Two dice thrown, one comes up 6?

Question

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).

Answer 1

Assuming the dice are fair dice, so the one you see will not affect the one you don't see. so the chance to get a 6 on the other dice is 1/6. Hence, in general, when you throw two dice at once, getting 6s on both is (1/6)*(1/6) = 1/36

Answer 2

The way you pose the question it should be 1/6. That is, you observe one die and do not observe the other. So whether or not the unobserved die comes out 6 does not affect what you observe on the other die and vice versa. We have simple independent random variables.

We have 2 dice, one which we observe and one which we do not observe, with 36 possible roll combinations. We see that the observed die comes out 6, and there are only 6 possible roll combinations where that happens; and only one of those has the other die come out 6.

If the question was posed differently so that you find out that at least one of the 2 dice comes out 6, then the probabilities are different. Out of 36 possible roll combinations, we find ourselves in one of the 11 of them with a at least one six: (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (5,6), (4,6), (3,6), (2,6), (1,6).

Out of those 11 possibilities, only one has both dies rolling a 6. So the probability of both dies coming out 6, given this information is 1/11. We don't get to count (6*,6) and (6,6*) 2 separate states since they're the same roll combination. Or, if you like, you could say there's a 50% chance that you saw (6*,6) and a 50% chance of (6,6*), so each of those count as a half a state.

Answer 3

Either you OBSERVE the outcome or you don't.

There is no in between.

There are 36 possible outcomes when rolling a pair of dice.

If you want one 6, that narrows it down to 1/6.

If you want both 6, that is 1/36.

That's it.

Nothing more... nothing less