Roll 6-sided die x times, chance of 6 occurring 10 times given that 1 occurs 0 times.
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I roll a 6-sided die an unknown number of times. The number 1 occurs no times, what is the chance that the number six occurs exactly 10 times?
I got as far as figuring out that I need the the probability that the die was rolled $x$ times given that a number occurred exactly $n$ times, however I can't figure out how to create an expression for this.
Edit:
To make this absolutely clear, I will define the problem like this. We roll a die $x$ times, where $x$ is a uniformly distributed random number between $0$ and $n$, and count the number of times that each number appears, $n_1$ is the number of ones, $n_2$ is the number of twos, etc. We want to calculate the probability $P(n_1 = 0 | n_6 = 10)$. We can use the binomial theorem to find a value for $P(n_1 = 0 | x=y)$ and $P(n_6 = 0 | x=y)$ for any $y$. By adding these together for all $x$, it should be possible to to find that number. I believe that the limit as $nrightarrow infty$ exists, but I might be wrong.
probability sequences-and-series
asked Aug 11 at 1:28
Information Aether
876
 |Â
show 1 more comment
up vote
1
down vote
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I roll a 6-sided die an unknown number of times. The number 1 occurs no times, what is the chance that the number six occurs exactly 10 times?
I got as far as figuring out that I need the the probability that the die was rolled $x$ times given that a number occurred exactly $n$ times, however I can't figure out how to create an expression for this.
Edit:
To make this absolutely clear, I will define the problem like this. We roll a die $x$ times, where $x$ is a uniformly distributed random number between $0$ and $n$, and count the number of times that each number appears, $n_1$ is the number of ones, $n_2$ is the number of twos, etc. We want to calculate the probability $P(n_1 = 0 | n_6 = 10)$. We can use the binomial theorem to find a value for $P(n_1 = 0 | x=y)$ and $P(n_6 = 0 | x=y)$ for any $y$. By adding these together for all $x$, it should be possible to to find that number. I believe that the limit as $nrightarrow infty$ exists, but I might be wrong.
probability sequences-and-series
asked Aug 11 at 1:28
Information Aether
876
This is equivalent to ignoring any rolls for which a 1 comes up, which effectively makes your die 5-sided.
– amd
Aug 11 at 2:43
The core of the question is that the number of rolls is unknown, but given that six occurred exactly 10 times, we can guess that the number of rolls is probably roughly 60. However, I need a more rigorous way of doing this to answer the question.
– Information Aether
Aug 11 at 3:36
That’s not what you’ve written in your question, though. You ask for the probability of rolling 6 exactly 10 times, not for the probability distribution of the number of rolls needed to get ten sixes, which is rather different.
– amd
Aug 11 at 3:46
@InformationAether Your edit does not really clarify your question. Make it more specific. What is does $n_1$ or $x$ mean in this context?
– callculus
Aug 11 at 3:59
1
what is given? $n_6=10$ or is it $n_1=0$?
– Siong Thye Goh
Aug 11 at 4:02
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I roll a 6-sided die an unknown number of times. The number 1 occurs no times, what is the chance that the number six occurs exactly 10 times?
I got as far as figuring out that I need the the probability that the die was rolled $x$ times given that a number occurred exactly $n$ times, however I can't figure out how to create an expression for this.
Edit:
To make this absolutely clear, I will define the problem like this. We roll a die $x$ times, where $x$ is a uniformly distributed random number between $0$ and $n$, and count the number of times that each number appears, $n_1$ is the number of ones, $n_2$ is the number of twos, etc. We want to calculate the probability $P(n_1 = 0 | n_6 = 10)$. We can use the binomial theorem to find a value for $P(n_1 = 0 | x=y)$ and $P(n_6 = 0 | x=y)$ for any $y$. By adding these together for all $x$, it should be possible to to find that number. I believe that the limit as $nrightarrow infty$ exists, but I might be wrong.
probability sequences-and-series
asked Aug 11 at 1:28
Information Aether
876
I roll a 6-sided die an unknown number of times. The number 1 occurs no times, what is the chance that the number six occurs exactly 10 times?
