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3

$begingroup$

We have an 8-sided die. Throwing it 20 times,
what is the probability of rolling a number larger than 6 at least 12 times and at most 16 times?

If you know of an efficient program to write so that I won't have to type in the maths manually, I would be so happy.

r probability dice

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asked Jun 2 at 6:41

Fluffy WombatFluffy Wombat

182

New contributor

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Check out our Code of Conduct.

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closed as off-topic by COOLSerdash, mkt, Siong Thye Goh, mdewey, Glen_b♦ Jun 2 at 16:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question appears to be off-topic because EITHER it is not about statistics, machine learning, data analysis, data mining, or data visualization, OR it focuses on programming, debugging, or performing routine operations within a statistical computing platform. If the latter, you could try the support links we maintain." – COOLSerdash, mkt, Siong Thye Goh, mdewey, Glen_b

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

3

$begingroup$

We have an 8-sided die. Throwing it 20 times,
what is the probability of rolling a number larger than 6 at least 12 times and at most 16 times?

If you know of an efficient program to write so that I won't have to type in the maths manually, I would be so happy.

r probability dice

share|cite|improve this question

asked Jun 2 at 6:41

Fluffy WombatFluffy Wombat

182

New contributor

Fluffy Wombat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

closed as off-topic by COOLSerdash, mkt, Siong Thye Goh, mdewey, Glen_b♦ Jun 2 at 16:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question appears to be off-topic because EITHER it is not about statistics, machine learning, data analysis, data mining, or data visualization, OR it focuses on programming, debugging, or performing routine operations within a statistical computing platform. If the latter, you could try the support links we maintain." – COOLSerdash, mkt, Siong Thye Goh, mdewey, Glen_b

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

We have an 8-sided die. Throwing it 20 times,
what is the probability of rolling a number larger than 6 at least 12 times and at most 16 times?

If you know of an efficient program to write so that I won't have to type in the maths manually, I would be so happy.

r probability dice

share|cite|improve this question

asked Jun 2 at 6:41

Fluffy WombatFluffy Wombat

182

New contributor

Fluffy Wombat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Jun 2 at 6:41

Fluffy WombatFluffy Wombat

182

New contributor

Fluffy Wombat 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 Jun 2 at 6:41

Fluffy WombatFluffy Wombat

182

New contributor

Fluffy Wombat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 2 at 6:41

Fluffy WombatFluffy Wombat

182

New contributor

Fluffy Wombat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 2 at 6:41

Fluffy WombatFluffy Wombat

182

2 Answers
2

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3

$begingroup$

The answer by CoolSerdash shows you how to simulate the probability of interest. Here is a direct probability calculation without simulation:

sides <- 8 # Number of sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Cut-off value (must roll above this)
n_lower <- 12 # Lower bound
n_upper <- 16 # Upper bound

prob <- (sides - no_cut)/sides;
pbinom(n_upper, throws, prob) - pbinom(n_lower-1, throws, prob);

[1] 0.000935362

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answered Jun 2 at 7:52

BenBen

31.6k236137

$endgroup$

add a comment | 

2

$begingroup$

Here is a quick implementation. I'm sure there are more efficient ways.

d <- 8 # Sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Number at least on die
n_lower <- 12 # Lower cutoff
n_upper <- 16 # Upper cutoff

n_sim <- 1000000 # Number of simulations

res <- NULL

res <- replicate(n_sim, 
 x <- sample.int(d, size = throws, replace = TRUE)
 s <- sum(x > no_cut) 
 (s >= n_lower && s <= n_upper)
)

sum(res)/n_sim # simulated probability

share|cite|improve this answer

answered Jun 2 at 7:00

COOLSerdashCOOLSerdash

16.8k75395

$endgroup$

add a comment | 

3

$begingroup$

The answer by CoolSerdash shows you how to simulate the probability of interest. Here is a direct probability calculation without simulation:

sides <- 8 # Number of sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Cut-off value (must roll above this)
n_lower <- 12 # Lower bound
n_upper <- 16 # Upper bound

prob <- (sides - no_cut)/sides;
pbinom(n_upper, throws, prob) - pbinom(n_lower-1, throws, prob);

