Approach 1: First Principles (Law of Total Expectation / Recursive Method)
Let $E$ be the expected (average) number of rolls required to get a $5$.
When we roll the fair $6$-sided die for the first time, there are two mutually exclusive scenarios:
- We roll a 5:
- Probability: $P(\text{Rolling } 5) = \frac{1}{6}$
- Number of rolls taken: $1$
- We do not roll a 5: (we roll a $1, 2, 3, 4,$ or $6$)
- Probability: $P(\text{Not rolling } 5) = \frac{5}{6}$
- Number of rolls taken: $1$ roll is spent, and due to the memoryless property of independent die rolls, the expected number of remaining rolls required resets back to $E$. Thus, total expected rolls in this branch is $(1 + E)$.
Setting up the equation using the Law of Total Expectation:
$$E = \left(\frac{1}{6} \times 1\right) + \left(\frac{5}{6} \times (1 + E)\right)$$
Now, let's solve for $E$:
$$E = \frac{1}{6} + \frac{5}{6} + \frac{5}{6}E$$
$$E = 1 + \frac{5}{6}E$$
Subtract $\frac{5}{6}E$ from both sides:
$$E - \frac{5}{6}E = 1$$
$$\frac{1}{6}E = 1$$
$$E = 6$$
Approach 2: Using Geometric Distribution Framework
Let $X$ be a discrete random variable representing the number of independent trials (rolls) until the first success (getting a $5$).
- Success event ($S$): Rolling a $5 \implies p = \frac{1}{6}$
- Failure event ($F$): Rolling any other number $\implies q = 1 - p = \frac{5}{6}$
Since each roll is independent and has a constant probability of success, $X$ follows a Geometric Distribution:
$$X \sim \text{Geometric}\left(p = \frac{1}{6}\right)$$
The probability mass function (PMF) for the number of trials up to and including the first success is given by:
$$P(X = k) = q^{k-1}p = \left(\frac{5}{6}\right)^{k-1}\left(\frac{1}{6}\right) \quad \text{for } k = 1, 2, 3, \dots$$
The expected value $E[X]$ of a geometric random variable is defined as:
$$E[X] = \sum_{k=1}^{\infty} k \cdot P(X = k) = \frac{1}{p}$$
Substituting our value of $p$:
$$E[X] = \frac{1}{\frac{1}{6}} = 6$$
Correct Answer:
The expected number of rolls until he gets a $5$ is $6$.
