Probability of rolling $T$ or higher on at least $M$ of $R$ dice of $S$ sides












0












$begingroup$


I am trying to figure out how to write an equation to solve the following problem:




  • S = # of sides on a die

  • R = # of dice being rolled

  • T = Minimum result to count as a success event on each die (Aka "Threshold")

  • M = Minimum # of success events desired


For example, if I want to know the probability (P) of getting at least two "5's" or higher when rolling three 6-sided dice, then:
begin{align}
S &= 6\
R &= 3\
T &= 5\
M &= 2
end{align}



Is there a formula I could use to determine P?



Thanks!






probability dice







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by “minimum result?” Does it refer to the value rolled for each individual die or the total of those values? If the former, look up “binomial distribution.” There are formulas for its CDF, which is what you’re looking for here.
    $endgroup$
    – amd
    Dec 8 '18 at 1:04










  • $begingroup$
    Minimum result is the lowest value to be counted as a success on any die. Also called the "threshold." (T) In the example above, I want to know the odds of rolling at least two "5s" or higher. Thus, T = 5. In such a case, a 5 or a 6 on a die counts as a success. If T = 5 then the chances of one success on one die are 1/3 since both a 5 and a 6 count as a success on each die.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:09






  • 1




    $begingroup$
    In that case, look up the binomial distribution, as I said. You’re asking about its cumulative distribution function, for which there are standard formulas.
    $endgroup$
    – amd
    Dec 8 '18 at 1:11










  • $begingroup$
    I edited OP to make that part clearer. Basically, I am trying to develop a formula or chart that lets me answer questions like "What are the odds if you roll three six sided dice that you will get at least to dice with a 5 or higher?" But I am trying to abstract it so I can plug in the 4 variables of # of sides on the dice, # of dice being rolled, the minimum value to be counted as a success on each die, and the minimum # of "successful" dice desired.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:12










  • $begingroup$
    Thank you! I'll check it out.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:13
















0












$begingroup$


I am trying to figure out how to write an equation to solve the following problem:




  • S = # of sides on a die

  • R = # of dice being rolled

  • T = Minimum result to count as a success event on each die (Aka "Threshold")

  • M = Minimum # of success events desired


For example, if I want to know the probability (P) of getting at least two "5's" or higher when rolling three 6-sided dice, then:
begin{align}
S &= 6\
R &= 3\
T &= 5\
M &= 2
end{align}



Is there a formula I could use to determine P?



Thanks!






probability dice







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by “minimum result?” Does it refer to the value rolled for each individual die or the total of those values? If the former, look up “binomial distribution.” There are formulas for its CDF, which is what you’re looking for here.
    $endgroup$
    – amd
    Dec 8 '18 at 1:04










  • $begingroup$
    Minimum result is the lowest value to be counted as a success on any die. Also called the "threshold." (T) In the example above, I want to know the odds of rolling at least two "5s" or higher. Thus, T = 5. In such a case, a 5 or a 6 on a die counts as a success. If T = 5 then the chances of one success on one die are 1/3 since both a 5 and a 6 count as a success on each die.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:09






  • 1




    $begingroup$
    In that case, look up the binomial distribution, as I said. You’re asking about its cumulative distribution function, for which there are standard formulas.
    $endgroup$
    – amd
    Dec 8 '18 at 1:11










  • $begingroup$
    I edited OP to make that part clearer. Basically, I am trying to develop a formula or chart that lets me answer questions like "What are the odds if you roll three six sided dice that you will get at least to dice with a 5 or higher?" But I am trying to abstract it so I can plug in the 4 variables of # of sides on the dice, # of dice being rolled, the minimum value to be counted as a success on each die, and the minimum # of "successful" dice desired.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:12










  • $begingroup$
    Thank you! I'll check it out.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:13














0












0








0





$begingroup$


I am trying to figure out how to write an equation to solve the following problem:




  • S = # of sides on a die

  • R = # of dice being rolled

  • T = Minimum result to count as a success event on each die (Aka "Threshold")

  • M = Minimum # of success events desired


For example, if I want to know the probability (P) of getting at least two "5's" or higher when rolling three 6-sided dice, then:
begin{align}
S &= 6\
R &= 3\
T &= 5\
M &= 2
end{align}



Is there a formula I could use to determine P?



