Two dice (with numbers 1 to 6 on the faces) are rolled.
One die rolls a 6.
What is the probability of rolling a double 6?
One solution is to say that P(2 sixes) = $\frac{1}{6}$ since the first die gives a 6, so the only way to get a double six is by rolling a six on the other die (which has a 1 in 6 chance).
Another solution is to say that there are 11 possible combinations if one die rolls a six i.e. (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2) and (6, 1). So the probability of rolling a double six if one six has already been rolled is $\frac{1}{11}$.
Which answer is correct and why?
Best Answer
There is confusion between two questions.
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 $1/6.$
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.