# Roll a fair 6 sided die twice, What is the probability that one or both rolls are 6?

Roll a fair 6 sided die twice, What is the probability that one or both rolls are 6?

This question is confusing me on a conceptual level, specifically the "or both rolls."

With out that, the question is what is the probability of getting a 6 by rolling a die twice, which is of course 11/36.

The problem I am having is I don't understand how the "or both" factors in. If you roll a 6 on the first and second roll (1/36 chance) is that counted separately than getting a 6 on the first or second roll?

And if so, would the answer be 1/6 + 1/6 + 1/36 = 13/36?

And that leads into my second question:

What is the probability that one or both rolls are even numbers (2, 4 or 6’s)?

IF the above work is correct, then by the same logic the chance of the first roll being even is 1/2 + the chance of the second roll being even which is 1/2 + the chance of them both being even which is 1/4. This adds up to >1 which is obviously impossible, but I don't know what is going wrong.

Thanks.

• An equivalent way to phrase the question is “what is the probability that any $6$ is rolled.” This should lead you to compute $1-P(\text{no sixes are rolled})$.
– amd
Dec 1, 2015 at 22:31

Almost. As you argued, probability to hit a 6 on each roll is 1/6, but the sum overcounts rolling 6 twice, so the actual probability of rolling 6 on either one or both roll is $$\frac{1}{6} + \frac{1}{6} - \frac{1}{36} = \frac{11}{36}.$$ Can you take it from here?

• Hmm, alright I understand that, and therefore the answer to my second would be 3/4. Just to make sure, the next question is: What is the probability that the sum of the two rolls is less than 6 or at least one of the rolls is an even number? I am getting 1/9 (prob of sum <6) + 3/4 (prob of at least 1 even) - 7/36 (pairs that sum < 6 and have at least 1 even) which comes out to 2/3. Is that right?
– Luke
Dec 1, 2015 at 21:33
• @BNSlug seems right to me Dec 1, 2015 at 21:55
• @BNSlug This is an example of the Inclusion-Exclusion formula Dec 2, 2015 at 0:34

I think your confusion is the use of the word "or" and how it relates to probability computations. In particular, it seems like your reasoning can be represented by the following computation:

\begin{align*} P(\text{One or both rolls are 6}) &= P(\text{The first die is 6 OR the second die is 6 OR both dice are 6}) \\ &= P(\text{The first die is 6}) + P(\text{The second die is 6}) + P(\text{Both dice are 6}) \\ &= \frac{1}{6} + \frac{1}{6} + \frac{1}{36} = \frac{13}{36} \end{align*}

The error in this computation is that it is NOT true that $P(\text{A OR B}) = P(A) + P(B)$ for events $A$, $B$, in general. This is only true when $A$ and $B$ are so-called `disjoint' events, that is, it is impossible for both to happen. The formula that works in general is:

$$P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B})$$

Also, the statement "The first die is 6 OR the second die is 6 OR both dice are 6" is redundant- it means the exact same thing as "The first die is 6 OR the second die is 6", and this is the exact same thing as "At least one die is 6".

• Further to that, the proper way to partition the event into disjoint sets, is: \begin{align}\mathsf P(\textsf{One of both rolls are }6) = & \mathsf P(\textsf{the first roll is 6 AND the second roll is not 6}) \\[0ex] & + \mathsf P(\textsf{the first roll is not 6 AND the second roll is 6})\\[0ex] & +\mathsf P(\textsf{the first roll is 6 AND the second roll is 6}) \\[1ex] = & \frac{5}{36}+\frac{5}{36}+\frac{1}{36}\end{align} Dec 2, 2015 at 3:14

If you role two dice, then there are 36 equally likely outcomes, (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Of those, 11, the last of each of those six lines and all of the last, have at least one 6 so the probability that "one or both rolls is a 6" is 11/36. Your "13/36" is wrong because both of the "1/6" In your sum include "(6/6)". It should be 1/6+ 1/6- 1/36 where the 1/36 being subtracted is to take away one of the (6, 6)s.

The general rule for "Probability of A or B" is "Probability of A"+ "Probability of B" minus "Probability of A and B".

When you roll two dice, the probability the first die is even is 1/2, the probability the second die is 1/2, and the probability both are even is (1/2)(1/2)= 1/4 (the results of the two rolls are independent) so the probability that either one or both are even is 1/2+ 1/2- 1/4= 3/4.

(Of the 36 outcomes above, (1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) have "either one or both even". There are 27 of them so the probability is 27/36= 3/4.)

You can decompose the problem as follows:

$$P(Atleast\;one\;roll\:is\;6)$$ = $$1\;-\;P(No\; roll \; is \; 6)$$
$$P(No\; roll \; is \; 6)\;=\;\frac{5}{6}*\frac{5}{6}=\frac{25}{36}$$

Hence, $$P(Atleast\;one\;roll\:is\;6)$$ = $$1\;-\frac{25}{36} = \frac{11}{36}$$

Your doubt is based on the following property:

$$P(A\;\cup B) =P(A)+P(B)$$

This property is true only when A and B are independent.
$$P(First\;die\;6)$$ and $$P(Second\;die\;6)$$ both contain the event $$E=Both \;die\;6$$
Hence, they are not truly independent and the event $$E\;$$gets counted twice.
Therefore, in this case it's $$P(A\;\cup B) =P(A)+P(B)-P(A\cap B)$$
Where the events are $$A=First\;roll\;is\;6$$ and $$B=Second\;roll\;is\;6$$

• How is it different from accepted solution? Jul 15, 2020 at 6:46
• @Sagar Chand This explains it in detail and provides an alternative solution approach too Jul 15, 2020 at 6:51