Chances of rolling "Snake Eyes" at least once in a series of rolls. 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.
 A: If you roll $n$ times, then the probability of rolling snake eyes at least once is $1-\left(\frac{35}{36}\right)^n$, 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.
A: The probability of hitting it at least once is $1$ minus the probabilty of never hitting it.
Every time you roll the dice, you have a $35/36$ chance of not hitting it. If you roll the dice $n$ 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 $35/36\times 35/36$, the probabilty of not hitting with $3$ rolls is $35/36\times 35/36\times 35/36=(35/36)^3$ 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 $99\%$
A: 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 $1-\big(1-\frac{1}{k}\big)^k$, and $\big(1-\frac{1}{k}\big)^k$ converges to $e^{-1}$. So the probability will be about $1-e^{-1}\approx 63.2\%$, and this approximation will get better the larger $k$ is. (For $k=36$ the real answer is $63.7\%$.)
