Probability of rolling six 1's, twice in 3 rolls [closed]

Question

I was playing the dice game "10,000" and rolled six 1's twice, in 3 rolls.

What is the probability of doing that? I found that the probability of doing it once is ~1/7700, but doing it twice in 3 rolls?

Answer 1

Let us consider one trial: roll six dice. In order to roll 6 ones, you need to roll $$1\>1\>1\>1\,1\>1$$

Using independence (the outcome of one dice roll does not affect the chances in the next roll), the chance of this is $$\frac{1}{6}\cdot\frac16\cdot\frac16\cdot\frac16\cdot\frac16\cdot\frac16 = \left(\frac{1}{6}\right)^6$$

Call this $p$. Now, this was just for one trial. There are three trials in total. I will call getting all ones a success (S), otherwise it was a failure (F). Notice that we can count how many ways we can succeed twice: $$\text{SSF, SFS, FSS}$$

So, in order to succeed twice, we must fail once. There are $3$ different ways this can happen. And finally, $p+q=1$ since the events "get all ones" and "not get all ones" in one trial are mutually exclusive. Since the outcomes don't affect the chances of in the other trials, we have that the chance that you get all ones twice in three trials is $$3p^2(1-p) = 3\left(\left(\frac16\right)^6\right)^2\left(1-\left(\frac16\right)^6\right)\approx 1.378\times 10^{-9}$$

Answer 2

To find the probability that you get $\textbf{at least}$ two rolls that are all 6's, you can take the probability of getting all 6's exactly twice and add the probability of getting all 6's all three times.

This gives $3\cdot\big(\frac{1}{6}\big)^6\cdot\big(\frac{1}{6}\big)^6\cdot\left(1-\big(\frac{1}{6}\big)^6\right)+\big(\frac{1}{6}\big)^6\cdot\big(\frac{1}{6}\big)^6\cdot\big(\frac{1}{6}\big)^6=3\left(\frac{1}{6}\right)^{12}-2\left(\frac{1}{6}\right)^{18}\approx 1.38\times10^{-9}$