Problem 948

We want to determine the probability that Carina wins 3 games
before she loses 2 games.

This means that she either wins 3 and loses 0, or wins 3 and loses
1.

If Carina wins her first three games, we do not need to consider the
case of Carina losing her fourth game, because we can stop after she
wins 3 games.

Putting this another way, once Carina has won her third game, the
outcomes of any later games do not affect the probability because wins
or losses at that stage will not affect the question that is being
asked.

Using W to represent a win and L to represent a loss, the possible
sequence of wins and losses that we need to examine are WWW, LWWW, WLWW,
and WWLW.

In the case of WWW, the probabilities of the specific outcome in each
of the three games are $\frac{1}{2}$,
$\frac{3}{4}$, $\frac{3}{4}$, because the probability of a
win after a win is $\frac{3}{4}$.

Therefore, the probability of WWW is $\$\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{32}\$.$

In the case of LWWW, the probabilities of the specific outcome in
each of the four games are $\frac{1}{2}$, $\frac{1}{3}$, $\frac{3}{4}$, $\frac{3}{4}$, because the probability of a
loss in the first game is $\frac{1}{2}$, the probability of a win
after a loss is $\frac{1}{3}$, and
the probability of a win after a win is $\frac{3}{4}$.

Therefore, the probability of LWWW is $\$\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{96} = \frac{3}{32}\$.$

Using similar arguments, the probability of WLWW is $\$\frac{1}{2} \times \frac{1}{4} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{96} = \frac{1}{32}\$.$

Here, we used the fact that the probability of a loss after a win is
$1 - \frac{3}{4} = \frac{1}{4}$.

Finally, the probability of WWLW is $\$\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4} \times \frac{1}{3} = \frac{3}{96} = \frac{1}{32}\$.$

Therefore, the probability that Carina wins 3 games before she loses
2 games is $\$\frac{9}{32} + \frac{3}{32} + \frac{1}{32} + \frac{1}{32} = \frac{14}{32} = \frac{7}{16}\$,$
which is in lowest terms.

The sum of the numerator and denominator of this fraction is 23.