Dice Maxxing

Solution

Answer: 4.47

Let the two dice rolls be \(X_1\) and \(X_2\). We keep the larger value \[ M = \max(X_1, X_2). \] There are \[ 6^2 = 36 \] equally likely outcomes. First find the distribution of the maximum. For \(k = 1,\dots,6\), \[ P(M \le k) = \left(\frac{k}{6}\right)^2 \] since both dice must be at most \(k\). Thus \[ P(M = k) = P(M \le k) - P(M \le k-1) = \left(\frac{k}{6}\right)^2 - \left(\frac{k-1}{6}\right)^2. \] Now we can compute these probabilities as such: \[ P(M=1) = \frac{1}{36} \] \[ P(M=2) = \frac{4}{36} - \frac{1}{36} = \frac{3}{36} \] \[ P(M=3) = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \] \[ P(M=4) = \frac{16}{36} - \frac{9}{36} = \frac{7}{36} \] \[ P(M=5) = \frac{25}{36} - \frac{16}{36} = \frac{9}{36} \] \[ P(M=6) = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}. \] Expected value is thus given by \[ E(M) = \sum k P(M=k). \] \[ E(M) = 1\left(\frac{1}{36}\right) + 2\left(\frac{3}{36}\right) + 3\left(\frac{5}{36}\right) + 4\left(\frac{7}{36}\right) + 5\left(\frac{9}{36}\right) + 6\left(\frac{11}{36}\right). \] \[ E(M) = \frac{161}{36} = \approx 4.472. \]