I got as far as figuring out that I need the the probability that the die was rolled $x$ times given that a number occurred exactly $n$ times, however I can't figure out how to create an expression for this.
Edit:
To make this absolutely clear, I will define the problem like this. We roll a die $x$ times, where $x$ is a uniformly distributed random number between $0$ and $n$, and count the number of times that each number appears, $n_1$ is the number of ones, $n_2$ is the number of twos, etc. We want to calculate the probability $P(n_1 = 0 | n_6 = 10)$. We can use the binomial theorem to find a value for $P(n_1 = 0 | x=y)$ and $P(n_6 = 0 | x=y)$ for any $y$. By adding these together for all $x$, it should be possible to to find that number. I believe that the limit as $nrightarrow infty$ exists, but I might be wrong.
probability sequences-and-series
asked Aug 11 at 1:28
Information Aether
876
edited Aug 11 at 19:05
asked Aug 11 at 1:28
Information Aether
876
asked Aug 11 at 1:28
Information Aether
876
asked Aug 11 at 1:28
Information Aether
876
876
This is equivalent to ignoring any rolls for which a 1 comes up, which effectively makes your die 5-sided.
– amd
Aug 11 at 2:43
The core of the question is that the number of rolls is unknown, but given that six occurred exactly 10 times, we can guess that the number of rolls is probably roughly 60. However, I need a more rigorous way of doing this to answer the question.
– Information Aether
Aug 11 at 3:36
That’s not what you’ve written in your question, though. You ask for the probability of rolling 6 exactly 10 times, not for the probability distribution of the number of rolls needed to get ten sixes, which is rather different.
– amd
Aug 11 at 3:46
@InformationAether Your edit does not really clarify your question. Make it more specific. What is does $n_1$ or $x$ mean in this context?
– callculus
Aug 11 at 3:59
1
what is given? $n_6=10$ or is it $n_1=0$?
– Siong Thye Goh
Aug 11 at 4:02
 |Â
show 1 more comment
This is equivalent to ignoring any rolls for which a 1 comes up, which effectively makes your die 5-sided.
– amd
Aug 11 at 2:43
The core of the question is that the number of rolls is unknown, but given that six occurred exactly 10 times, we can guess that the number of rolls is probably roughly 60. However, I need a more rigorous way of doing this to answer the question.
– Information Aether
Aug 11 at 3:36
That’s not what you’ve written in your question, though. You ask for the probability of rolling 6 exactly 10 times, not for the probability distribution of the number of rolls needed to get ten sixes, which is rather different.
– amd
Aug 11 at 3:46
@InformationAether Your edit does not really clarify your question. Make it more specific. What is does $n_1$ or $x$ mean in this context?
– callculus
Aug 11 at 3:59
1
what is given? $n_6=10$ or is it $n_1=0$?
– Siong Thye Goh
Aug 11 at 4:02
This is equivalent to ignoring any rolls for which a 1 comes up, which effectively makes your die 5-sided.
– amd
Aug 11 at 2:43
This is equivalent to ignoring any rolls for which a 1 comes up, which effectively makes your die 5-sided.
– amd
Aug 11 at 2:43
The core of the question is that the number of rolls is unknown, but given that six occurred exactly 10 times, we can guess that the number of rolls is probably roughly 60. However, I need a more rigorous way of doing this to answer the question.
– Information Aether
Aug 11 at 3:36
The core of the question is that the number of rolls is unknown, but given that six occurred exactly 10 times, we can guess that the number of rolls is probably roughly 60. However, I need a more rigorous way of doing this to answer the question.
– Information Aether
Aug 11 at 3:36
That’s not what you’ve written in your question, though. You ask for the probability of rolling 6 exactly 10 times, not for the probability distribution of the number of rolls needed to get ten sixes, which is rather different.
