On the morning of September $30,1982$, the won–lost records of the three leading baseball teams in the Western Division of the National League were as follows:

Each team had $3$games remaining. All $3$of the Giants’ games were with the Dodgers, and the $3$remaining games of the Braves were against the San Diego Padres. Suppose that the outcomes of all remaining games are independent and each game is equally likely to be won by either participant. For each team, what is the probability that it will win the division title? If two teams tie for first place, they have a playoff game, which each team has an equal chance of winning.

Braves - $87$ wins

Giants - $86$ wins

Dodgers - $86$ wins

Events:

$B,G,D$- events that Braves, Giants, Dodgers have most wins in the division, respectively

$B_i$- event that the Braves won $i$of their $3$remaining games $i=0,1,2,3$

$G_i$- event that the Giants won $i$of their $3$remaining games $i=0,1,2,3$

If Giants won $i$games, Dodgers won $3-i$

Probabilities:

Since all three games are independent, and both sides are equally likely to win, each result has probability $\frac{1}{2}·\frac{1}{2}·\frac{1}{2}=\frac{1}{8}$ of occuring

$P\left(B_0\right)=P\left(B_3\right)=P\left(G_0\right)=P\left(G_3\right)=\frac{1}{8}$

$P\left(B_1\right)=P\left(B_2\right)=P\left(G_1\right)=P\left(G_2\right)=\frac{3}{8}$

The probabilites in the second row are greater because those events happen in three different ways regarding the order of wins/losses.

$P\left(B\right)$

Rewrite event $B$ using $B_0,B_1,B_2,B_3$

Since the Braves are in the lead if they win all three games they certainly win $-P\left(B{B}_3\right)=P\left(B_3\right)=\frac{1}{8}$

If the Braves win $0$ of the $3$remaining games, they will certainly lose the title, because one of the other teams will win at least two games -role="math" localid="1647918458782" $P\left(B{B}_0\right)=0$.

Now condition upon which is the result of the games between the other two teams

$P\left(B{B}_1\right)=P\left(B{B}_1{G}_0\right)+P\left(B{B}_1{G}_1\right)+P\left(B{B}_1{G}_2\right)+P\left(B{B}_1{G}_3\right)$

$=0+P\left(B\mid B_1{G}_1\right)P\left(B_1\right)P\left(G_1\right)+P\left(B\mid B_1{G}_2\right)P\left(B_1\right)P\left(G_2\right)+0$

$=\frac{1}{2}\frac{3}{8}\frac{3}{8}+\frac{1}{2}\frac{3}{8}\frac{3}{8}$

$=\frac{9}{64}$

In cases $B_1{G}_1$and $B_1{G}_2$, the Braves are tied with Dodgers and Giants, respectively, and the probability of winning is the probability of winning the playoffs $-1/2$.

$P\left(B{B}_2\right)$

localid="1647918820164" $P\left(B{B}_2\right)=P\left(B{B}_2{G}_0\right)+P\left(B{B}_2{G}_1\right)+P\left(B{B}_2{G}_2\right)+P\left(B{B}_2{G}_3\right)$

$=P\left(B\mid B_2{G}_0\right)P\left(B_2\right)P\left(G_0\right)+P\left(B\mid B_2{G}_1\right)P\left(B_2\right)P\left(G_1\right)+P\left(B\mid B_2{G}_2\right)P\left(B_2\right)P\left(G_2\right)+P\left(B\mid B_2{G}_3\right)P\left(B_2\right)P\left(G_3\right)$

$=\frac{1}{2}\frac{3}{8}\frac{1}{8}+1·\frac{3}{8}\frac{3}{8}+1·\frac{3}{8}\frac{3}{8}+\frac{1}{2}\frac{3}{8}\frac{1}{8}$

$=\frac{21}{64}$

Returning to the starting equation:

$P\left(B\right)=0+\frac{9}{64}+\frac{21}{64}+\frac{1}{8}=\frac{19}{32}$.

Since the Giants and the Dodgers are in identical positions (they play from the same result with probability$1/2$ of winning each of the next three games), they should have equal probability of winning the division title.

$P\left(G\right)=P\left(D\right)=\frac{13}{64}$

The condition upon the number of games won between the Giants and Dodgers, Braves and Padres.