Basic Probability Die Roll Game

I am not sure if I got the correct answers to these basic probability questions.

You and I play a die rolling game, with a fair die. The die is equally likely to land on any of its $6$ faces. We take turns rolling the die, as follows.

At each round, the player rolling the die wins (and the game stops) if (s)he rolls either "$3$", "$4$","$5$", or "$6$". Otherwise, the next player has a chance to roll.

1. Suppose I roll 1st. What is the probability that I win in the 1st round?


2. Suppose I roll in the 1st round, and we play the game for at most three rounds. What is the probability that
   a. I win?
   b. You win?
   c. No one wins? (i.e., I roll "$1$" or "$2$" in the 3rd round?)

My answers:

1. $4/6$


2. a. I think it is $(4/6)^2$ . Can someone explain why it isn't $4/6 + 4/6$ ?
   b. I think this is $4/6$.
   c. I think it is $(2/6)^3$

Answer

What is the probability that the player who rolls first wins in the first round?

Your answer of $4/6 = 2/3$ is correct.

What is the probability that the player who rolls first wins the game if the game lasts at most three rounds?

The player who rolls first if he or she wins during the first round, second round, or third round. We know that the probability that the player who rolls first has probability $2/3$ of winning in the first round.

For the player who rolls first to win in the second round, both players must roll a 1 or 2 in the first round, then the first player must roll a 3, 4, 5, or 6 in the second round. The probability that this event occurs is
$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{27}$$

For the first player to win in the third round, both players must roll a 1 or 2 for the first two rounds, then the first player must roll a 3, 4, 5, or 6 in the third round. The probability that this occurs is
$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{243}$$

Since these three events are mutually exclusive, the probability that the player who rolls first wins the game is
$$\frac{2}{3} + \frac{2}{27} + \frac{2}{243} = \frac{162}{243} + \frac{18}{243} + \frac{2}{243} = \frac{182}{243}$$

The probability that the first player wins cannot be
$$\frac{4}{6} + \frac{4}{6} = \frac{8}{6} = \frac{4}{3} > 1$$
since it is impossible for a probability to exceed $1$.

What is the probability that the player who rolls second wins the game if the game lasts at most three rounds.

The player who rolls second can win in the first round if the first player rolls a 1 or 2, then the second player rolls a 3, 4, 5, or 6. The probability that this event occurs is
$$\frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$$
For the second player to win in the second round, the first and second players must both roll a 1 or 2 in the first round, the first player must roll a 1 or 2 in the second round, then the second player must roll a 3, 4, 5, or 6 in the second round. The probability that this event occurs is
$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{81}$$
For the second player to win in the third round, both players must roll a 1 or 2 in the first two rounds, the first player must roll a 1 or 2 in the third round, then the second player must roll a 3, 4, 5, or 6 in the third round. The probability that this event occurs is

$$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{729}$$
Since these three events are mutually exclusive, the probability that the second player wins is
$$\frac{2}{9} + \frac{2}{81} + \frac{2}{729} = \frac{162}{729} + \frac{18}{729} + \frac{2}{729} = \frac{182}{729}$$

What is the probability that neither player wins a game that lasts at most three rounds?

For this event to occur, both players must roll a 1 or 2 in all three rounds. The probability that this event occurs is
$$\left(\frac{2}{6}\right)^6 = \left(\frac{1}{3}\right)^6 = \frac{1}{729}$$

Check: There are three possibilities: The first player wins, the second player wins, or neither player wins. Therefore, the probabilities of these three events should add up to $1$. Observe that
$$\frac{182}{243} + \frac{182}{729} + \frac{1}{729} = \frac{546}{729} + \frac{182}{729} + \frac{1}{729} = 1$$