We are tasked with finding the expected number of different numbers rolled when three standard fair 6-sided dice are rolled. To solve this, we will calculate the expected number of distinct outcomes by considering the probabilities of different cases (i.e., how many distinct numbers show up).

Step 1: Define the indicator variables

Let the three dice rolls be denoted by $X_1, X_2, X_3$, where each die shows a number between 1 and 6. We define indicator random variables $I_k$ for $k = 1, 2, 3, 4, 5, 6$, which are:

\begin{cases} 1 & \text{if the number } k \text{ appears on at least one of the dice}, \\ 0 & \text{otherwise}. \end{cases}$$ The total number of distinct numbers rolled is then given by the sum of the indicators: $$D = \sum_{k=1}^{6} I_k.$$ Thus, the expected number of distinct numbers is: \[ E[D] = E\left[ \sum_{k=1}^{6} I_k \right]. \] By the linearity of expectation, this becomes: \[ E[D] = \sum_{k=1}^{6} E[I_k]. \] ### Step 2: Calculate $$E[I_k]$$ The expectation $$E[I_k]$$ is the probability that the number $$k$$ appears on at least one of the dice. For each die, the probability that it does **not** show $$k$$ is $$\frac{5}{6}$$, since there are 5 other outcomes. Thus, the probability that none of the three dice shows $$k$$ is: $$\left( \frac{5}{6} \right)^3.$$ Therefore, the probability that $$k$$ appears on at least one die is: $$P(\text{at least one } k) = 1 - \left( \frac{5}{6} \right)^3.$$ Thus, we have: $$E[I_k] = 1 - \left( \frac{5}{6} \right)^3 = 1 - \frac{125}{216} = \frac{91}{216}.$$ ### Step 3: Calculate the total expected number of distinct numbers Since $$E[I_k] = \frac{91}{216}$$ for each $$k$$, the expected number of distinct numbers rolled is: $$E[D] = \sum_{k=1}^{6} \frac{91}{216} = 6 \times \frac{91}{216} = \frac{546}{216} = \frac{91}{36}.$$ Thus, the expected number of different numbers rolled is: $$\boxed{\frac{91}{36}}.$$ ### Further Exploration: 1. What happens if we roll two dice instead of three? How does the expected number of distinct numbers change? 2. If we roll four dice, how does the expected number of distinct numbers compare to when we roll three dice? 3. How does the probability of having at least one $$k$$ on the dice change if more than three dice are rolled? 4. What would the expected number of distinct outcomes be if the dice had more than 6 sides, say 10 sides? 5. How can this technique be generalized to any number of dice rolled and any number of sides on each die? **Tip:** The linearity of expectation allows us to calculate expected values in problems with complex counting, like this one, by breaking the problem down into manageable pieces using indicator variables.