The exercise can be solved by using the definition of the conditional probability, or we can also compute the probability of the “6 on second role” event directly over the sample space using the fundamental axioms and theorems of probability.

Without conditional probabiilty

The sample space consists of the single rolls and the double rolls. Representing single rolls as numbers and double rolls as tuples of numbers, we have:

\[\Omega = \lbrace 1,3,5\rbrace \cup \lbrace 2,4,6\rbrace \times \lbrace 1,2,3,4,5,6\rbrace\]

The total number of outcomes \(\card{\Omega} = 21\).

There are 3 outcomes that have a 6 in the second roll, namely \((2,6), (4,6), (6,6)\). Therefore, the probability of rolling a 6 on the second roll is: \(\frac{3}{21}= \frac{1}{7}\), one might think. But this is not correct, because this reasoning assumes that all the 21 outcomes are equally likely, but they are not. Why so?

Concentrate on the first roll. The event of having an odd number must share the same amount of cake (probability mass) with having an even number. This is because our dice is fair. Therefore, what is left from the entire cake to the set of tuples is only the half. As there is no reason to think that the outcomes in the tuples region of our sample space are not equally likely, each and every tuple should share the half of the cake equally among themselves. This means \(1/2\) total mass of probability will be shared between 18 tuples, leaving \(1/36\) to each tuple. As there are \(3\) tuples with a \(6\) in the second roll, the probability of rolling a 6 on the second roll is \(3 \times 1/36 = 1/12\).

With conditional probability

First the definition:

The conditional probability of an event \(A\) given an event \(B\) is the probability of \(A\) occurring given that \(B\) has occurred. It is denoted as \(P(A\given B)\) and can be calculated using the formula:

\[P(A\given B) = \frac{P(A \cap B)}{P(B)}\]

Let \(A=\)“6 on second roll” and \(B=\)“even on first roll”. We are interested in finding \(P(A\cap B)\). Rearranging we get:

\[P(A\cap B) = P(A\given B)P(B)\]

\(P(B)=1/2\) is not hard to find. What about \(P(A\given B)\)? If we know that the first roll was even, then we now that there is a second roll, and that the second role has equally likely \(6\) outcomes. Therefore, \(P(A\given B) = 1/6\). Putting everything together,

\[P(A\cap B)= \frac{1}{2}\times\frac{1}{6}=\frac{1}{12}\]

See total probability theorem