Monopoly Dice

It is easy to check that the expected number of squares moved on a single die roll, given that that roll is a double, is $\tfrac16(2 + 4 + 6 + 8 + 10 + 12) = 7$ It is also possible to show that the expected number of squares moved on a single die roll, given that that roll is not a double, is also $7$ . Thus the expected number of squares moved on a single die roll is $7$ , independently of whether a double is rolled or not. Thus the expected number of squares moved in a Monopoly move is equal to $7$ times the expected number of rolls in that move that result in a move of some sort. If $N$ is that number of rolls, then $P(N=1) = \tfrac56$ , while $P(N=2) = \tfrac{5}{36} + \tfrac{1}{216} = \tfrac{31}{216}$ and $P(N=3) = \tfrac{5}{216}$ . Recall that the probability of rolling a double is $\tfrac16$ ,and $N$ can be equal to $2$ if three doubles are rolled.

Thus the expected number of moves is $7\left(\frac56 \times 1 + \frac{31}{216} \times 2 + \frac{5}{216} \times3\right) \; = \; \frac{1799}{216}$ making the answer $1799 + 216 = \boxed{2015}$ .

The expected value of rolling one die is $\frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \frac{7}{2}$ , so the expected value of rolling two dice is $2 \cdot \frac{7}{2} = 7$ .

Of the $6^2 = 36$ combinations of rolling two dice, there are $6$ possible doubles, for a probability of $\frac{6}{36} = \frac{1}{6}$ .

Therefore, the first roll you can expect a $7$ , and $\frac{1}{6}$ of the time you will have doubles and expect another $7$ , and $\frac{1}{6}$ of that $\frac{1}{6}$ of the time you will have two doubles at a $\frac{5}{6}$ probability that you won't have a third double and expect another $7$ , so that

$E = 7 + \frac{1}{6} \cdot 7 + \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \cdot 7 = \frac{1799}{216}$

This means that $p = 1799$ , $q = 216$ , and $p + q = \boxed{2015}$ .