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How many times must I roll a die to confidently assess its fairness?
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(Apologies in advance for use of lay language rather than statistical language.)
If I want to measure the odds of rolling each side of a specific physical six-sided die to within about +/- 2% with a reasonable confidence of certainty, how many sample die rolls would be needed?
i.e. How many times would I need to roll a die, counting each result, to be 98% sure that the chances it rolls each side are within 14.6% - 18.7%? (Or some similar criteria where one would be about 98% sure the die is fair to within 2%.)
(This is a real-world concern for simulation games using dice and wanting to be sure certain dice designs are acceptably close to 1/6 chance of rolling each number. There are claims that many common dice designs have been measured rolling 29% 1's by rolling several such dice 1000 times each.)
probability inference pdf dice
edited 28 mins ago
user1205901
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asked 6 hours ago
Dronz
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Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
6
down vote
favorite
(Apologies in advance for use of lay language rather than statistical language.)
If I want to measure the odds of rolling each side of a specific physical six-sided die to within about +/- 2% with a reasonable confidence of certainty, how many sample die rolls would be needed?
i.e. How many times would I need to roll a die, counting each result, to be 98% sure that the chances it rolls each side are within 14.6% - 18.7%? (Or some similar criteria where one would be about 98% sure the die is fair to within 2%.)
(This is a real-world concern for simulation games using dice and wanting to be sure certain dice designs are acceptably close to 1/6 chance of rolling each number. There are claims that many common dice designs have been measured rolling 29% 1's by rolling several such dice 1000 times each.)
probability inference pdf dice
edited 28 mins ago
user1205901
3,259144495
asked 6 hours ago
Dronz
1335
New contributor
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
(Apologies in advance for use of lay language rather than statistical language.)
If I want to measure the odds of rolling each side of a specific physical six-sided die to within about +/- 2% with a reasonable confidence of certainty, how many sample die rolls would be needed?
i.e. How many times would I need to roll a die, counting each result, to be 98% sure that the chances it rolls each side are within 14.6% - 18.7%? (Or some similar criteria where one would be about 98% sure the die is fair to within 2%.)
(This is a real-world concern for simulation games using dice and wanting to be sure certain dice designs are acceptably close to 1/6 chance of rolling each number. There are claims that many common dice designs have been measured rolling 29% 1's by rolling several such dice 1000 times each.)
probability inference pdf dice
edited 28 mins ago
user1205901
3,259144495
asked 6 hours ago
Dronz
1335
New contributor
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
(Apologies in advance for use of lay language rather than statistical language.)
If I want to measure the odds of rolling each side of a specific physical six-sided die to within about +/- 2% with a reasonable confidence of certainty, how many sample die rolls would be needed?
i.e. How many times would I need to roll a die, counting each result, to be 98% sure that the chances it rolls each side are within 14.6% - 18.7%? (Or some similar criteria where one would be about 98% sure the die is fair to within 2%.)
(This is a real-world concern for simulation games using dice and wanting to be sure certain dice designs are acceptably close to 1/6 chance of rolling each number. There are claims that many common dice designs have been measured rolling 29% 1's by rolling several such dice 1000 times each.)
probability inference pdf dice
probability inference pdf dice
edited 28 mins ago
user1205901
3,259144495
asked 6 hours ago
Dronz
1335
New contributor
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 28 mins ago
user1205901
3,259144495
asked 6 hours ago
Dronz
1335
New contributor
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 28 mins ago
user1205901
3,259144495
edited 28 mins ago
user1205901
3,259144495
edited 28 mins ago
user1205901
3,259144495
3,259144495
asked 6 hours ago
Dronz
1335
New contributor
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Dronz
1335
asked 6 hours ago
Dronz
1335
1335
New contributor
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dronz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Let $n$ be the number of rolls and $X$ the number of rolls that land on some specified side. Then $X$ follows a Binomial(n,p) distribution where $p$ is the probability of getting that specified side.
