You are watching: Probability of rolling snake eyes

The probability that hitting that at least once is $1$ minus the probabilty the

**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, then the only case where you have actually never struggle it, is when you have actually not hit that every solitary time.

The probabilty of not hitting through $2$ rolls is therefore $35/36\times 35/36$, the probabilty of not hitting v $3$ roll is $35/36\times 35/36\times 35/36=(35/36)^3$ and also so top top till $(35/36)^n$.

Thus the probability of hitting it at least once is $1-(35/36)^n$ whereby $n$ is the variety 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

J. M. Ain't a ugandan-news.comematician

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answered Oct 3 "18 in ~ 13:19

b00n heTb00n heT

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The other answers describe the general formula for the probability of never rolling line eyes in a collection of $n$ rolls.

However, you also ask specifically about the situation $n=36$, i.e. If you have a $1$ in $k$ possibility of success, what is your opportunity of obtaining at the very least one success in $k$ trials? It transforms out that the answer come this concern is quite comparable for any reasonably huge value that $k$.

It is $1-\big(1-\frac1k\big)^k$, and also $\big(1-\frac1k\big)^k$ converges come $e^-1$. For this reason the probability will certainly be about $1-e^-1\approx 63.2\%$, and 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

specifically LimeEspecially Lime

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If you roll $n$ times, then the probability of rolling snake eyes at least once is $1-\left(\frac3536\right)^n$, as you either roll snake eye at the very least once or no at all (so the probability of these two occasions should amount to $1$), and the probability of never ever rolling line eyes is the exact same as requiring the you roll one of the other $35$ possible outcomes on each roll.

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reply Oct 3 "18 at 13:22

Sam StreeterSam Streeter

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