for this reason I recognize that if you role a typical pair the dice, your opportunities of acquiring Snake eyes (double 1s) is $1$ in $36$. What I"m not sure of is just how to carry out the tennis2007.orgematics to number out your opportunities of rolling snake Eyes at least once throughout a collection of rolls. I know if I role 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 favor to figure out exactly how i can not qualify it is.

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The probability that hitting it at least once is $1$ minus the probabilty the never hitting it.

Every time you roll the dice, you have a $35/36$ opportunity of not hitting it. If you role the dice $n$ times, then the only case where you have actually never hit it, is once you have actually not hit it every single time.

The probabilty of no hitting v $2$ rolls is for this reason $35/36 imes 35/36$, the probabilty of not hitting v $3$ roll is $35/36 imes 35/36 imes 35/36=(35/36)^3$ and so ~ above till $(35/36)^n$.

Thus the probability the hitting the at the very least once is $1-(35/36)^n$ where $n$ is the number of throws.

After $164$ throws, the probability of hitting the at least once is $99\%$


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edited Oct 4 "18 at 13:00
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J. M. Ain't a tennis2007.orgematician
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b00n heTb00n heT
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The various other answers describe the basic formula because that the probability of never ever rolling snake eyes in a collection of $n$ rolls.

However, you additionally ask specifically about the instance $n=36$, i.e. If you have a $1$ in $k$ chance of success, what is your chance of gaining at the very least one success in $k$ trials? It turns out the the answer come this inquiry is quite similar for any reasonably large value that $k$.

It is $1-ig(1-frac1kig)^k$, and also $ig(1-frac1kig)^k$ converges come $e^-1$. So the probability will certainly be about $1-e^-1approx 63.2\%$, and also this approximation will certainly get better the larger $k$ is. (For $k=36$ the genuine answer is $63.7\%$.)


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answered Oct 3 "18 at 13:40
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particularly LimeEspecially Lime
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If you role $n$ times, climate the probability the rolling snake eyes at least once is $1-left(frac3536 ight)^n$, together you either roll snake eyes at least once or not at every (so the probability of these two events should amount to $1$), and also the probability of never ever rolling line eyes is the same as requiring the you roll one of the other $35$ possible outcomes on every roll.


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answer Oct 3 "18 at 13:22
Sam StreeterSam Streeter
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