10
$\begingroup$

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.

$\endgroup$

3 Answers 3

42
$\begingroup$

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 $99\%$

$\endgroup$
1
  • 1
    $\begingroup$ Editing in the actual value for 36 rolls(as specified in the question) would make this answer slightly better $\endgroup$
    – Quintec
    Oct 5, 2018 at 1:19
32
$\begingroup$

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 $63.7\%$.)

$\endgroup$
8
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.