for this reason I recognize that if you role a traditional pair that dice, your possibilities of acquiring Snake eye (double 1s) is $1$ in $36$. What I"m not certain of is how to carry out the londonchinatown.orgematics to figure out your possibilities of rolling snake Eyes at least once throughout a collection of rolls. I know if I role the dice $36$ time it won"t result in a $100\%$ possibility of rolling line Eyes, and also while i imagine it"s in the upper nineties, I"d choose to figure out exactly how i can not qualify it is.

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

Every time you role the dice, you have actually a $35/36$ chance of not hitting it. If you role the dice $n$ times, climate the only situation where you have never struggle it, is as soon as you have not hit it every solitary time.

The probabilty of not hitting through $2$ rolls is hence $35/36 imes 35/36$, the probabilty of no hitting through $3$ rolls is $35/36 imes 35/36 imes 35/36=(35/36)^3$ and also so top top till $(35/36)^n$.

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

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

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edited Oct 4 "18 at 13:00

J. M. Ain't a londonchinatown.orgematician
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answered Oct 3 "18 at 13:19

b00n heTb00n heT
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The various other answers explain the basic formula for the probability of never ever rolling snake eyes in a collection of $n$ rolls.

However, you additionally ask specifically around the situation $n=36$, i.e. If you have actually a $1$ in $k$ possibility of success, what is your opportunity of acquiring at the very least one success in $k$ trials? It transforms out that the answer to this concern is quite similar for any kind of reasonably big value of $k$.

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 this approximation will get much better the bigger $k$ is. (For $k=36$ the genuine answer is $63.7\%$.)

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answer Oct 3 "18 at 13:40

specifically LimeEspecially Lime
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If you roll $n$ times, then the probability of rolling line eyes at the very least once is $1-left(frac3536 ight)^n$, together you either role snake eyes at the very least once or no at every (so the probability of these two events should sum to $1$), and also the probability of never ever rolling snake eyes is the very same as requiring the you roll among the various other $35$ feasible outcomes on every roll.

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reply Oct 3 "18 in ~ 13:22
Sam StreeterSam Streeter
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