Probability of rolling a double $6$ with two dice

Solution 1:

There is confusion between two questions.

1. You have rolled a six. Now for a double-six, you need to get a 6 on the second roll. The second roll is independent of the first. So the probability of a 6 again on the second roll is $1/6.$

2. If both dice have already been rolled out of your sight, and you are told that there is at least one 6, then conditional on that information, what is the probability that the dice actually show a double-6. Then the analysis of @Shayne2020 (+1) leading to the answer $1/11$ is correct.

Solution 2:

This can be phrased as

Given that at least one die rolls a 6, what is the probability of rolling a double 6?

We can use the conditional probability formula: $$P(A|B) = \frac{P(A\ \mathrm{and}\ B)}{P(B)}$$ This means "the probability of event A given event B, is the probability of A and B divided by the probability of B".

\begin{align} P(\mathrm{double\ 6}|\mathrm{at\ least\ one\ 6}) &= \frac{P(\mathrm{double\ 6}\ \mathrm{and}\ \mathrm{at\ least\ one\ 6})}{P(\mathrm{at\ least\ one\ 6})}\\ &= \frac{P(\mathrm{double\ 6})}{P(\mathrm{at\ least\ one\ 6})}\\ &= \frac{1/36}{11/36}\\ &= \frac{1}{11} \end{align}

Solution 3:

One die rolls a 6.

This doesn't clarify which one. Was it the first die or the second one? (Note that the dice are distinguishable by the turn of their throws.)

So, the latter logic is correct. The conditional probability reduces the sample space $S$ into ${(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2) (6, 1)}$

Now, $|S|= 11$

And the only favorable outcome (rolling two 6's ) is $(6,6)$.

Thus, the probability is $\frac{1}{11}$