[1] 0.000935362

share|cite|improve this answer

answered Jun 2 at 7:52

BenBen

31.6k236137

$endgroup$

add a comment | 

3

$begingroup$

The answer by CoolSerdash shows you how to simulate the probability of interest. Here is a direct probability calculation without simulation:

sides <- 8 # Number of sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Cut-off value (must roll above this)
n_lower <- 12 # Lower bound
n_upper <- 16 # Upper bound

prob <- (sides - no_cut)/sides;
pbinom(n_upper, throws, prob) - pbinom(n_lower-1, throws, prob);

[1] 0.000935362

share|cite|improve this answer

answered Jun 2 at 7:52

BenBen

31.6k236137

$endgroup$

add a comment | 

$begingroup$

The answer by CoolSerdash shows you how to simulate the probability of interest. Here is a direct probability calculation without simulation:

sides <- 8 # Number of sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Cut-off value (must roll above this)
n_lower <- 12 # Lower bound
n_upper <- 16 # Upper bound

prob <- (sides - no_cut)/sides;
pbinom(n_upper, throws, prob) - pbinom(n_lower-1, throws, prob);

[1] 0.000935362

share|cite|improve this answer

answered Jun 2 at 7:52

BenBen

31.6k236137

$endgroup$

sides <- 8 # Number of sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Cut-off value (must roll above this)
n_lower <- 12 # Lower bound
n_upper <- 16 # Upper bound

prob <- (sides - no_cut)/sides;
pbinom(n_upper, throws, prob) - pbinom(n_lower-1, throws, prob);

[1] 0.000935362

share|cite|improve this answer

answered Jun 2 at 7:52

BenBen

31.6k236137

answered Jun 2 at 7:52

BenBen

31.6k236137

answered Jun 2 at 7:52

BenBen

31.6k236137

2

$begingroup$

Here is a quick implementation. I'm sure there are more efficient ways.

d <- 8 # Sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Number at least on die
n_lower <- 12 # Lower cutoff
n_upper <- 16 # Upper cutoff

n_sim <- 1000000 # Number of simulations

res <- NULL

res <- replicate(n_sim, 
 x <- sample.int(d, size = throws, replace = TRUE)
 s <- sum(x > no_cut) 
 (s >= n_lower && s <= n_upper)
)

sum(res)/n_sim # simulated probability

share|cite|improve this answer

answered Jun 2 at 7:00

COOLSerdashCOOLSerdash

16.8k75395

$endgroup$

add a comment | 

2

$begingroup$

Here is a quick implementation. I'm sure there are more efficient ways.

d <- 8 # Sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Number at least on die
n_lower <- 12 # Lower cutoff
n_upper <- 16 # Upper cutoff

n_sim <- 1000000 # Number of simulations

res <- NULL

res <- replicate(n_sim, 
 x <- sample.int(d, size = throws, replace = TRUE)
 s <- sum(x > no_cut) 
 (s >= n_lower && s <= n_upper)
)

sum(res)/n_sim # simulated probability

share|cite|improve this answer

answered Jun 2 at 7:00

COOLSerdashCOOLSerdash

16.8k75395

$endgroup$

add a comment | 

$begingroup$

Here is a quick implementation. I'm sure there are more efficient ways.

d <- 8 # Sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Number at least on die
n_lower <- 12 # Lower cutoff
n_upper <- 16 # Upper cutoff

n_sim <- 1000000 # Number of simulations

res <- NULL

res <- replicate(n_sim, 
 x <- sample.int(d, size = throws, replace = TRUE)
 s <- sum(x > no_cut) 
 (s >= n_lower && s <= n_upper)
)

sum(res)/n_sim # simulated probability

share|cite|improve this answer

answered Jun 2 at 7:00

COOLSerdashCOOLSerdash

16.8k75395

$endgroup$

d <- 8 # Sides of the die
throws <- 20 # Number of throws
no_cut <- 6 # Number at least on die
n_lower <- 12 # Lower cutoff
n_upper <- 16 # Upper cutoff

n_sim <- 1000000 # Number of simulations

res <- NULL

res <- replicate(n_sim, 
 x <- sample.int(d, size = throws, replace = TRUE)
 s <- sum(x > no_cut) 
 (s >= n_lower && s <= n_upper)
)

sum(res)/n_sim # simulated probability

share|cite|improve this answer

answered Jun 2 at 7:00

COOLSerdashCOOLSerdash

16.8k75395

answered Jun 2 at 7:00

COOLSerdashCOOLSerdash

16.8k75395

answered Jun 2 at 7:00

COOLSerdashCOOLSerdash

16.8k75395