Stopping Six

Solution

Answer: 6

Let \(X\) be the number of rolls needed until the first \(6\) appears. Each roll is independent, and the probability of getting a \(6\) on any roll is \[ p = \frac{1}{6}. \] So \(X\) has a geometric distribution with parameter \(p=\frac16\). For a geometric random variable, the expected value is \[ E(X) = \frac{1}{p}. \] Substituting \(p=\frac16\), \[ E(X) = \frac{1}{1/6} = 6. \]