Two dice thrown, one comes up 6 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).
 A: The possibility of one die rolling a six is totally unrelated to the other die.
Imagine rolling one die after the other. I think we all agree it's the same as rolling them at the same time.


*

*The first one can be 1,2,3,4,5,6 so the chance is 1/6.

*The second one can also be 1,2,3,4,5,6, this is totally unrelated to what we rolled with the first die, so the chance is also 1/6


As you already confirmed one being 6, the other one being a 6 still has a chance of 1/6.
I think the confusion lies within the double-six chance. It's 1/6 for the first die, and out of this only 1/6 of times the second die happens to be a 6 too, so it's 1/6 * 1/6 = 1/36. But this doesn't count if you can already confirm one die being a 6, thus the first die has a chance of 1/1 the second still 1/6.
Note that even if the dice are coloured (green and red) that doesn't matter in the case for a double six. Just because the green die was a six doesn't influence the chance of the red die being a six, it is always 1/6.
Coloured dice make only a difference if we are talking about two different numbers. As an example, chances to have a (3-green, 6-red) is 1/36 whereas having a 3 and a 6 on non coloured dice has a chance of 2/36.
NOTE:

but I see that one of them was a 6

isn't a precise description and leaves room for interpretation, it could mean "but I see one of them, and it was a 6" or "but I see that exactly one was a 6", given the question, i assume the former was meant.
A: There are two ways to pose the condition of the question, which lead to different answers:
(1) I happen to catch sight of the first-rolled die; it shows a $6$.
(2) I know only that at least one of the dice is a $6$.
In the first case, the answer is 1/6; in the second, it is 1/11.
If we modify the condition of case 1, to specify the second-rolled die rather than the first-rolled, or the die rolled with the left hand (say), or the red die (assuming that one die is red and the other blue), it does not change the probability from 1/6. In fact, any condition that tends to specify the prior identity of the die you spot tends to move the probability from 1/11 toward 1/6.
A: 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.
A: If the intuition is not yet clear, perhaps one can do a formal conditional probability calculation. Let $A$ be the event "at least one $6$" and $D$ the event "double $6$." We want $\Pr(D|A)$. By the definition of conditional probability this is $\frac{\Pr(A\cap D)}{\Pr(A)}$. 
The event $A\cap D$ is just the event $D$, and has probability $\frac{1}{36}$.
Now there are $11$ outcomes in which there is at least one $6$, so $\Pr(A)=\frac{11}{36}$.
Now we can compute the conditional probability. 
A: There are a total of $36$ outcomes, consisting of ordered pairs $(a,b)$ where $a,b\in \{1,\cdots,6\}$. Of those outcomes, the following 11 are possible given your knowledge:
$(1,6),\cdots,(6,6)$ and $(6,1),\cdots,(6,5)$ (we already counted $(6,6)$).
Among those 11, there is 1 corresponding to both 6's. Thus your answer is $1/11$.
A: Simple answer: 1/6.
The probability of one dice does not affect the other.
That 1 was a 6 can be ignored; it happens to be true in this case, but is the same as saying "I saw that the wall of the room we were in was orange; given the wall was orange, what is the probability of the dice being a six?".
So assuming the dice are regular 6 sided, unbiased die, the probability of the die being a 6 is 1/6.
A: Probability is another term for relative frequency. If I repeat an experiment $n$ times and observe that a chosen event occurs $k$ times then its probability $p(n)$ is $\frac{k}{n}$. Due to the law of large numbers $p(n)$ converges against a value $p$. Then $p$ is the value we tend to speak of when we refer to the probability of an experiment and a chosen event.
You describe a situation and not an experiment. Therefore one could argue that the idea of probability/relative frequency doesn't make sense at all from a formal point of view. However, the description of a situation can unambigously induce an experiment, e.g. when I throw a dice what is the probability that an odd number occurs? It is clear how this can be repeated and what counts as $n$ and $k$.
Unfortunately, your situation does not unambigously describe an experiment. Let us assume that one dice is red and the other one is green. Then I can think of at least two possible experiments which seem to reflect your situation:


*

*Your friend throws both dices and you always happen to see the value of the red die.  If the red die shows a 6 you increase $n$ by one. If the green die shows a 6 as well you increase $k$ by one.