– amd
Aug 11 at 3:46
That’s not what you’ve written in your question, though. You ask for the probability of rolling 6 exactly 10 times, not for the probability distribution of the number of rolls needed to get ten sixes, which is rather different.
– amd
Aug 11 at 3:46
@InformationAether Your edit does not really clarify your question. Make it more specific. What is does $n_1$ or $x$ mean in this context?
– callculus
Aug 11 at 3:59
@InformationAether Your edit does not really clarify your question. Make it more specific. What is does $n_1$ or $x$ mean in this context?
– callculus
Aug 11 at 3:59
1
1
what is given? $n_6=10$ or is it $n_1=0$?
– Siong Thye Goh
Aug 11 at 4:02
what is given? $n_6=10$ or is it $n_1=0$?
– Siong Thye Goh
Aug 11 at 4:02
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
Guide:
If a particular number $j$ occured exactly $n_j=n$ times out of $x$ times, then for the remaining $x-n$ times, the other number occur equally likely. We already know that for those tosses, the outcome can't be $j$, but for it take other values that is not equal to $j$, it occur with probability $frac15$ each.
Let $X sim Unif 1, 6$, let $j, k in 1, ldots, 6, j ne k, $then we have
$$P(X=j|X ne k)=fracP(X=j, Xne k)P(X ne k)=fracP(X=j)1-P(X=k)=fracfrac16frac56=frac15$$
- Binomial distribution might be helpful for your question.
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
Sure, given that a certain number occurred n times, the average number of times that all other numbers occurred is n. But that doesn't mean a different number can't occur a different number of times. The greater the distance between between numbers, the lower the probability.
– Information Aether
Aug 11 at 2:08
yes, they can occcur a different number of time, just saying that their distribution is now independently uniform, at a particular toss, each number happens with probability $frac15$.
– Siong Thye Goh
Aug 11 at 2:10
That doesn't answer my question.
– Information Aether
Aug 11 at 2:28
hmmm... can you clarify what is your question?
– Siong Thye Goh
Aug 11 at 2:30
"Given that a certain number offcured $n$ times, the average number of times that all numbers occured is $n$" doesn't make sense, it depends on how many tosses remain. For example if out of $6$ tosses, $1$ occur exactly $5$ times, then for the remaining one toss, the probability of getting $2$ is equal to $frac15$, same for probability of getting $3$.
– Siong Thye Goh
Aug 11 at 2:32
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Guide:
If a particular number $j$ occured exactly $n_j=n$ times out of $x$ times, then for the remaining $x-n$ times, the other number occur equally likely. We already know that for those tosses, the outcome can't be $j$, but for it take other values that is not equal to $j$, it occur with probability $frac15$ each.
Let $X sim Unif 1, 6$, let $j, k in 1, ldots, 6, j ne k, $then we have
$$P(X=j|X ne k)=fracP(X=j, Xne k)P(X ne k)=fracP(X=j)1-P(X=k)=fracfrac16frac56=frac15$$
- Binomial distribution might be helpful for your question.
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
Sure, given that a certain number occurred n times, the average number of times that all other numbers occurred is n. But that doesn't mean a different number can't occur a different number of times. The greater the distance between between numbers, the lower the probability.
– Information Aether
Aug 11 at 2:08
yes, they can occcur a different number of time, just saying that their distribution is now independently uniform, at a particular toss, each number happens with probability $frac15$.
– Siong Thye Goh
Aug 11 at 2:10
That doesn't answer my question.
– Information Aether
Aug 11 at 2:28
hmmm... can you clarify what is your question?
– Siong Thye Goh
Aug 11 at 2:30
"Given that a certain number offcured $n$ times, the average number of times that all numbers occured is $n$" doesn't make sense, it depends on how many tosses remain. For example if out of $6$ tosses, $1$ occur exactly $5$ times, then for the remaining one toss, the probability of getting $2$ is equal to $frac15$, same for probability of getting $3$.
