If my friend throws two dice, and covers them up, but I see that one of them was a 6, what's the probability that they were both 6s given this knowledge?
I'm under the impression that the answer is 2/7, because the other die could be any of the other numbers, but if he really did roll double sixes you could have seen either one, so there are two ways for that to happen. That makes seven equally likely possibilities: (6*,1) (6*,2) (6*,3) (6*,4) (6*,5) (6*,6) and (6,6*), where * represents the one you saw.
My question is whether the answer should really be 2/12 = 1/6 since you might think you ought to count the cases (1,6*) (2,6*) etc. as separate—that is, the case in which the other die comes up as a 6 and you see it. You could distinguish the dice by painting one red, for example.
I hope the question is well posed. Let me know if you think it should be clarified.
EDIT: Thanks for the speedy responses everyone. One way I thought about the question is that instead of the 36 outcomes we typically think of for two dice, there are now 72 possible outcomes—for each roll there are two events corresponding to seeing die A or die B. In this case when we condition on the fact that we saw one of the dice to be a 6 we've restricted our sample space in the way I've described above.
For clarity, this means we now have the following possibilities:
(6*,6) (6,6*) (6*,5) (6*,4) (6*,3) (6*,2) (6*,1)
I'm not sure whether to include the remaining possibilities or not:
(1,6*) (2,6*) (3,6*) (4,6*) (5,6*)
Clearly the answer depends highly on the interpretation of the wording of the question. I'm interpreting it to mean you're equally likely to spot one die or the other. I'm fairly sure this situation is different than being given the information that at least one of the dice is a six.
Can anyone convince me why this isn't a legitimate way to interpret the question, or otherwise she'd some light on which restricted sample space is the correct one? I feel like it has something to do with this indistinguishable to of the two sixes (so maybe painting one red would ruin it).
Best Answer
The answer depends on the details of just how you obtained your information. If somebody who saw both dice gives you as sole information that at least one of them is$~6$, then that limits the possible outcomes to $11$ (out of originally $36$) possibilities, just one of which is double sixes. The probability you asked for would then be$~\frac1{11}$ as in the answer by pre-kidney.
However, as you formulated the question you saw one of the dice yourself. It is virtually impossible to look at both dice and to just obtain an image telling you that one of them is$~6$. (If the question were about slides which could be either transparent or black, then superposing them and seeing that the pair is not transparent would give this kind of observation, telling you at least one slide is black; however no such devious way of observing dice seems possible.) So you may mentally label the die you saw as $A$ and the one you didn't see as$~B$. Your observation told you nothing about $B$, so it has $\frac16$ probability of being a $6$ (too), presuming it is fair. So given the way you stated the problem, I feel that $\frac16$ is the correct answer to the question.