*You happen to randomly see either the value of the green or the red die. If this value is 6 you increase $n$. If the other value is 6 as well then you increase $k$.


Obviously, the first situation reduces to simply throwing one die and therefore $p=\frac{1}{6}$. In the second case it is $p=\frac{1}{11}$ as already explained.
A: Just to add a different perspective, one could think about this question in terms of "How would you simulate the process by which you came to know that there is one 6?"
For example, from the description, we can infer the following generative process/simulation algorithm:


*

*Roll two die

*If neither are 6's then return to step 1, ELSE continue to Step 3

*IF there is only one 6, then show that to the other player ELSE randomly pick one of the two 6's to show.


So, that is the generative process. What can we say of the probabilities?
First, we know that there must be at least one 6 for this experiment to get off the ground. So that is our "Base" probability. Let $N=$number of 6's, then $P(N=0)=\frac{25}{36}\rightarrow\;\;P(N>0)=\frac{11}{36}$
Second, of these, how often will only one 6 turn up? We want $P(N=1|N>0)=\large\frac{2\left(\frac{1}{6}\cdot\frac{5}{6}\right)}{\frac{11}{36}}=\frac{10}{11}$
Now, let $H=$ the face value of the unseen die, then $P(H=6|N=1)=0$
However, if $N=2$ then it is irrelevant which die we are shown, as we know $P(H=6|N=2)=1$
Thus, $\small P(H=6|N>0)=P(N=1|N>0)P(H=6|N=1)+P(N=2|N>0)P(H=6|N=2)=\frac{10}{11}\cdot 0+\frac{1}{11}\cdot 1=\frac{1}{11}$
Ok, so that's my interpretation. What if we challenge steps 2 and 3 in the algorithm?


*

*Step 2: If we don't have at least 1 six, he wouldn't be able to show me a six. So that is justified.

*Step 3: Again, if there is only 1 six, then the person doesn't have a choice of which one to show me in order to make the outcome match our knowledge, so again, this step is justified.


Therefore, it appears that the answer is $\frac{1}{11}$
HOWEVER
What if we don't care that it's a 6, per say, then we want to know the probability that both dice come up with the same number. In this case, there are six pairs that satisfy this, of 36 total pairs. In this case, we get the probability being $\frac{1}{6}$
It seems like the latter scenario is more likely, since why would the person be determined to show a six? More likely, the algorithm can be modified to read:


*

*Roll two die

*Randomly cover one up; show the other to the player.


In this case, it is mere coincidence that we saw a 6 vs a 3 or a 2 etc.
This boils down to the reference class problem. We don't know if the person had a particular number in mind or some non-random method for choosing which one to cover up, so we're stuck with ambiguity.
A: If you choose the die to see (e.g. show me the first die), then trivially it is 1/6 for the other die.This applies even if you randomly choose which die to see. This is because you ll be in one of the two cases:
$P_{(6,6)|(\text{first was 6})}=P_{(6,6)|(\text{second was 6})}=1/6$
( e.g. see first one with 70% probability: $\frac{70}{100}*\frac{1}{6}+\frac{30}{100}*\frac{1}{6}=\frac{1}{6}$)
If your friend chooses which die to reveal, then all you know is there is at least one 6 rolled (you are already in one of the yellow+red cases). You want the red case, so its probability happening is for fair dice $P_{(6,6)|(\text{at least one 6 rolled})} = 1/11$

A: I'll have a bash at the 72 possibility approach.
Borrowing the image from ntg: 

You can see that there are:
25 situation where there is no six. We're not interested in those.
10 situations where there is one six.
1 situation where there is a double six.
Here then is the different approach.
For each of the 10 "one six" situations, there is a 50/50 chances that I get to see the six. Clearly, we're only interested in those times when we do. 
So if you look again at the image above and split each cell in half, then there are 12 half cells which represent a six in a situation where you can see a six. Two of them are in the double six.
So if you can see a six, then there must be a $2/12$ or $1/6$ chance that you're looking at half of a double six.
