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).
Solution 1:
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.
Solution 2:
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.
Solution 3:
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.
Solution 4:
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.
Solution 5:
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 $1/11$.