By the central limit theorem, we know that
$$sqrtn (X/n - p) to N(0,p(1-p))$$
Since $X/n$ is the sample mean of $n$ Bernoulli$(p)$ random variables. Hence for large $n$, confidence intervals for $p$ can be constructed as
$$fracXn pm Z sqrtfracp(1-p)n$$
Since $p$ is unknown, we can replace it with the sample average $hatp = X/n$, and by various convergence theorems, we know the resulting confidence interval will be asymptotically valid. So we get confidence intervals of the form
$$hatp pm Z sqrtfrachatp(1-hatp)n$$
with $hatp = X/n$. I'm going to assume you know what $Z$-scores are. For example, if you want a 95% confidence interval, you take $Z=1.96$. So for a given confidence level $alpha$ we have
$$hatp pm Z_alpha sqrtfrachatp(1-hatp)n$$
Now let's say you want this confidence interval to be of length less than $C_alpha$, and want to know how big a sample we need to make this case. Well this is equivelant to asking what $n_alpha$ satisfies
$$Z_alpha sqrtfrachatp(1-hatp)n_alpha leq fracC_alpha2$$
Which is then solved to obtain
$$n_alpha geq left(frac2 Z_alphaC_alpharight)^2 hatp(1-hatp)$$
So plug in your values for $Z_alpha$, $C_alpha$, and estimated $hatp$ to obtain an estimate for $n_alpha$. Note that since $p$ is unknown this is only an estimate, but asymptotically (as $n$ gets larger) it should be accurate.
answered 6 hours ago
Xiaomi
50111
1
Thanks. As I have not done college-type math in decades, could I trouble you to plug in the numbers and actually give me a ballpark number of times I'd need to roll a die, as an integer?
– Dronz
5 hours ago
4
if $p = 1/6$ and you want to know how large $n$ needs to be 98% sure the dice is fair to within 2%, $n$ needs to be at least $n geq 766$. Ignore my last comment, used incorrect $C_alpha$.
– Xiaomi
4 hours ago
Super, thank you! (So the 1000-roll tests were a reasonable number to use, particularly when done on several dice each. Interesting.)
– Dronz
2 hours ago
add a comment |Â
up vote
1
down vote
I started looking into this and it turns out that it's a very interesting question. It's a lot trickier than finding the confidence interval for a binomial, since you'd want to keep all probabilities in check. Have a look at this paper on simultaneous confidence intervals for multinomial distributions.
You can find some code in this blog post, which also gives a quick summary on some of the work that's been done on this.
answered 6 hours ago
idnavid
3778
4
Really, it's better to summarize the findings in your answer and include supplemental links than post what is, in effect, a link-only answer. Links rot, and then your answer becomes useless!
– jbowman
6 hours ago
1
@jbowman Noted. Thanks!
– idnavid
5 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Let $n$ be the number of rolls and $X$ the number of rolls that land on some specified side. Then $X$ follows a Binomial(n,p) distribution where $p$ is the probability of getting that specified side.
By the central limit theorem, we know that
$$sqrtn (X/n - p) to N(0,p(1-p))$$
Since $X/n$ is the sample mean of $n$ Bernoulli$(p)$ random variables. Hence for large $n$, confidence intervals for $p$ can be constructed as
$$fracXn pm Z sqrtfracp(1-p)n$$
Since $p$ is unknown, we can replace it with the sample average $hatp = X/n$, and by various convergence theorems, we know the resulting confidence interval will be asymptotically valid. So we get confidence intervals of the form
$$hatp pm Z sqrtfrachatp(1-hatp)n$$
with $hatp = X/n$. I'm going to assume you know what $Z$-scores are. For example, if you want a 95% confidence interval, you take $Z=1.96$. So for a given confidence level $alpha$ we have
$$hatp pm Z_alpha sqrtfrachatp(1-hatp)n$$
Now let's say you want this confidence interval to be of length less than $C_alpha$, and want to know how big a sample we need to make this case. Well this is equivelant to asking what $n_alpha$ satisfies
$$Z_alpha sqrtfrachatp(1-hatp)n_alpha leq fracC_alpha2$$
Which is then solved to obtain
$$n_alpha geq left(frac2 Z_alphaC_alpharight)^2 hatp(1-hatp)$$
So plug in your values for $Z_alpha$, $C_alpha$, and estimated $hatp$ to obtain an estimate for $n_alpha$. Note that since $p$ is unknown this is only an estimate, but asymptotically (as $n$ gets larger) it should be accurate.
answered 6 hours ago
Xiaomi
50111
1
Thanks. As I have not done college-type math in decades, could I trouble you to plug in the numbers and actually give me a ballpark number of times I'd need to roll a die, as an integer?
