Mike and Alain play a game in which each player is equally likely to win. The first player to win three games becomes the champion, and no further games are played. If Mike has won the first game, what is the probability that Mike becomes the champion?
(A) $\frac{1}{4}$
(B) $\frac{5}{8}$
(C) $\frac{11}{16}$
(D) $\frac{3}{5}$
(E) $\frac{3}{4}$
We introduce the letter M to represent a game that Mike has won, and the letter A to represent a game that Alain has won.

We construct a tree diagram to show all possible outcomes.

The $\mathrm{M}$ at the far left of the tree represents the fact that Mike won the first game.

The second "column" shows the two possible outcomes for the second game - Mike could win (M), or Alain could win (A).

The third, fourth and fifth columns show the possible outcomes of games 3,4 and 5 , respectively.

A branch of the tree is linked by arrows and each branch gives the

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-144.jpg?height=442&width=418&top_left_y=1676&top_left_x=1515)
outcomes of the games which lead to one of the players becoming the champion.

Once Mike has won 3 games, or Alain has won 3 games, the branch ends and the outcome of that final game is circled.

Since the first player to win 3 games becomes the champion, one way that Mike could become the champion is to win the second and third games (since he has already won the first game). We represent this possibility as MMM, as shown along the top branch of the tree diagram. In this MMM possibility, the final $\mathrm{M}$ is circled in the diagram, meaning that Mike has become the champion.

Since we are asked to determine the probability that Mike becomes the champion, we search
for all branches through the tree diagram which contain $3 \mathrm{Ms}$ (paths ending with a circled M). The tree diagram shows six such branches: MMM, MMAM, MMAAM, MAMM, MAMAM, and MAAMM.

All other branches end with a circled A, meaning that Alain has won 3 games and becomes the champion.

Since each player is equally likely to win a game, then Mike wins a game with probability $\frac{1}{2}$, and Alain wins a game with probability $\frac{1}{2}$.

Of the six ways that Mike can win (listed above), only one of these ends after 3 games (MMM). The probability that Mike wins in exactly 3 games is equal to the probability that Mike wins the second game, which is $\frac{1}{2}$, multiplied by the probability that Mike wins the third game, which is also $\frac{1}{2}$.

That is, the probability that Mike becomes the champion by winning the first 3 games is $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$.

Of the six ways that Mike can win, two of these end after 4 games (MMAM, MAMM).

The probability that Mike wins games two and four but Alain wins game three (MMAM) is equal to the probability that Mike wins the second game, which is $\frac{1}{2}$, multiplied by the probability that Alain wins the third game, which is also $\frac{1}{2}$, multiplied by probability that Mike wins the fourth game, $\frac{1}{2}$.

In this case, Mike becomes the champion with probability $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}$.

Similary, the probability that Mike becomes the champion by winning games three and four, but loses game two, is also $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}$.

Finally, we determine the probabilty that Mike becomes the champion by winning in exactly 5 games (there are three possibilities: MMAAM, MAMAM and MAAMM).

Each of these three possibilities happens with probability $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{16}$.

Therefore, if Mike has won the first game, then the probability that he becomes champion is $\frac{1}{4}+2 \times \frac{1}{8}+3 \times \frac{1}{16}=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}=\frac{4+4+3}{16}=\frac{11}{16}$.

ANSWER: (C)