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Chances of rolling “Snake Eyes†at least once in a series of rolls.
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So I know that if you roll a standard pair of dice, your chances of getting Snake Eyes (double 1s) is $1$ in $36$. What I'm not sure of is how to do the math to figure out your chances of rolling Snake Eyes at least once during a series of rolls. I know if I roll the dice $36$ times it won't lead to a $100%$ chance of rolling Snake Eyes, and while I imagine it's in the upper nineties, I'd like to figure out exactly how unlikely it is.
probability dice
edited 28 mins ago
Especially Lime
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So I know that if you roll a standard pair of dice, your chances of getting Snake Eyes (double 1s) is $1$ in $36$. What I'm not sure of is how to do the math to figure out your chances of rolling Snake Eyes at least once during a series of rolls. I know if I roll the dice $36$ times it won't lead to a $100%$ chance of rolling Snake Eyes, and while I imagine it's in the upper nineties, I'd like to figure out exactly how unlikely it is.
probability dice
edited 28 mins ago
Especially Lime
19.8k22353
asked 54 mins ago
aslum
1637
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So I know that if you roll a standard pair of dice, your chances of getting Snake Eyes (double 1s) is $1$ in $36$. What I'm not sure of is how to do the math to figure out your chances of rolling Snake Eyes at least once during a series of rolls. I know if I roll the dice $36$ times it won't lead to a $100%$ chance of rolling Snake Eyes, and while I imagine it's in the upper nineties, I'd like to figure out exactly how unlikely it is.
probability dice
edited 28 mins ago
Especially Lime
19.8k22353
asked 54 mins ago
aslum
1637
So I know that if you roll a standard pair of dice, your chances of getting Snake Eyes (double 1s) is $1$ in $36$. What I'm not sure of is how to do the math to figure out your chances of rolling Snake Eyes at least once during a series of rolls. I know if I roll the dice $36$ times it won't lead to a $100%$ chance of rolling Snake Eyes, and while I imagine it's in the upper nineties, I'd like to figure out exactly how unlikely it is.
probability dice
probability dice
edited 28 mins ago
Especially Lime
19.8k22353
asked 54 mins ago
aslum
1637
edited 28 mins ago
Especially Lime
19.8k22353
asked 54 mins ago
aslum
1637
edited 28 mins ago
Especially Lime
19.8k22353
edited 28 mins ago
Especially Lime
19.8k22353
edited 28 mins ago
Especially Lime
19.8k22353
19.8k22353
asked 54 mins ago
aslum
1637
asked 54 mins ago
aslum
1637
asked 54 mins ago
aslum
1637
1637
add a comment |Â
add a comment |Â
3 Answers
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If you roll $n$ times, then the probability of rolling snake eyes at least once is $1-left(frac3536right)^n$, as you either roll snake eyes at least once or not at all (so the probability of these two events should sum to $1$), and the probability of never rolling snake eyes is the same as requiring that you roll one of the other $35$ possible outcomes on each roll.
answered 46 mins ago
Sam Streeter
313
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up vote
2
down vote
The probability of hitting it at least once is $1$ minus the probabilty of never hitting it.
Every time you roll the dices you have a $35/36$ chance of not hitting it. If you roll the dices $n$ times, then the only case where you have never hit it, is when you have not hit it every single time.
The probabilty of not hitting with $2$ rolls is thus $35/36times 35/36$, the probabilty of not hitting with $3$ rolls is $35/36times 35/36times 35/36=(35/36)^3$ and so on till $(35/36)^n$.
Thus the probability of hitting it at least once is $1-(35/36)^n$ where $n$ is the number of throws.
After $164$ throws, the probability of hitting it at least once is $99%$
answered 49 mins ago
b00n heT
9,11011833
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up vote
2
down vote
The other answers explain the general formula for the probability of never rolling snake eyes in a series of $n$ rolls.
However, you also ask specifically about the case $n=36$, i.e. if you have a $1$ in $k$ chance of success, what is your chance of getting at least one success in $k$ trials? It turns out that the answer to this question is quite similar for any reasonably large value of $k$.
