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Odds of rolling m or more 6's if you roll a 6 sided die n times

0

$begingroup$

If you roll a six-sided die $n$ times the odds of never rolling a 6 are $(frac 56)^n$ and so, the odds of rolling one or more 6s are $1-(frac56)^n$.

Can this be generalized to a situation where you desire $m$ 'hits'?

The specific situation I started thinking about this is one where the odds of success are $1$ in $10$, and $3$ successes are needed. I assumed that determining the odds of 3 hits after $n$ attempts would be fairly simple but I hit a wall pretty quickly (I have a pretty poor stats background).

I'm less concerned about that specific case, but more interested in: is there a simple solution to this problem in general?

probability

share|cite|improve this question

edited Mar 26 at 13:37

Ernie060

2,950719

asked Mar 26 at 1:02

BrianBrian

225

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. This is just a binomial process with success probability $frac 16$ so the standard formulas apply. If your numbers are large, then the usual normal approximation can be quite accurate.
  $endgroup$
  – lulu
  Mar 26 at 1:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow thanks, I love math but have never driven deep into the stat side!
  $endgroup$
  – Brian
  Mar 26 at 2:13
  

  
  
  
  

add a comment | 

0

$begingroup$

If you roll a six-sided die $n$ times the odds of never rolling a 6 are $(frac 56)^n$ and so, the odds of rolling one or more 6s are $1-(frac56)^n$.

Can this be generalized to a situation where you desire $m$ 'hits'?

The specific situation I started thinking about this is one where the odds of success are $1$ in $10$, and $3$ successes are needed. I assumed that determining the odds of 3 hits after $n$ attempts would be fairly simple but I hit a wall pretty quickly (I have a pretty poor stats background).

I'm less concerned about that specific case, but more interested in: is there a simple solution to this problem in general?

probability

share|cite|improve this question

edited Mar 26 at 13:37

Ernie060

2,950719

asked Mar 26 at 1:02

BrianBrian

225

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. This is just a binomial process with success probability $frac 16$ so the standard formulas apply. If your numbers are large, then the usual normal approximation can be quite accurate.
  $endgroup$
  – lulu
  Mar 26 at 1:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow thanks, I love math but have never driven deep into the stat side!
  $endgroup$
  – Brian
  Mar 26 at 2:13
  

  
  
  
  

add a comment | 

$begingroup$

If you roll a six-sided die $n$ times the odds of never rolling a 6 are $(frac 56)^n$ and so, the odds of rolling one or more 6s are $1-(frac56)^n$.

Can this be generalized to a situation where you desire $m$ 'hits'?

The specific situation I started thinking about this is one where the odds of success are $1$ in $10$, and $3$ successes are needed. I assumed that determining the odds of 3 hits after $n$ attempts would be fairly simple but I hit a wall pretty quickly (I have a pretty poor stats background).

I'm less concerned about that specific case, but more interested in: is there a simple solution to this problem in general?

probability

share|cite|improve this question

edited Mar 26 at 13:37

Ernie060

2,950719

asked Mar 26 at 1:02

BrianBrian

225

$endgroup$

share|cite|improve this question

edited Mar 26 at 13:37

Ernie060

2,950719

asked Mar 26 at 1:02

BrianBrian

225

share|cite|improve this question

edited Mar 26 at 13:37

Ernie060

2,950719

asked Mar 26 at 1:02

BrianBrian

225

edited Mar 26 at 13:37

Ernie060

2,950719

edited Mar 26 at 13:37

Ernie060

2,950719

asked Mar 26 at 1:02

BrianBrian

225

asked Mar 26 at 1:02

BrianBrian

225

1 Answer
1

active

oldest

votes

1

$begingroup$

as mentioned in the comments, this is a typical application of the Binomial distribution.

The Binomial distribution has two parameters:

• 
  $n$ is the number of trials, e.g. how often you throw the die


• 
  $p$ is the success probability of one trial, e.g. the probability of throwing a six.

Given these two parameters you can easily compute the probability of $m$ successes in $n$ trials by:

$$
Bbb P (text{exactly} m text{successes}) = {{n}choose {m}} p^m (1 -p)^{n-m}.
$$

share|cite|improve this answer

answered Mar 26 at 13:39

CetttCettt

2,010623

$endgroup$

add a comment | 

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1

$begingroup$

as mentioned in the comments, this is a typical application of the Binomial distribution.

The Binomial distribution has two parameters:

• 
  $n$ is the number of trials, e.g. how often you throw the die


• 
  $p$ is the success probability of one trial, e.g. the probability of throwing a six.

Given these two parameters you can easily compute the probability of $m$ successes in $n$ trials by:

$$
Bbb P (text{exactly} m text{successes}) = {{n}choose {m}} p^m (1 -p)^{n-m}.
$$

share|cite|improve this answer

answered Mar 26 at 13:39

CetttCettt

2,010623

$endgroup$

add a comment | 

1

$begingroup$

as mentioned in the comments, this is a typical application of the Binomial distribution.

The Binomial distribution has two parameters:

• 
  $n$ is the number of trials, e.g. how often you throw the die


• 
  $p$ is the success probability of one trial, e.g. the probability of throwing a six.

Given these two parameters you can easily compute the probability of $m$ successes in $n$ trials by:

$$
Bbb P (text{exactly} m text{successes}) = {{n}choose {m}} p^m (1 -p)^{n-m}.
$$

share|cite|improve this answer

answered Mar 26 at 13:39

CetttCettt

2,010623

$endgroup$

add a comment | 

$begingroup$

as mentioned in the comments, this is a typical application of the Binomial distribution.

The Binomial distribution has two parameters:

• 
  $n$ is the number of trials, e.g. how often you throw the die


• 
  $p$ is the success probability of one trial, e.g. the probability of throwing a six.

Given these two parameters you can easily compute the probability of $m$ successes in $n$ trials by:

$$
Bbb P (text{exactly} m text{successes}) = {{n}choose {m}} p^m (1 -p)^{n-m}.
$$

share|cite|improve this answer

answered Mar 26 at 13:39

CetttCettt

2,010623

$endgroup$

share|cite|improve this answer

answered Mar 26 at 13:39

CetttCettt

2,010623

answered Mar 26 at 13:39

CetttCettt

2,010623

answered Mar 26 at 13:39

CetttCettt

2,010623