– Siong Thye Goh
Aug 11 at 2:32
 |Â
show 2 more comments
up vote
1
down vote
Guide:
If a particular number $j$ occured exactly $n_j=n$ times out of $x$ times, then for the remaining $x-n$ times, the other number occur equally likely. We already know that for those tosses, the outcome can't be $j$, but for it take other values that is not equal to $j$, it occur with probability $frac15$ each.
Let $X sim Unif 1, 6$, let $j, k in 1, ldots, 6, j ne k, $then we have
$$P(X=j|X ne k)=fracP(X=j, Xne k)P(X ne k)=fracP(X=j)1-P(X=k)=fracfrac16frac56=frac15$$
- Binomial distribution might be helpful for your question.
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
Sure, given that a certain number occurred n times, the average number of times that all other numbers occurred is n. But that doesn't mean a different number can't occur a different number of times. The greater the distance between between numbers, the lower the probability.
– Information Aether
Aug 11 at 2:08
yes, they can occcur a different number of time, just saying that their distribution is now independently uniform, at a particular toss, each number happens with probability $frac15$.
– Siong Thye Goh
Aug 11 at 2:10
That doesn't answer my question.
– Information Aether
Aug 11 at 2:28
hmmm... can you clarify what is your question?
– Siong Thye Goh
Aug 11 at 2:30
"Given that a certain number offcured $n$ times, the average number of times that all numbers occured is $n$" doesn't make sense, it depends on how many tosses remain. For example if out of $6$ tosses, $1$ occur exactly $5$ times, then for the remaining one toss, the probability of getting $2$ is equal to $frac15$, same for probability of getting $3$.
– Siong Thye Goh
Aug 11 at 2:32
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
Guide:
If a particular number $j$ occured exactly $n_j=n$ times out of $x$ times, then for the remaining $x-n$ times, the other number occur equally likely. We already know that for those tosses, the outcome can't be $j$, but for it take other values that is not equal to $j$, it occur with probability $frac15$ each.
Let $X sim Unif 1, 6$, let $j, k in 1, ldots, 6, j ne k, $then we have
$$P(X=j|X ne k)=fracP(X=j, Xne k)P(X ne k)=fracP(X=j)1-P(X=k)=fracfrac16frac56=frac15$$
- Binomial distribution might be helpful for your question.
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
Guide:
If a particular number $j$ occured exactly $n_j=n$ times out of $x$ times, then for the remaining $x-n$ times, the other number occur equally likely. We already know that for those tosses, the outcome can't be $j$, but for it take other values that is not equal to $j$, it occur with probability $frac15$ each.
Let $X sim Unif 1, 6$, let $j, k in 1, ldots, 6, j ne k, $then we have
$$P(X=j|X ne k)=fracP(X=j, Xne k)P(X ne k)=fracP(X=j)1-P(X=k)=fracfrac16frac56=frac15$$
- Binomial distribution might be helpful for your question.
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
edited Aug 11 at 2:44
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
answered Aug 11 at 1:44
Siong Thye Goh
78.6k134997
78.6k134997
Sure, given that a certain number occurred n times, the average number of times that all other numbers occurred is n. But that doesn't mean a different number can't occur a different number of times. The greater the distance between between numbers, the lower the probability.
– Information Aether
Aug 11 at 2:08
yes, they can occcur a different number of time, just saying that their distribution is now independently uniform, at a particular toss, each number happens with probability $frac15$.
– Siong Thye Goh
Aug 11 at 2:10
That doesn't answer my question.
– Information Aether
Aug 11 at 2:28
hmmm... can you clarify what is your question?
– Siong Thye Goh
Aug 11 at 2:30
"Given that a certain number offcured $n$ times, the average number of times that all numbers occured is $n$" doesn't make sense, it depends on how many tosses remain. For example if out of $6$ tosses, $1$ occur exactly $5$ times, then for the remaining one toss, the probability of getting $2$ is equal to $frac15$, same for probability of getting $3$.