It is $1-big(1-frac1kbig)^k$, and $big(1-frac1kbig)^k$ converges to $e^-1$. So the probability will be about $1-e^-1approx 63.2%$, and this approximation will get better the larger $k$ is. (For $k=36$ the real answer is $63.7%$.)
answered 29 mins ago
Especially Lime
19.8k22353
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If you roll $n$ times, then the probability of rolling snake eyes at least once is $1-left(frac3536right)^n$, as you either roll snake eyes at least once or not at all (so the probability of these two events should sum to $1$), and the probability of never rolling snake eyes is the same as requiring that you roll one of the other $35$ possible outcomes on each roll.
answered 46 mins ago
Sam Streeter
313
New contributor
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
If you roll $n$ times, then the probability of rolling snake eyes at least once is $1-left(frac3536right)^n$, as you either roll snake eyes at least once or not at all (so the probability of these two events should sum to $1$), and the probability of never rolling snake eyes is the same as requiring that you roll one of the other $35$ possible outcomes on each roll.
answered 46 mins ago
Sam Streeter
313
New contributor
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you roll $n$ times, then the probability of rolling snake eyes at least once is $1-left(frac3536right)^n$, as you either roll snake eyes at least once or not at all (so the probability of these two events should sum to $1$), and the probability of never rolling snake eyes is the same as requiring that you roll one of the other $35$ possible outcomes on each roll.
answered 46 mins ago
Sam Streeter
313
New contributor
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you roll $n$ times, then the probability of rolling snake eyes at least once is $1-left(frac3536right)^n$, as you either roll snake eyes at least once or not at all (so the probability of these two events should sum to $1$), and the probability of never rolling snake eyes is the same as requiring that you roll one of the other $35$ possible outcomes on each roll.
answered 46 mins ago
Sam Streeter
313
New contributor
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 46 mins ago
Sam Streeter
313
New contributor
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 46 mins ago
Sam Streeter
313
answered 46 mins ago
Sam Streeter
313
313
New contributor
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sam Streeter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
2
down vote
The probability of hitting it at least once is $1$ minus the probabilty of never hitting it.
Every time you roll the dices you have a $35/36$ chance of not hitting it. If you roll the dices $n$ times, then the only case where you have never hit it, is when you have not hit it every single time.
The probabilty of not hitting with $2$ rolls is thus $35/36times 35/36$, the probabilty of not hitting with $3$ rolls is $35/36times 35/36times 35/36=(35/36)^3$ and so on till $(35/36)^n$.
Thus the probability of hitting it at least once is $1-(35/36)^n$ where $n$ is the number of throws.
After $164$ throws, the probability of hitting it at least once is $99%$
answered 49 mins ago
b00n heT
9,11011833
add a comment |Â
up vote
2
down vote
The probability of hitting it at least once is $1$ minus the probabilty of never hitting it.
Every time you roll the dices you have a $35/36$ chance of not hitting it. If you roll the dices $n$ times, then the only case where you have never hit it, is when you have not hit it every single time.
The probabilty of not hitting with $2$ rolls is thus $35/36times 35/36$, the probabilty of not hitting with $3$ rolls is $35/36times 35/36times 35/36=(35/36)^3$ and so on till $(35/36)^n$.
Thus the probability of hitting it at least once is $1-(35/36)^n$ where $n$ is the number of throws.
After $164$ throws, the probability of hitting it at least once is $99%$
answered 49 mins ago
b00n heT
9,11011833
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The probability of hitting it at least once is $1$ minus the probabilty of never hitting it.
Every time you roll the dices you have a $35/36$ chance of not hitting it. If you roll the dices $n$ times, then the only case where you have never hit it, is when you have not hit it every single time.
The probabilty of not hitting with $2$ rolls is thus $35/36times 35/36$, the probabilty of not hitting with $3$ rolls is $35/36times 35/36times 35/36=(35/36)^3$ and so on till $(35/36)^n$.
Thus the probability of hitting it at least once is $1-(35/36)^n$ where $n$ is the number of throws.
