this exercise requires an enhanced for loop, which is present. Running this in my ide gives the correct information.

public class ScrabblePlayer {
  // A String representing all of the tiles that this player has
  private String tiles;

  public void ScrabblePlayer() {
    tiles = "";
  }

  public String getTiles() {
    return tiles;
  }

  public void addTile(char tile) {
    tiles += tile;
  }

  public boolean hasTile(char tile) {
    return tiles.indexOf(tile) != -1;
  }

  public int getCountOfLetter(char letter) {
    int count = 0;
    char[] ch = new char[tiles.length()];
    ch = tiles.toCharArray();
    for(char c : ch){
      if(c == letter){
        count++;
      }
    }

    return count;
  }
}

// This code is here for example purposes only
public class Example {

  public static void main(String[] args) {
    ScrabblePlayer player1 = new ScrabblePlayer();
    player1.addTile('d');
    player1.addTile('d');
    player1.addTile('p');
    player1.addTile('e');
    player1.addTile('l');
    player1.addTile('u');

    ScrabblePlayer player2 = new ScrabblePlayer();
    player2.addTile('z');
    player2.addTile('z');
    player2.addTile('y');
    player2.addTile('f');
    player2.addTile('u');
    player2.addTile('z');

    int count = 0;
    // This would set count to 1 because player1 has 1 'p' tile in her collection of tiles
    count = player1.getCountOfLetter('p');
    // This would set count to 2 because player1 has 2 'd'' tiles in her collection of tiles
    count = player1.getCountOfLetter('d');
    // This would set 0, because there isn't an 'a' tile in player1's tiles
    count = player1.getCountOfLetter('a');

    // This will return 3 because player2 has 3 'z' tiles in his collection of tiles
    count = player2.getCountOfLetter('z');
    // This will return 1 because player2 has 1 'f' tiles in his collection of tiles
    count = player2.getCountOfLetter('f');
  }
}