# 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).

• Note you want to think Bayes' Theorem here. See Nicolas' answer. Note that the probability you don't roll a six on both dice is $(5/6)^2 = 25/36$ and so the probability of the event occuring is $1 - 25/36 = 11/36$. Aug 1, 2014 at 5:25
• I'm not sure I understand the question. If you know that one is a 6, then the question becomes: "what's the change that the other is a 6", right? So it would be 1/6. Aug 1, 2014 at 5:55
• Aug 1, 2014 at 11:01

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.

• +1, "The answer depends on the details of just how you obtained your information." Too many probability questions are phrased ambiguously.
– JiK
Aug 1, 2014 at 8:39
• +1: I read this and I went through several stages of grief: Denial ("Surely not! It's just 1/6!"), Anger ("But... But...! That's clearly not what was being asked!"), and finally Acceptance as I realised you are 100% correct. Aug 1, 2014 at 12:30

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.

• The other die may depend on the visible one if his friend chose which one to show. Aug 1, 2014 at 10:59
• @Cthulhu Right, if the friends rules were: "If there rolls exactly 1 6, cover the other die. If there rolls 2 6's, then cover one arbitrarily" then the odds would in fact not be 1/6, it would actually be 1/36. These questions are of course all about what the rules were. This is what most people's issue with monty hall is as well. They don't understand that the rules that the host plays by, reveals information. Aug 1, 2014 at 14:13
• @Cthulhu you're right, as i added in the note, i assume he (accidentally) sees one (not both) of the dice and it was a six. Aug 2, 2014 at 20:42

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.

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.

• No, we want $\: P(D\hspace{.05 in}|\hspace{.02 in}\text{ he saw a 6}) \;$. $\;\;\;\;$
– user57159
Aug 1, 2014 at 6:24
• The event $A$ is my interpretation of "he saw a $6$." One could interpret it differently, for example that he saw a $6$ on the green die, and then the answer would be different. Aug 1, 2014 at 6:27
• "At least one 6" and "he saw a 6" are quite different, I think. If he saw a 6, then he also saw the colour of the die that had the 6. If he is only told that there is a 6, he has no way of knowing which one it is. Aug 1, 2014 at 7:20

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 $1/11$.

• This might be a stupid question, but if we are not counting $(6,6)$ twice, then why are we counting $(1,6)$ twice? ($(1,6)$ and $(6,1)$) Aug 1, 2014 at 5:36
• It's a good question, maybe a practical example will help. Suppose my procedure is as follows: I throw the two dice, and write down $(a,b)$ if the one on the left is $a$ and the one on the right is $b$. Thus the two outcomes $(1,6)$ and $(6,1)$ come from different events. However, both the outcomes $(6,6)$ and $(6,6)$ represent the same event. Aug 1, 2014 at 5:50
• Among those 11, there is 1 corresponding to both 6's. $\:$ For the outcome (6,6), the die he saw would have probability one of showing a 6. $\:$ For the other 10, the die he saw would have probability half of showing a 6. $\:$ (1*1)/((1*1)+(10*(1/2)) = 1/(1+(10/2)) = 1/(1+5) = 1/6 . $\:$ Thus his answer is 1/6. $\;\;\;\;$
– user57159
Aug 1, 2014 at 6:21
• IMHO, if I understand this “left die” / “right die” nomenclature correctly, it’s not necessarily illuminating or clarifying. We could have a situation like “My friend rolled two dice, and cupped her hand to protect them from my view, but one of them rolled further than she expected. It rolled to a place where I could see it, and I saw that it was a 6.” It may be better to think of differently colored dice. (red die = 6, blue die = 6) is clearly one possible outcome, while (red = 1, blue = 6) and (red = 6, blue = 1) are (clearly?) distinct. Aug 1, 2014 at 16:57
• My point is that the answer to the question "if we are not counting (6,6) twice, then why are we counting (1,6) twice? ((1,6) and (6,1))" becomes clearer if you establish (even if only conceptually) a way of distinguishing between the two dice that's independent of the dice-rolling event. Aug 1, 2014 at 17:20

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.

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

1. Two rolled dice contain at least one six (11/36)
2. We then happen to see that 6, if we only look at exactly one die (6/11)
3. 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.

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:

1. 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.
2. 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.

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:

1. Roll two die
2. If neither are 6's then return to step 1, ELSE continue to Step 3
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?

1. 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.
2. 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:

1. Roll two die
2. 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.

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 $P_{(6,6)|(\text{at least one 6 rolled})} = 1/11$ 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.