How many times must I roll a die to confidently assess its fairnes?

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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

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asked 1 hour ago

Dronz

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up vote
2
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

share|cite|improve this question

asked 1 hour ago

Dronz

1114

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 | 

up vote
2
down vote

favorite

up vote
2
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

share|cite|improve this question

asked 1 hour ago

Dronz

1114

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

share|cite|improve this question

asked 1 hour ago

Dronz

1114

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.

share|cite|improve this question

asked 1 hour ago

Dronz

1114

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.

share|cite|improve this question

share|cite|improve this question

asked 1 hour ago

Dronz

1114

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 1 hour ago

Dronz

1114

asked 1 hour ago

Dronz

1114

1114

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 | 

add a comment | 

2 Answers
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up vote
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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.

share|cite|improve this answer

answered 35 mins ago

Xiaomi

44611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  24 mins ago
  

  
  
  
  

add a comment | 

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.

share|cite|improve this answer

answered 39 mins ago

idnavid

3778

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jbowman Noted. Thanks!
  – idnavid
  15 mins ago
  

  
  
  
  

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

up vote
2
down vote

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.

share|cite|improve this answer

answered 35 mins ago

Xiaomi

44611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  24 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

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.

share|cite|improve this answer

answered 35 mins ago

Xiaomi

44611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  24 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

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.

share|cite|improve this answer

answered 35 mins ago

Xiaomi

44611

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.

share|cite|improve this answer

answered 35 mins ago

Xiaomi

44611

share|cite|improve this answer

share|cite|improve this answer

answered 35 mins ago

Xiaomi

44611

answered 35 mins ago

Xiaomi

44611

answered 35 mins ago

Xiaomi

44611

44611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  24 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  24 mins 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
24 mins 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
24 mins ago

add a comment | 

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.

share|cite|improve this answer

answered 39 mins ago

idnavid

3778

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jbowman Noted. Thanks!
  – idnavid
  15 mins ago
  

  
  
  
  

add a comment | 

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.

share|cite|improve this answer

answered 39 mins ago

idnavid

3778

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jbowman Noted. Thanks!
  – idnavid
  15 mins ago
  

  
  
  
  

add a comment | 

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.

share|cite|improve this answer

answered 39 mins 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.

share|cite|improve this answer

answered 39 mins ago

idnavid

3778

share|cite|improve this answer

share|cite|improve this answer

answered 39 mins ago

idnavid

3778

answered 39 mins ago

idnavid

3778

answered 39 mins ago

idnavid

3778

3778

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jbowman Noted. Thanks!
  – idnavid
  15 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  38 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jbowman Noted. Thanks!
  – idnavid
  15 mins 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
38 mins 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
38 mins ago

@jbowman Noted. Thanks!
– idnavid
15 mins ago

@jbowman Noted. Thanks!
– idnavid
15 mins ago

add a comment | 

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