– Dronz
5 hours ago
4
if $p = 1/6$ and you want to know how large $n$ needs to be 98% sure the dice is fair to within 2%, $n$ needs to be at least $n geq 766$. Ignore my last comment, used incorrect $C_alpha$.
– Xiaomi
4 hours ago
Super, thank you! (So the 1000-roll tests were a reasonable number to use, particularly when done on several dice each. Interesting.)
– Dronz
2 hours ago
add a comment |Â
up vote
6
down vote
accepted
Let $n$ be the number of rolls and $X$ the number of rolls that land on some specified side. Then $X$ follows a Binomial(n,p) distribution where $p$ is the probability of getting that specified side.
By the central limit theorem, we know that
$$sqrtn (X/n - p) to N(0,p(1-p))$$
Since $X/n$ is the sample mean of $n$ Bernoulli$(p)$ random variables. Hence for large $n$, confidence intervals for $p$ can be constructed as
$$fracXn pm Z sqrtfracp(1-p)n$$
Since $p$ is unknown, we can replace it with the sample average $hatp = X/n$, and by various convergence theorems, we know the resulting confidence interval will be asymptotically valid. So we get confidence intervals of the form
$$hatp pm Z sqrtfrachatp(1-hatp)n$$
with $hatp = X/n$. I'm going to assume you know what $Z$-scores are. For example, if you want a 95% confidence interval, you take $Z=1.96$. So for a given confidence level $alpha$ we have
$$hatp pm Z_alpha sqrtfrachatp(1-hatp)n$$
Now let's say you want this confidence interval to be of length less than $C_alpha$, and want to know how big a sample we need to make this case. Well this is equivelant to asking what $n_alpha$ satisfies
$$Z_alpha sqrtfrachatp(1-hatp)n_alpha leq fracC_alpha2$$
Which is then solved to obtain
$$n_alpha geq left(frac2 Z_alphaC_alpharight)^2 hatp(1-hatp)$$
So plug in your values for $Z_alpha$, $C_alpha$, and estimated $hatp$ to obtain an estimate for $n_alpha$. Note that since $p$ is unknown this is only an estimate, but asymptotically (as $n$ gets larger) it should be accurate.
answered 6 hours ago
Xiaomi
50111
1
Thanks. As I have not done college-type math in decades, could I trouble you to plug in the numbers and actually give me a ballpark number of times I'd need to roll a die, as an integer?
– Dronz
5 hours ago
4
if $p = 1/6$ and you want to know how large $n$ needs to be 98% sure the dice is fair to within 2%, $n$ needs to be at least $n geq 766$. Ignore my last comment, used incorrect $C_alpha$.
– Xiaomi
4 hours ago
Super, thank you! (So the 1000-roll tests were a reasonable number to use, particularly when done on several dice each. Interesting.)
– Dronz
2 hours ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Let $n$ be the number of rolls and $X$ the number of rolls that land on some specified side. Then $X$ follows a Binomial(n,p) distribution where $p$ is the probability of getting that specified side.
By the central limit theorem, we know that
$$sqrtn (X/n - p) to N(0,p(1-p))$$
Since $X/n$ is the sample mean of $n$ Bernoulli$(p)$ random variables. Hence for large $n$, confidence intervals for $p$ can be constructed as
$$fracXn pm Z sqrtfracp(1-p)n$$
Since $p$ is unknown, we can replace it with the sample average $hatp = X/n$, and by various convergence theorems, we know the resulting confidence interval will be asymptotically valid. So we get confidence intervals of the form
$$hatp pm Z sqrtfrachatp(1-hatp)n$$
with $hatp = X/n$. I'm going to assume you know what $Z$-scores are. For example, if you want a 95% confidence interval, you take $Z=1.96$. So for a given confidence level $alpha$ we have
$$hatp pm Z_alpha sqrtfrachatp(1-hatp)n$$
Now let's say you want this confidence interval to be of length less than $C_alpha$, and want to know how big a sample we need to make this case. Well this is equivelant to asking what $n_alpha$ satisfies
$$Z_alpha sqrtfrachatp(1-hatp)n_alpha leq fracC_alpha2$$
Which is then solved to obtain
$$n_alpha geq left(frac2 Z_alphaC_alpharight)^2 hatp(1-hatp)$$
So plug in your values for $Z_alpha$, $C_alpha$, and estimated $hatp$ to obtain an estimate for $n_alpha$. Note that since $p$ is unknown this is only an estimate, but asymptotically (as $n$ gets larger) it should be accurate.
