endgroup$
– saulspatz
Feb 3 at 1:53
$begingroup$
! means factorial For $100$ hats the probability of no duplicate is about $6.666times10^{-14}$ For $1000$ hats, it's about $7.691times10^{-134}$
$endgroup$
– saulspatz
Feb 3 at 1:53
$begingroup$
How did you even manage to get those answers? Google calculator wouldnt even work with such numbers.
$endgroup$
– R Z
Feb 3 at 1:58
$begingroup$
How did you even manage to get those answers? Google calculator wouldnt even work with such numbers.
$endgroup$
– R Z
Feb 3 at 1:58
$begingroup$
I did it in python.
$endgroup$
– saulspatz
Feb 3 at 1:58
$begingroup$
I did it in python.
$endgroup$
– saulspatz
Feb 3 at 1:58
add a comment |
$begingroup$
This is actually the generalized birthday problem. $n$ is the number of days. The number of items you have to mark to get a $50%$ chance of a match is about $sqrt {2 ln 2} sqrt n$. For other probabilities the constant $sqrt {2 ln 2}$ changes, but the $sqrt n$ behavior does not. Roughly speaking, the chance of any pair of marks going on the same item is about $frac 1n$ so you need the number of pairs of marks to be about $n$. The number of pairs is about half the square of the number of marks.
This means that if you choose a fixed fraction of $n$ to be the number of marks, eventually the probability of a remark will approach $1$ as $n$ gets large.
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks for you reply, but i dont understand this. Imagine you have numbers 0-9 (10 number in total) whats the probability to create a 5 digit number where all the numbers are different? Like 12345, 13589 but not 11081 22222 etc
$endgroup$
– R Z
Feb 2 at 16:42
$begingroup$
There are $10$ choices for the first digit, $9$ for the second, and so on, so $10cdot 9cdot 8cdot 7 cdot 6/10^5$
$endgroup$
– Ross Millikan
Feb 2 at 17:17
$begingroup$
So the answer will be 30,24 percent of it being all different numbers? I got there manualy before by 10/10 * 9/10 * 8/10 * 7/10 * 6/10... haha thanks for this simple formula.
$endgroup$
– R Z
Feb 3 at 0:59
$begingroup$
Yes, that is correct.
$endgroup$
– Ross Millikan
Feb 3 at 1:47
add a comment |
$begingroup$
This is actually the generalized birthday problem. $n$ is the number of days. The number of items you have to mark to get a $50%$ chance of a match is about $sqrt {2 ln 2} sqrt n$. For other probabilities the constant $sqrt {2 ln 2}$ changes, but the $sqrt n$ behavior does not. Roughly speaking, the chance of any pair of marks going on the same item is about $frac 1n$ so you need the number of pairs of marks to be about $n$. The number of pairs is about half the square of the number of marks.
This means that if you choose a fixed fraction of $n$ to be the number of marks, eventually the probability of a remark will approach $1$ as $n$ gets large.
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks for you reply, but i dont understand this. Imagine you have numbers 0-9 (10 number in total) whats the probability to create a 5 digit number where all the numbers are different? Like 12345, 13589 but not 11081 22222 etc
$endgroup$
– R Z
Feb 2 at 16:42
$begingroup$
There are $10$ choices for the first digit, $9$ for the second, and so on, so $10cdot 9cdot 8cdot 7 cdot 6/10^5$
$endgroup$
– Ross Millikan
Feb 2 at 17:17
$begingroup$
So the answer will be 30,24 percent of it being all different numbers? I got there manualy before by 10/10 * 9/10 * 8/10 * 7/10 * 6/10... haha thanks for this simple formula.
$endgroup$
– R Z
Feb 3 at 0:59
$begingroup$
Yes, that is correct.
$endgroup$
– Ross Millikan
Feb 3 at 1:47
add a comment |
$begingroup$
This is actually the generalized birthday problem. $n$ is the number of days. The number of items you have to mark to get a $50%$ chance of a match is about $sqrt {2 ln 2} sqrt n$. For other probabilities the constant $sqrt {2 ln 2}$ changes, but the $sqrt n$ behavior does not. Roughly speaking, the chance of any pair of marks going on the same item is about $frac 1n$ so you need the number of pairs of marks to be about $n$. The number of pairs is about half the square of the number of marks.
This means that if you choose a fixed fraction of $n$ to be the number of marks, eventually the probability of a remark will approach $1$ as $n$ gets large.