After $164$ throws, the probability of hitting it at least once is $99%$
answered 49 mins ago
b00n heT
9,11011833
The probability of hitting it at least once is $1$ minus the probabilty of never hitting it.
Every time you roll the dices you have a $35/36$ chance of not hitting it. If you roll the dices $n$ times, then the only case where you have never hit it, is when you have not hit it every single time.
The probabilty of not hitting with $2$ rolls is thus $35/36times 35/36$, the probabilty of not hitting with $3$ rolls is $35/36times 35/36times 35/36=(35/36)^3$ and so on till $(35/36)^n$.
Thus the probability of hitting it at least once is $1-(35/36)^n$ where $n$ is the number of throws.
After $164$ throws, the probability of hitting it at least once is $99%$
answered 49 mins ago
b00n heT
9,11011833
edited 41 mins ago
answered 49 mins ago
b00n heT
9,11011833
answered 49 mins ago
b00n heT
9,11011833
answered 49 mins ago
b00n heT
9,11011833
9,11011833
add a comment |Â
add a comment |Â
up vote
2
down vote
The other answers explain the general formula for the probability of never rolling snake eyes in a series of $n$ rolls.
However, you also ask specifically about the case $n=36$, i.e. if you have a $1$ in $k$ chance of success, what is your chance of getting at least one success in $k$ trials? It turns out that the answer to this question is quite similar for any reasonably large value of $k$.
It is $1-big(1-frac1kbig)^k$, and $big(1-frac1kbig)^k$ converges to $e^-1$. So the probability will be about $1-e^-1approx 63.2%$, and this approximation will get better the larger $k$ is. (For $k=36$ the real answer is $63.7%$.)
answered 29 mins ago
Especially Lime
19.8k22353
add a comment |Â
up vote
2
down vote
The other answers explain the general formula for the probability of never rolling snake eyes in a series of $n$ rolls.
However, you also ask specifically about the case $n=36$, i.e. if you have a $1$ in $k$ chance of success, what is your chance of getting at least one success in $k$ trials? It turns out that the answer to this question is quite similar for any reasonably large value of $k$.
It is $1-big(1-frac1kbig)^k$, and $big(1-frac1kbig)^k$ converges to $e^-1$. So the probability will be about $1-e^-1approx 63.2%$, and this approximation will get better the larger $k$ is. (For $k=36$ the real answer is $63.7%$.)
answered 29 mins ago
Especially Lime
19.8k22353
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The other answers explain the general formula for the probability of never rolling snake eyes in a series of $n$ rolls.
However, you also ask specifically about the case $n=36$, i.e. if you have a $1$ in $k$ chance of success, what is your chance of getting at least one success in $k$ trials? It turns out that the answer to this question is quite similar for any reasonably large value of $k$.
It is $1-big(1-frac1kbig)^k$, and $big(1-frac1kbig)^k$ converges to $e^-1$. So the probability will be about $1-e^-1approx 63.2%$, and this approximation will get better the larger $k$ is. (For $k=36$ the real answer is $63.7%$.)
answered 29 mins ago
Especially Lime
19.8k22353
The other answers explain the general formula for the probability of never rolling snake eyes in a series of $n$ rolls.
However, you also ask specifically about the case $n=36$, i.e. if you have a $1$ in $k$ chance of success, what is your chance of getting at least one success in $k$ trials? It turns out that the answer to this question is quite similar for any reasonably large value of $k$.
It is $1-big(1-frac1kbig)^k$, and $big(1-frac1kbig)^k$ converges to $e^-1$. So the probability will be about $1-e^-1approx 63.2%$, and this approximation will get better the larger $k$ is. (For $k=36$ the real answer is $63.7%$.)
answered 29 mins ago
Especially Lime
19.8k22353
answered 29 mins ago
Especially Lime
19.8k22353
answered 29 mins ago
Especially Lime
19.8k22353
answered 29 mins ago
Especially Lime
19.8k22353
19.8k22353
add a comment |Â
add a comment |Â
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