answered 6 hours ago
Xiaomi
50111
Let $n$ be the number of rolls and $X$ the number of rolls that land on some specified side. Then $X$ follows a Binomial(n,p) distribution where $p$ is the probability of getting that specified side.
By the central limit theorem, we know that
$$sqrtn (X/n - p) to N(0,p(1-p))$$
Since $X/n$ is the sample mean of $n$ Bernoulli$(p)$ random variables. Hence for large $n$, confidence intervals for $p$ can be constructed as
$$fracXn pm Z sqrtfracp(1-p)n$$
Since $p$ is unknown, we can replace it with the sample average $hatp = X/n$, and by various convergence theorems, we know the resulting confidence interval will be asymptotically valid. So we get confidence intervals of the form
$$hatp pm Z sqrtfrachatp(1-hatp)n$$
with $hatp = X/n$. I'm going to assume you know what $Z$-scores are. For example, if you want a 95% confidence interval, you take $Z=1.96$. So for a given confidence level $alpha$ we have
$$hatp pm Z_alpha sqrtfrachatp(1-hatp)n$$
Now let's say you want this confidence interval to be of length less than $C_alpha$, and want to know how big a sample we need to make this case. Well this is equivelant to asking what $n_alpha$ satisfies
$$Z_alpha sqrtfrachatp(1-hatp)n_alpha leq fracC_alpha2$$
Which is then solved to obtain
$$n_alpha geq left(frac2 Z_alphaC_alpharight)^2 hatp(1-hatp)$$
So plug in your values for $Z_alpha$, $C_alpha$, and estimated $hatp$ to obtain an estimate for $n_alpha$. Note that since $p$ is unknown this is only an estimate, but asymptotically (as $n$ gets larger) it should be accurate.
answered 6 hours ago
Xiaomi
50111
answered 6 hours ago
Xiaomi
50111
answered 6 hours ago
Xiaomi
50111
answered 6 hours ago
Xiaomi
50111
50111
1
Thanks. As I have not done college-type math in decades, could I trouble you to plug in the numbers and actually give me a ballpark number of times I'd need to roll a die, as an integer?
– Dronz
5 hours ago
4
if $p = 1/6$ and you want to know how large $n$ needs to be 98% sure the dice is fair to within 2%, $n$ needs to be at least $n geq 766$. Ignore my last comment, used incorrect $C_alpha$.
– Xiaomi
4 hours ago
Super, thank you! (So the 1000-roll tests were a reasonable number to use, particularly when done on several dice each. Interesting.)
– Dronz
2 hours ago
add a comment |Â
1
Thanks. As I have not done college-type math in decades, could I trouble you to plug in the numbers and actually give me a ballpark number of times I'd need to roll a die, as an integer?
– Dronz
5 hours ago
4
if $p = 1/6$ and you want to know how large $n$ needs to be 98% sure the dice is fair to within 2%, $n$ needs to be at least $n geq 766$. Ignore my last comment, used incorrect $C_alpha$.
– Xiaomi
4 hours ago
Super, thank you! (So the 1000-roll tests were a reasonable number to use, particularly when done on several dice each. Interesting.)
– Dronz
2 hours ago
1
1
Thanks. As I have not done college-type math in decades, could I trouble you to plug in the numbers and actually give me a ballpark number of times I'd need to roll a die, as an integer?
– Dronz
5 hours ago
Thanks. As I have not done college-type math in decades, could I trouble you to plug in the numbers and actually give me a ballpark number of times I'd need to roll a die, as an integer?
– Dronz
5 hours ago
4
4
if $p = 1/6$ and you want to know how large $n$ needs to be 98% sure the dice is fair to within 2%, $n$ needs to be at least $n geq 766$. Ignore my last comment, used incorrect $C_alpha$.
