A turn consists of rolling a standard die and tossing a fair coin. The game is won when the die shows a 1 or a 6 and the coin shows heads. What is the probability the game will be won before the fourth turn? Express your answer as a common fraction.
Let's put it into this way.
If a player wins game 1: (1/3)*(1/2)=1/6
If the player wins before the 4th turn: P(wins game 1)+P(lose game 1, wins game 2)+P(loses game 1 and 2, but wins game 3)
Player loses game 1 but wins game 2: (5/6)*(1/6)=5/36
Player loses game 1 and 2, but wins game 3: [(5/6)^2](1/6)=25/216 (very confusing, I'm sorry)
Player wins before 4th turn: 1/5+5/36+25/216=91/216