Thanks!






probability dice







share|cite|improve this question











$endgroup$




I am trying to figure out how to write an equation to solve the following problem:




  • S = # of sides on a die

  • R = # of dice being rolled

  • T = Minimum result to count as a success event on each die (Aka "Threshold")

  • M = Minimum # of success events desired


For example, if I want to know the probability (P) of getting at least two "5's" or higher when rolling three 6-sided dice, then:
begin{align}
S &= 6\
R &= 3\
T &= 5\
M &= 2
end{align}



Is there a formula I could use to determine P?



Thanks!






probability dice




probability dice






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 3:05









David K

54.1k342116




54.1k342116










asked Dec 8 '18 at 0:48









Graham MartinGraham Martin

32




32












  • $begingroup$
    What do you mean by “minimum result?” Does it refer to the value rolled for each individual die or the total of those values? If the former, look up “binomial distribution.” There are formulas for its CDF, which is what you’re looking for here.
    $endgroup$
    – amd
    Dec 8 '18 at 1:04










  • $begingroup$
    Minimum result is the lowest value to be counted as a success on any die. Also called the "threshold." (T) In the example above, I want to know the odds of rolling at least two "5s" or higher. Thus, T = 5. In such a case, a 5 or a 6 on a die counts as a success. If T = 5 then the chances of one success on one die are 1/3 since both a 5 and a 6 count as a success on each die.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:09






  • 1




    $begingroup$
    In that case, look up the binomial distribution, as I said. You’re asking about its cumulative distribution function, for which there are standard formulas.
    $endgroup$
    – amd
    Dec 8 '18 at 1:11










  • $begingroup$
    I edited OP to make that part clearer. Basically, I am trying to develop a formula or chart that lets me answer questions like "What are the odds if you roll three six sided dice that you will get at least to dice with a 5 or higher?" But I am trying to abstract it so I can plug in the 4 variables of # of sides on the dice, # of dice being rolled, the minimum value to be counted as a success on each die, and the minimum # of "successful" dice desired.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:12










  • $begingroup$
    Thank you! I'll check it out.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:13


















  • $begingroup$
    What do you mean by “minimum result?” Does it refer to the value rolled for each individual die or the total of those values? If the former, look up “binomial distribution.” There are formulas for its CDF, which is what you’re looking for here.
    $endgroup$
    – amd
    Dec 8 '18 at 1:04










  • $begingroup$
    Minimum result is the lowest value to be counted as a success on any die. Also called the "threshold." (T) In the example above, I want to know the odds of rolling at least two "5s" or higher. Thus, T = 5. In such a case, a 5 or a 6 on a die counts as a success. If T = 5 then the chances of one success on one die are 1/3 since both a 5 and a 6 count as a success on each die.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:09






  • 1




    $begingroup$
    In that case, look up the binomial distribution, as I said. You’re asking about its cumulative distribution function, for which there are standard formulas.
    $endgroup$
    – amd
    Dec 8 '18 at 1:11










  • $begingroup$
    I edited OP to make that part clearer. Basically, I am trying to develop a formula or chart that lets me answer questions like "What are the odds if you roll three six sided dice that you will get at least to dice with a 5 or higher?" But I am trying to abstract it so I can plug in the 4 variables of # of sides on the dice, # of dice being rolled, the minimum value to be counted as a success on each die, and the minimum # of "successful" dice desired.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:12










  • $begingroup$
    Thank you! I'll check it out.
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 1:13
















