Gary and Deep play a game in which there are no ties. Each player is equally likely to win each game. The first player to win 4 games becomes the champion, and no further games are played. Gary wins the first two games. What is the probability that Deep becomes the champion?
Since the first player to win 4 games becomes the champion, then Gary and Deep play at most 7 games. (The maximum number of games comes when the two players have each won 3 games and then one player becomes the champion on the next (7th) game.)

We are told that Gary wins the first two games.

For Deep to become the champion, the two players must thus play 6 or 7 games, because Deep wins 4 games and loses at least 2 games. We note that Deep cannot lose 4 games, otherwise Gary would become the champion.

If Deep wins and the two players play a total of 6 games, then the sequence of wins must be GGDDDD. (Here D stands for a win by Deep and G stands for a win by Gary.)

If Deep wins and the two players play a total of 7 games, then Deep wins 4 of the last 5 matches and must win the last (7th) match since he is the champion.

Therefore, the sequence of wins must be GGGDDDD or GGDGDDD or GGDDGDD or GGDDDGD. (In other words, Gary can win the 3rd, 4th, 5th, or 6th game.)

The probability of the sequence GGDDDD occurring after Gary has won the first 2 games is $\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$. This is because the probability of a specific outcome in any specific game is $\frac{1}{2}$, because each player is equally likely to win each game, and there are 4 games with undetermined outcome.

Similarly, the probability of each of the sequences GGGDDDD, GGDGDDD, GGDDGDD, and GGDDDGD occurring is $\left(\frac{1}{2}\right)^{5}=\frac{1}{32}$.

Therefore, the probability that Gary wins the first two games and then Deep becomes the champion is $\frac{1}{16}+4 \cdot \frac{1}{32}=\frac{6}{32}=\frac{3}{16}$.

ANSWER: $\frac{3}{16}$