The elegant solution is to randomly assign a number from between 1 and 90 inclusive to Jane, where groups are defined by the numbers (1-30, 31-60, 61-90). Now the probability that Joe is assigned to the same group as Jane is \(\frac{29}{89}\).

The more complicated solution is to think of groups one at a time. The probability that Joe and Jane are both in the first group is to divide the number of ways that this can happen by the total number of ways that the first group can be formed. There are \({88 \choose 28}\) ways to form the first group with Joe and Jane, and \({90 \choose 30}\) ways to form the first group without any restrictions. Therefore the probability that Joe and Jane both fall into the first group is \({88 \choose 28} / {90 \choose 30}\). By symmetry, the probability that Joe and Jane both fall into the second or third group is the same, so the total probability that Joe and Jane end up in the same group is \(3 \cdot {88 \choose 28} / {90 \choose 30}\).