Ryan stands on the bottom-left square of a 2017 by 2017 grid of squares, where each square is colored either black, gray, or white according to the pattern as depicted to the right. Each second he moves either one square up, one square to the right, or both one up and to the right, selecting between these three options uniformly and independently. Noting that he begins on a black square, find the probability that Ryan is still on a black square after 2017 seconds.
[center][img]http://i.imgur.com/WNp59XW.png[/img][/center]
1. **Understanding the Grid and Movement:**
   - Ryan starts at the bottom-left corner of a 2017 by 2017 grid.
   - Each square is colored black, gray, or white in a repeating pattern based on the sum of its coordinates modulo 3.
   - Ryan can move up, right, or diagonally up-right each second.

2. **Color Pattern Analysis:**
   - The color of a square at coordinates \((x, y)\) is determined by \((x + y) \mod 3\):
     - Black if \((x + y) \equiv 0 \mod 3\)
     - Gray if \((x + y) \equiv 1 \mod 3\)
     - White if \((x + y) \equiv 2 \mod 3\)
   - Ryan starts at \((0, 0)\), which is black since \(0 + 0 \equiv 0 \mod 3\).

3. **Movement and Coordinate Sum:**
   - Each move changes the sum of the coordinates:
     - Moving up or right increases the sum by 1.
     - Moving diagonally up-right increases the sum by 2.
   - The probability of each move:
     - Up or right: \(\frac{2}{3}\)
     - Diagonally up-right: \(\frac{1}{3}\)

4. **Sum of Coordinates After 2017 Moves:**
   - Let \(n\) be the number of moves that increase the sum by 1.
   - Let \(2017 - n\) be the number of moves that increase the sum by 2.
   - The total sum after 2017 moves is \(n + 2(2017 - n) = 4034 - n\).

5. **Condition for Ending on a Black Square:**
   - Ryan ends on a black square if \(4034 - n \equiv 0 \mod 3\).
   - Simplifying, \(n \equiv 2 \mod 3\).

6. **Probability Calculation:**
   - We need to find the probability that \(n \equiv 2 \mod 3\).
   - The number of valid \(n\) values is \(\sum_{i=0}^{671} \binom{2017}{3i+2}\).
   - The probability of each sequence of moves is \(\left(\frac{2}{3}\right)^n \left(\frac{1}{3}\right)^{2017-n}\).

7. **Using Binomial Theorem and Roots of Unity Filter:**
   - The sum can be computed using the binomial theorem and roots of unity filter.
   - The result is close to \(\frac{1}{3}\).

The final answer is \(\boxed{\frac{1}{3}}\).