On January 15 in the stormy town of Stormville, there is a $50\%$ chance of rain. Every day, the probability of it raining has a $50\%$ chance of being $\frac{2017}{2016}$ times that of the previous day (or $100\%$ if this new quantity is over $100\%$) and a $50\%$ chance of being $\frac{1007}{2016}$ times that of the previous day. What is the probability that it rains on January 20?

[i]2018 CCA Math Bonanza Lightning Round #3.3[/i]
1. **Initial Probability on January 15:**
   The probability of rain on January 15 is given as \( P_0 = 0.5 \).

2. **Daily Probability Adjustment:**
   Each day, the probability of rain can change in one of two ways:
   - With a \(50\%\) chance, the probability is multiplied by \(\frac{2017}{2016}\).
   - With a \(50\%\) chance, the probability is multiplied by \(\frac{1007}{2016}\).

3. **General Probability Expression:**
   Let \( P_n \) be the probability of rain on the \(n\)-th day. The probability on January 20 (5 days after January 15) can be expressed as:
   \[
   P_5 = P_0 \cdot \left( \frac{2017}{2016} \right)^k \cdot \left( \frac{1007}{2016} \right)^{5-k}
   \]
   where \( k \) is the number of times the probability is multiplied by \(\frac{2017}{2016}\) over the 5 days.

4. **Summing Over All Possible Paths:**
   The total probability of rain on January 20 is the sum of the probabilities for all possible values of \( k \) from 0 to 5, weighted by the number of ways each scenario can occur (binomial coefficient) and the probability of each scenario:
   \[
   P_5 = \sum_{k=0}^5 \binom{5}{k} \left( \frac{1}{2} \right)^5 \left( \frac{2017}{2016} \right)^k \left( \frac{1007}{2016} \right)^{5-k} \cdot P_0
   \]

5. **Simplifying the Expression:**
   Since \( P_0 = 0.5 \), we can factor this out:
   \[
   P_5 = 0.5 \cdot \sum_{k=0}^5 \binom{5}{k} \left( \frac{1}{2} \right)^5 \left( \frac{2017}{2016} \right)^k \left( \frac{1007}{2016} \right)^{5-k}
   \]

6. **Recognizing the Binomial Theorem:**
   The sum inside the expression is a binomial expansion:
   \[
   \sum_{k=0}^5 \binom{5}{k} \left( \frac{2017}{2016} \cdot \frac{1}{2} \right)^k \left( \frac{1007}{2016} \cdot \frac{1}{2} \right)^{5-k} = \left( \frac{2017}{4032} + \frac{1007}{4032} \right)^5
   \]
   Simplifying the terms inside the parentheses:
   \[
   \frac{2017}{4032} + \frac{1007}{4032} = \frac{3024}{4032} = \frac{3}{4}
   \]

7. **Final Calculation:**
   Thus, the expression simplifies to:
   \[
   P_5 = 0.5 \cdot \left( \frac{3}{4} \right)^5
   \]
   Calculating \(\left( \frac{3}{4} \right)^5\):
   \[
   \left( \frac{3}{4} \right)^5 = \frac{3^5}{4^5} = \frac{243}{1024}
   \]
   Therefore:
   \[
   P_5 = 0.5 \cdot \frac{243}{1024} = \frac{243}{2048}
   \]

The final answer is \(\boxed{\frac{243}{2048}}\).