You are watching: Probability of rolling snake eyes

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

**never**hitting it.

Every time you roll the dice, you have actually a $35/36$ possibility of **not** hitting it. If you roll the dice $n$ times, then the only situation where you have actually never fight it, is once you have not hit that every solitary time.

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

Thus the probability of hitting it at the very 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 occupychristmas.orgematician

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answer Oct 3 "18 in ~ 13:19

b00n heTb00n heT

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

However, you likewise ask specifically about the situation $n=36$, i.e. If you have actually a $1$ in $k$ possibility of success, what is your opportunity of getting at least one success in $k$ trials? It turns out the the answer come this question is quite comparable for any kind of reasonably large value of $k$.

It is $1-\big(1-\frac1k\big)^k$, and also $\big(1-\frac1k\big)^k$ converges come $e^-1$. Therefore the probability will be about $1-e^-1\approx 63.2\%$, and also this approximation will certainly get much better the bigger $k$ is. (For $k=36$ the genuine answer is $63.7\%$.)

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reply Oct 3 "18 in ~ 13:40

specifically LimeEspecially Lime

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

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reply Oct 3 "18 in ~ 13:22

Sam StreeterSam Streeter

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