$begingroup$
What do you mean by “minimum result?” Does it refer to the value rolled for each individual die or the total of those values? If the former, look up “binomial distribution.” There are formulas for its CDF, which is what you’re looking for here.
$endgroup$
– amd
Dec 8 '18 at 1:04




$begingroup$
What do you mean by “minimum result?” Does it refer to the value rolled for each individual die or the total of those values? If the former, look up “binomial distribution.” There are formulas for its CDF, which is what you’re looking for here.
$endgroup$
– amd
Dec 8 '18 at 1:04












$begingroup$
Minimum result is the lowest value to be counted as a success on any die. Also called the "threshold." (T) In the example above, I want to know the odds of rolling at least two "5s" or higher. Thus, T = 5. In such a case, a 5 or a 6 on a die counts as a success. If T = 5 then the chances of one success on one die are 1/3 since both a 5 and a 6 count as a success on each die.
$endgroup$
– Graham Martin
Dec 8 '18 at 1:09




$begingroup$
Minimum result is the lowest value to be counted as a success on any die. Also called the "threshold." (T) In the example above, I want to know the odds of rolling at least two "5s" or higher. Thus, T = 5. In such a case, a 5 or a 6 on a die counts as a success. If T = 5 then the chances of one success on one die are 1/3 since both a 5 and a 6 count as a success on each die.
$endgroup$
– Graham Martin
Dec 8 '18 at 1:09




1




1




$begingroup$
In that case, look up the binomial distribution, as I said. You’re asking about its cumulative distribution function, for which there are standard formulas.
$endgroup$
– amd
Dec 8 '18 at 1:11




$begingroup$
In that case, look up the binomial distribution, as I said. You’re asking about its cumulative distribution function, for which there are standard formulas.
$endgroup$
– amd
Dec 8 '18 at 1:11












$begingroup$
I edited OP to make that part clearer. Basically, I am trying to develop a formula or chart that lets me answer questions like "What are the odds if you roll three six sided dice that you will get at least to dice with a 5 or higher?" But I am trying to abstract it so I can plug in the 4 variables of # of sides on the dice, # of dice being rolled, the minimum value to be counted as a success on each die, and the minimum # of "successful" dice desired.
$endgroup$
– Graham Martin
Dec 8 '18 at 1:12




$begingroup$
I edited OP to make that part clearer. Basically, I am trying to develop a formula or chart that lets me answer questions like "What are the odds if you roll three six sided dice that you will get at least to dice with a 5 or higher?" But I am trying to abstract it so I can plug in the 4 variables of # of sides on the dice, # of dice being rolled, the minimum value to be counted as a success on each die, and the minimum # of "successful" dice desired.
$endgroup$
– Graham Martin
Dec 8 '18 at 1:12












$begingroup$
Thank you! I'll check it out.
$endgroup$
– Graham Martin
Dec 8 '18 at 1:13




$begingroup$
Thank you! I'll check it out.
$endgroup$
– Graham Martin
Dec 8 '18 at 1:13










1 Answer
1






active

oldest

votes


















2












$begingroup$

Let $A$ be the number of acceptable numbers, that is numbers at or above the threshold. Then $A=S-T+1$. In your example, $A=2$, since $5$ and $6$ are the only acceptable numbers. The probability that a particular die has an acceptable number is $A/S = 1 -(T-1)/S$ and the probability that it is not acceptable is, of course, $(T-1)/S.$



For a given value $k,$ the probability that exactly $k$ rolls out of $R$ are acceptable is $${Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$ since there are ${Rchoose k}$ ways to choose the $k$ acceptable dice. For a success, we must have $kge M$ so the formula you seek is$$P=sum_{k=M}^R{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$



P.S.



