Probability of rolling $T$ or higher on at least $M$ of $R$ dice of $S$ sides
$begingroup$
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
edited Dec 8 '18 at 3:05
David K
54.1k342116
asked Dec 8 '18 at 0:48
Graham MartinGraham Martin
32
$endgroup$
add a comment |
$begingroup$
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
edited Dec 8 '18 at 3:05
David K
54.1k342116
asked Dec 8 '18 at 0:48
Graham MartinGraham Martin
32
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 8 '18 at 3:05
David K
54.1k342116
asked Dec 8 '18 at 0:48
Graham MartinGraham Martin
32
$endgroup$
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
edited Dec 8 '18 at 3:05
David K
54.1k342116
asked Dec 8 '18 at 0:48
Graham MartinGraham Martin
32
edited Dec 8 '18 at 3:05
David K
54.1k342116
asked Dec 8 '18 at 0:48
Graham MartinGraham Martin
32
edited Dec 8 '18 at 3:05
David K
54.1k342116
edited Dec 8 '18 at 3:05
David K
54.1k342116
edited Dec 8 '18 at 3:05
David K
54.1k342116
54.1k342116
asked Dec 8 '18 at 0:48
Graham MartinGraham Martin
32
asked Dec 8 '18 at 0:48
Graham MartinGraham Martin
32
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
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,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$
answered Dec 8 '18 at 2:52
saulspatzsaulspatz
15.1k31331
$endgroup$
$begingroup$
Awesome! I will try this out. Thank you so much!
$endgroup$
– Graham Martin
Dec 8 '18 at 3:07
add a comment |
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$begingroup$
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,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$
answered Dec 8 '18 at 2:52
saulspatzsaulspatz
15.1k31331
$endgroup$
$begingroup$
Awesome! I will try this out. Thank you so much!
$endgroup$
– Graham Martin
Dec 8 '18 at 3:07
add a comment |
$begingroup$
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,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$
answered Dec 8 '18 at 2:52
saulspatzsaulspatz
15.1k31331
$endgroup$
$begingroup$
Awesome! I will try this out. Thank you so much!
$endgroup$
– Graham Martin
Dec 8 '18 at 3:07
add a comment |
$begingroup$
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,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$
answered Dec 8 '18 at 2:52
saulspatzsaulspatz
15.1k31331
$endgroup$
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,
$$P=1-sum_{k=0}^{M-1}{Rchoose k}left(1-{T-1over S}right)^kleft(T-1over Sright)^{R-k}$$
answered Dec 8 '18 at 2:52
saulspatzsaulspatz
15.1k31331
edited Dec 8 '18 at 3:15
answered Dec 8 '18 at 2:52
saulspatzsaulspatz
15.1k31331
answered Dec 8 '18 at 2:52
saulspatzsaulspatz
15.1k31331
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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