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
$endgroup$
This is actually the generalized birthday problem. $n$ is the number of days. The number of items you have to mark to get a $50%$ chance of a match is about $sqrt {2 ln 2} sqrt n$. For other probabilities the constant $sqrt {2 ln 2}$ changes, but the $sqrt n$ behavior does not. Roughly speaking, the chance of any pair of marks going on the same item is about $frac 1n$ so you need the number of pairs of marks to be about $n$. The number of pairs is about half the square of the number of marks.
This means that if you choose a fixed fraction of $n$ to be the number of marks, eventually the probability of a remark will approach $1$ as $n$ gets large.
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
answered Feb 2 at 14:39
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
Thanks for you reply, but i dont understand this. Imagine you have numbers 0-9 (10 number in total) whats the probability to create a 5 digit number where all the numbers are different? Like 12345, 13589 but not 11081 22222 etc
$endgroup$
– R Z
Feb 2 at 16:42
$begingroup$
There are $10$ choices for the first digit, $9$ for the second, and so on, so $10cdot 9cdot 8cdot 7 cdot 6/10^5$
$endgroup$
– Ross Millikan
Feb 2 at 17:17
$begingroup$
So the answer will be 30,24 percent of it being all different numbers? I got there manualy before by 10/10 * 9/10 * 8/10 * 7/10 * 6/10... haha thanks for this simple formula.
$endgroup$
– R Z
Feb 3 at 0:59
$begingroup$
Yes, that is correct.
$endgroup$
– Ross Millikan
Feb 3 at 1:47
add a comment |
$begingroup$
Thanks for you reply, but i dont understand this. Imagine you have numbers 0-9 (10 number in total) whats the probability to create a 5 digit number where all the numbers are different? Like 12345, 13589 but not 11081 22222 etc
$endgroup$
– R Z
Feb 2 at 16:42
$begingroup$
There are $10$ choices for the first digit, $9$ for the second, and so on, so $10cdot 9cdot 8cdot 7 cdot 6/10^5$
$endgroup$
– Ross Millikan
Feb 2 at 17:17
$begingroup$
So the answer will be 30,24 percent of it being all different numbers? I got there manualy before by 10/10 * 9/10 * 8/10 * 7/10 * 6/10... haha thanks for this simple formula.
$endgroup$
– R Z
Feb 3 at 0:59
$begingroup$
Yes, that is correct.
$endgroup$
– Ross Millikan
Feb 3 at 1:47
$begingroup$
Thanks for you reply, but i dont understand this. Imagine you have numbers 0-9 (10 number in total) whats the probability to create a 5 digit number where all the numbers are different? Like 12345, 13589 but not 11081 22222 etc
$endgroup$
– R Z
Feb 2 at 16:42
$begingroup$
Thanks for you reply, but i dont understand this. Imagine you have numbers 0-9 (10 number in total) whats the probability to create a 5 digit number where all the numbers are different? Like 12345, 13589 but not 11081 22222 etc
$endgroup$
– R Z
Feb 2 at 16:42
$begingroup$
There are $10$ choices for the first digit, $9$ for the second, and so on, so $10cdot 9cdot 8cdot 7 cdot 6/10^5$
$endgroup$
– Ross Millikan
Feb 2 at 17:17
$begingroup$
There are $10$ choices for the first digit, $9$ for the second, and so on, so $10cdot 9cdot 8cdot 7 cdot 6/10^5$
$endgroup$
– Ross Millikan
Feb 2 at 17:17
$begingroup$
So the answer will be 30,24 percent of it being all different numbers? I got there manualy before by 10/10 * 9/10 * 8/10 * 7/10 * 6/10... haha thanks for this simple formula.
$endgroup$
– R Z
Feb 3 at 0:59
$begingroup$
So the answer will be 30,24 percent of it being all different numbers? I got there manualy before by 10/10 * 9/10 * 8/10 * 7/10 * 6/10... haha thanks for this simple formula.
$endgroup$
– R Z
Feb 3 at 0:59
$begingroup$
Yes, that is correct.
$endgroup$
– Ross Millikan
Feb 3 at 1:47
$begingroup$
Yes, that is correct.
$endgroup$
– Ross Millikan
Feb 3 at 1:47
add a comment |
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$begingroup$
This isn't clear. How do you go about marking the hats? Do you mean if you choose a sample of $4$ of the $8$ with replacement, what is the probability that some hat is chosen at least twice? What have you done on this problem so far?
$endgroup$
– saulspatz
Feb 2 at 14:24
$begingroup$
You have 8 hats, you will choose/mark/ 4 times- what is the probability you choose 1 of them more than once! This is the base of the question. Or you have 20 hats and choose 10 times or 1000 hats and choose 500.....1:2 ratio.
$endgroup$
– R Z
Feb 2 at 16:33