If $M$ is less than half of $R,$ it will be more convenient to compute the probability of success as one minus the probability of failure, since there will be fewer term is the sum. That is,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Awesome! I will try this out. Thank you so much!
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 3:07











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1 Answer
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active

oldest

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active

oldest

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2












$begingroup$

Let $A$ be the number of acceptable numbers, that is numbers at or above the threshold. Then $A=S-T+1$. In your example, $A=2$, since $5$ and $6$ are the only acceptable numbers. The probability that a particular die has an acceptable number is $A/S = 1 -(T-1)/S$ and the probability that it is not acceptable is, of course, $(T-1)/S.$



For a given value $k,$ the probability that exactly $k$ rolls out of $R$ are acceptable is $${Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$ since there are ${Rchoose k}$ ways to choose the $k$ acceptable dice. For a success, we must have $kge M$ so the formula you seek is$$P=sum_{k=M}^R{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$



P.S.



If $M$ is less than half of $R,$ it will be more convenient to compute the probability of success as one minus the probability of failure, since there will be fewer term is the sum. That is,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Awesome! I will try this out. Thank you so much!
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 3:07
















2












$begingroup$

Let $A$ be the number of acceptable numbers, that is numbers at or above the threshold. Then $A=S-T+1$. In your example, $A=2$, since $5$ and $6$ are the only acceptable numbers. The probability that a particular die has an acceptable number is $A/S = 1 -(T-1)/S$ and the probability that it is not acceptable is, of course, $(T-1)/S.$



For a given value $k,$ the probability that exactly $k$ rolls out of $R$ are acceptable is $${Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$ since there are ${Rchoose k}$ ways to choose the $k$ acceptable dice. For a success, we must have $kge M$ so the formula you seek is$$P=sum_{k=M}^R{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$



P.S.



If $M$ is less than half of $R,$ it will be more convenient to compute the probability of success as one minus the probability of failure, since there will be fewer term is the sum. That is,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Awesome! I will try this out. Thank you so much!
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 3:07














2












2








2





$begingroup$

Let $A$ be the number of acceptable numbers, that is numbers at or above the threshold. Then $A=S-T+1$. In your example, $A=2$, since $5$ and $6$ are the only acceptable numbers. The probability that a particular die has an acceptable number is $A/S = 1 -(T-1)/S$ and the probability that it is not acceptable is, of course, $(T-1)/S.$



For a given value $k,$ the probability that exactly $k$ rolls out of $R$ are acceptable is $${Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$ since there are ${Rchoose k}$ ways to choose the $k$ acceptable dice. For a success, we must have $kge M$ so the formula you seek is$$P=sum_{k=M}^R{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$



P.S.



If $M$ is less than half of $R,$ it will be more convenient to compute the probability of success as one minus the probability of failure, since there will be fewer term is the sum. That is,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$






share|cite|improve this answer











$endgroup$



Let $A$ be the number of acceptable numbers, that is numbers at or above the threshold. Then $A=S-T+1$. In your example, $A=2$, since $5$ and $6$ are the only acceptable numbers. The probability that a particular die has an acceptable number is $A/S = 1 -(T-1)/S$ and the probability that it is not acceptable is, of course, $(T-1)/S.$



For a given value $k,$ the probability that exactly $k$ rolls out of $R$ are acceptable is $${Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$ since there are ${Rchoose k}$ ways to choose the $k$ acceptable dice. For a success, we must have $kge M$ so the formula you seek is$$P=sum_{k=M}^R{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$



P.S.



If $M$ is less than half of $R,$ it will be more convenient to compute the probability of success as one minus the probability of failure, since there will be fewer term is the sum. That is,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 3:15

























answered Dec 8 '18 at 2:52









saulspatzsaulspatz

15.1k31331




15.1k31331












  • $begingroup$
    Awesome! I will try this out. Thank you so much!
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 3:07


















  • $begingroup$
    Awesome! I will try this out. Thank you so much!
    $endgroup$
    – Graham Martin
    Dec 8 '18 at 3:07
















$begingroup$
Awesome! I will try this out. Thank you so much!
$endgroup$
– Graham Martin
Dec 8 '18 at 3:07




$begingroup$
Awesome! I will try this out. Thank you so much!
$endgroup$
– Graham Martin
Dec 8 '18 at 3:07


















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