TITLE: Two dice thrown, one comes up 6
QUESTION [27 upvotes]: 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).

REPLY [2 votes]: I'm posting this as a second answer, as it is a different, more visual approach to think about it
first of all, 36 already includes all possible outcomes, even if the dice are colored (red and green in this example).
this is what (6,1) should be interpreted for the following tables as
(Red-6, Green-1) or
(First-6, Second-1) or
(Left-6, Right-1)

all 36 possible outcomes while rolling two dice:
(1,1)  (2,1)  (3,1)  (4,1)  (5,1)  (6,1)

(1,2)  (2,2)  (3,2)  (4,2)  (5,2)  (6,2)

(1,3)  (2,3)  (3,3)  (4,3)  (5,3)  (6,3)

(1,4)  (2,4)  (3,4)  (4,4)  (5,4)  (6,4)

(1,5)  (2,5)  (3,5)  (4,5)  (5,5)  (6,5)

(1,6)  (2,6)  (3,6)  (4,6)  (5,6)  (6,6)

Following assumes we have seen exactly one die, not both, and that one die was a 6.
Now let's say we saw the Red/First/Left was a 6 that leaves us with:
Fig 1:
(---)  (---)  (---)  (---)  (---)  (6,1)

(---)  (---)  (---)  (---)  (---)  (6,2)

(---)  (---)  (---)  (---)  (---)  (6,3)

(---)  (---)  (---)  (---)  (---)  (6,4)

(---)  (---)  (---)  (---)  (---)  (6,5)

(---)  (---)  (---)  (---)  (---)  (6,6)

6 possible outcomes remain, thus the chance of the Green/Second/Right one being a 6 is 1/6
now if we looked at the Green/Second/Right die and it was a 6:
Fig 2:
(---)  (---)  (---)  (---)  (---)  (---)

(---)  (---)  (---)  (---)  (---)  (---)

(---)  (---)  (---)  (---)  (---)  (---)

(---)  (---)  (---)  (---)  (---)  (---)

(---)  (---)  (---)  (---)  (---)  (---)

(1,6)  (2,6)  (3,6)  (4,6)  (5,6)  (6,6) 

also 6 possible outcomes remain, the chance to the Red/First/Left die to be a 6 is also 1/6
I have shown above that the (6,6) is just one event and not 2 different, simply because we looked at exactly one die, either the Red or the Green one. Meaning either Fig.1 OR Fig.2
Now does it make any difference for a double-number case if the dice are uncolored/unordered ? No, it doesn't we still see exactly one die, not both. We just don't know if we should look at Fig.1 or Fig.2, but the chance is the same for both.
================================================================================

As a little extra, let's see what happens in the cases of

Someone else see's both dice and tells us that at least one die is a 6
Someone rolls the dice and reveals only the 6, never another number
We saw both dice, confirmed a 6, but unsee the other die (ok, this is getting ridiculous)

They are equivalent and the essence is, the pair must contain at least one six
The table now looks like this:
Fig 3:
(---)  (---)  (---)  (---)  (---)  (6,1)

(---)  (---)  (---)  (---)  (---)  (6,2)

(---)  (---)  (---)  (---)  (---)  (6,3)

(---)  (---)  (---)  (---)  (---)  (6,4)

(---)  (---)  (---)  (---)  (---)  (6,5)

(1,6)  (2,6)  (3,6)  (4,6)  (5,6)  (6,6)

11 possible outcomes that contain at least one six, but only one with a double six, therefore the chance would be 1/11, also note that (6,6) is still only one event.
To bring it all back together:
When rolling 2 dice 11 out of all 36 possible outcomes contain at least 1 six, the chance is 11/36
If we are looking at exactly one die (either Red or Green), only 6 out of those 11 outcomes we would happen to see the rolled 6, even if someone else told us it definitely contains at least one six. The chance is 6/11
Out of those 6, only once it would be a double-6, the chance is 1/6
So the chance that

Two rolled dice contain at least one six (11/36)
We then happen to see that 6, if we only look at exactly one die (6/11)
The other die would then be a 6 too. (1/6)

is (11/36) * (6/11) * (1/6) = 1/36
But by seeing one die, and it being a 6, the 1. and 2. event has already occurred. That leaves us with only 1/6.