– Siong Thye Goh
Aug 11 at 2:32
 |Â
show 2 more comments
Sure, given that a certain number occurred n times, the average number of times that all other numbers occurred is n. But that doesn't mean a different number can't occur a different number of times. The greater the distance between between numbers, the lower the probability.
– Information Aether
Aug 11 at 2:08
yes, they can occcur a different number of time, just saying that their distribution is now independently uniform, at a particular toss, each number happens with probability $frac15$.
– Siong Thye Goh
Aug 11 at 2:10
That doesn't answer my question.
– Information Aether
Aug 11 at 2:28
hmmm... can you clarify what is your question?
– Siong Thye Goh
Aug 11 at 2:30
"Given that a certain number offcured $n$ times, the average number of times that all numbers occured is $n$" doesn't make sense, it depends on how many tosses remain. For example if out of $6$ tosses, $1$ occur exactly $5$ times, then for the remaining one toss, the probability of getting $2$ is equal to $frac15$, same for probability of getting $3$.
– Siong Thye Goh
Aug 11 at 2:32
Sure, given that a certain number occurred n times, the average number of times that all other numbers occurred is n. But that doesn't mean a different number can't occur a different number of times. The greater the distance between between numbers, the lower the probability.
– Information Aether
Aug 11 at 2:08
Sure, given that a certain number occurred n times, the average number of times that all other numbers occurred is n. But that doesn't mean a different number can't occur a different number of times. The greater the distance between between numbers, the lower the probability.
– Information Aether
Aug 11 at 2:08
yes, they can occcur a different number of time, just saying that their distribution is now independently uniform, at a particular toss, each number happens with probability $frac15$.
– Siong Thye Goh
Aug 11 at 2:10
yes, they can occcur a different number of time, just saying that their distribution is now independently uniform, at a particular toss, each number happens with probability $frac15$.
– Siong Thye Goh
Aug 11 at 2:10
That doesn't answer my question.
– Information Aether
Aug 11 at 2:28
That doesn't answer my question.
– Information Aether
Aug 11 at 2:28
hmmm... can you clarify what is your question?
– Siong Thye Goh
Aug 11 at 2:30
hmmm... can you clarify what is your question?
– Siong Thye Goh
Aug 11 at 2:30
"Given that a certain number offcured $n$ times, the average number of times that all numbers occured is $n$" doesn't make sense, it depends on how many tosses remain. For example if out of $6$ tosses, $1$ occur exactly $5$ times, then for the remaining one toss, the probability of getting $2$ is equal to $frac15$, same for probability of getting $3$.
– Siong Thye Goh
Aug 11 at 2:32
"Given that a certain number offcured $n$ times, the average number of times that all numbers occured is $n$" doesn't make sense, it depends on how many tosses remain. For example if out of $6$ tosses, $1$ occur exactly $5$ times, then for the remaining one toss, the probability of getting $2$ is equal to $frac15$, same for probability of getting $3$.
– Siong Thye Goh
Aug 11 at 2:32
 |Â
show 2 more comments
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This is equivalent to ignoring any rolls for which a 1 comes up, which effectively makes your die 5-sided.
– amd
Aug 11 at 2:43
The core of the question is that the number of rolls is unknown, but given that six occurred exactly 10 times, we can guess that the number of rolls is probably roughly 60. However, I need a more rigorous way of doing this to answer the question.
– Information Aether
Aug 11 at 3:36
That’s not what you’ve written in your question, though. You ask for the probability of rolling 6 exactly 10 times, not for the probability distribution of the number of rolls needed to get ten sixes, which is rather different.
– amd
Aug 11 at 3:46
@InformationAether Your edit does not really clarify your question. Make it more specific. What is does $n_1$ or $x$ mean in this context?
– callculus
Aug 11 at 3:59
1
what is given? $n_6=10$ or is it $n_1=0$?
– Siong Thye Goh
Aug 11 at 4:02