Albert rolls a fair six-sided die thirteen times. For each time he rolls a number that is strictly greater than the previous number he rolled, he gains a point, where his first roll does not gain him a point. Find the expected number of points that Albert receives. 

[i]Proposed by Nathan Ramesh
1. **Define the problem**: Albert rolls a fair six-sided die thirteen times. For each roll after the first, he gains a point if the number rolled is strictly greater than the previous number. We need to find the expected number of points Albert receives.

2. **Calculate the probability of gaining a point on a single roll**: 
   - Let \(X_i\) be the outcome of the \(i\)-th roll.
   - For \(i \geq 2\), Albert gains a point if \(X_i > X_{i-1}\).
   - The probability that \(X_i > X_{i-1}\) can be calculated by considering all possible outcomes of \(X_{i-1}\) and \(X_i\):
     \[
     P(X_i > X_{i-1}) = \frac{1}{6} \cdot \frac{5}{6} + \frac{1}{6} \cdot \frac{4}{6} + \frac{1}{6} \cdot \frac{3}{6} + \frac{1}{6} \cdot \frac{2}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot 0
     \]
     \[
     = \frac{1}{6} \left( \frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + 0 \right)
     \]
     \[
     = \frac{1}{6} \cdot \frac{15}{6} = \frac{15}{36} = \frac{5}{12}
     \]

3. **Calculate the expected number of points**:
   - Since there are 12 comparisons (from the 2nd roll to the 13th roll), and each comparison has an expected value of \(\frac{5}{12}\) points:
     \[
     E(\text{total points}) = 12 \cdot \frac{5}{12} = 5
     \]

The final answer is \(\boxed{5}\).