Solution: Second throw
The solution is fairly straightforward by using conditional probability and the multiplication rule. However, we can also compute the probability directly over the sample space.
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 \(\mid\Omega\mid = 21\).
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, which is not the case. Why not?
Concentrate on the first roll. The event of having an odd number must share the same amount of cake (probability mass) with all the tuples that start with 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 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 we have only \(3\) tuples with a \(6\) in the second roll, the probability of rolling a 6 on the second roll is \(3 \times 1/36 = 1/12\).
If you have the notion of conditional probability and the multiplication rule in place, then the solution is easier to find. To get a \(6\) in the second roll, you need to have two things happening together: you roll an even number in the first roll, and you roll a \(6\) in the second roll. The probability of doing these at the same time is the product of their probabilities, which is \(1/2\times 1/6 = 1/12\).