– Xiaomi
4 hours ago
if $p = 1/6$ and you want to know how large $n$ needs to be 98% sure the dice is fair to within 2%, $n$ needs to be at least $n geq 766$. Ignore my last comment, used incorrect $C_alpha$.
– Xiaomi
4 hours ago
Super, thank you! (So the 1000-roll tests were a reasonable number to use, particularly when done on several dice each. Interesting.)
– Dronz
2 hours ago
Super, thank you! (So the 1000-roll tests were a reasonable number to use, particularly when done on several dice each. Interesting.)
– Dronz
2 hours ago
add a comment |Â
up vote
1
down vote
I started looking into this and it turns out that it's a very interesting question. It's a lot trickier than finding the confidence interval for a binomial, since you'd want to keep all probabilities in check. Have a look at this paper on simultaneous confidence intervals for multinomial distributions.
You can find some code in this blog post, which also gives a quick summary on some of the work that's been done on this.
answered 6 hours ago
idnavid
3778
4
Really, it's better to summarize the findings in your answer and include supplemental links than post what is, in effect, a link-only answer. Links rot, and then your answer becomes useless!
– jbowman
6 hours ago
1
@jbowman Noted. Thanks!
– idnavid
5 hours ago
add a comment |Â
up vote
1
down vote
I started looking into this and it turns out that it's a very interesting question. It's a lot trickier than finding the confidence interval for a binomial, since you'd want to keep all probabilities in check. Have a look at this paper on simultaneous confidence intervals for multinomial distributions.
You can find some code in this blog post, which also gives a quick summary on some of the work that's been done on this.
answered 6 hours ago
idnavid
3778
4
Really, it's better to summarize the findings in your answer and include supplemental links than post what is, in effect, a link-only answer. Links rot, and then your answer becomes useless!
– jbowman
6 hours ago
1
@jbowman Noted. Thanks!
– idnavid
5 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I started looking into this and it turns out that it's a very interesting question. It's a lot trickier than finding the confidence interval for a binomial, since you'd want to keep all probabilities in check. Have a look at this paper on simultaneous confidence intervals for multinomial distributions.
You can find some code in this blog post, which also gives a quick summary on some of the work that's been done on this.
answered 6 hours ago
idnavid
3778
I started looking into this and it turns out that it's a very interesting question. It's a lot trickier than finding the confidence interval for a binomial, since you'd want to keep all probabilities in check. Have a look at this paper on simultaneous confidence intervals for multinomial distributions.
You can find some code in this blog post, which also gives a quick summary on some of the work that's been done on this.
answered 6 hours ago
idnavid
3778
answered 6 hours ago
idnavid
3778
answered 6 hours ago
idnavid
3778
answered 6 hours ago
idnavid
3778
3778
4
Really, it's better to summarize the findings in your answer and include supplemental links than post what is, in effect, a link-only answer. Links rot, and then your answer becomes useless!
– jbowman
6 hours ago
1
@jbowman Noted. Thanks!
– idnavid
5 hours ago
add a comment |Â
4
Really, it's better to summarize the findings in your answer and include supplemental links than post what is, in effect, a link-only answer. Links rot, and then your answer becomes useless!
– jbowman
6 hours ago
1
@jbowman Noted. Thanks!
– idnavid
5 hours ago
4
4
Really, it's better to summarize the findings in your answer and include supplemental links than post what is, in effect, a link-only answer. Links rot, and then your answer becomes useless!
– jbowman
6 hours ago
Really, it's better to summarize the findings in your answer and include supplemental links than post what is, in effect, a link-only answer. Links rot, and then your answer becomes useless!
– jbowman
6 hours ago
1
1
@jbowman Noted. Thanks!
– idnavid
5 hours ago
@jbowman Noted. Thanks!
– idnavid
5 hours ago
add a comment |Â
Dronz is a new contributor. Be nice, and check out our Code of Conduct.
Dronz is a new contributor. Be nice, and check out our Code of Conduct.
Dronz is a new contributor. Be nice, and check out our Code of Conduct.
Dronz is a new contributor. Be nice, and check out our Code of Conduct.
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