A random walk is a process in which something moves from point to point, and where the direction of movement at each step is randomly chosen. Suppose that a person conducts a random walk on a line: he starts at $0$ and each minute randomly moves either $1$ unit in the positive direction or $1$ unit in the negative direction. What is his expected distance from the origin after $6$ moves?

[i]2017 CCA Math Bonanza Lightning Round #3.4[/i]
1. **Define the problem and the random walk process:**
   - The person starts at position \(0\).
   - Each minute, the person moves either \(+1\) unit or \(-1\) unit with equal probability.
   - We need to find the expected distance from the origin after \(6\) moves.

2. **Calculate the probabilities for each possible distance:**
   - After \(6\) moves, the possible distances from the origin are \(0, 2, 4, \) and \(6\) units.
   - The number of ways to achieve each distance can be calculated using binomial coefficients.

3. **Calculate the number of ways to achieve each distance:**
   - Distance \(6\) units: All moves must be in the same direction.
     \[
     \binom{6}{0} = 1 \quad \text{(all positive or all negative)}
     \]
   - Distance \(4\) units: \(5\) moves in one direction and \(1\) move in the opposite direction.
     \[
     \binom{6}{1} = 6
     \]
   - Distance \(2\) units: \(4\) moves in one direction and \(2\) moves in the opposite direction.
     \[
     \binom{6}{2} = 15
     \]
   - Distance \(0\) units: \(3\) moves in one direction and \(3\) moves in the opposite direction.
     \[
     \binom{6}{3} = 20
     \]

4. **Calculate the probabilities for each distance:**
   - Total number of possible sequences of moves:
     \[
     2^6 = 64
     \]
   - Probability of being \(6\) units away:
     \[
     \frac{2 \cdot 1}{64} = \frac{2}{64}
     \]
   - Probability of being \(4\) units away:
     \[
     \frac{2 \cdot 6}{64} = \frac{12}{64}
     \]
   - Probability of being \(2\) units away:
     \[
     \frac{2 \cdot 15}{64} = \frac{30}{64}
     \]
   - Probability of being \(0\) units away:
     \[
     \frac{20}{64}
     \]

5. **Calculate the expected distance:**
   - Expected distance \(E\) is given by:
     \[
     E = \left(\frac{2}{64} \cdot 6\right) + \left(\frac{12}{64} \cdot 4\right) + \left(\frac{30}{64} \cdot 2\right) + \left(\frac{20}{64} \cdot 0\right)
     \]
   - Simplify the expression:
     \[
     E = \frac{12}{64} + \frac{48}{64} + \frac{60}{64} + 0
     \]
     \[
     E = \frac{12 + 48 + 60}{64} = \frac{120}{64} = \frac{15}{8}
     \]

The final answer is \(\boxed{\frac{15}{8}}\)