# 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.

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 $$99\%$$

• Editing in the actual value for 36 rolls(as specified in the question) would make this answer slightly better Oct 5, 2018 at 1:19

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 $$63.7\%$$.)

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.