Question

The World Series, the annual championship series of Major League Baseball (MLB) in the US and Canada, is a seven-game series. The team that wins first four games is declared the champion. We assume that (i) each game is independent and that (ii) two teams are evenly matched.

 

a.    Calculate that the series last 4 games.

b.    Calculate that the series last 5 games.

c.    Calculate that the series last 6 games.

d.    Calculate that the series last 7 games.

Answer 1

Since two teams are evenly matched, let their be two teams, Team A and Team B. Each team has an equal probability of winning the game. The probability that team A wins is $p(A \space wins)=1/2$ and probability that team B wins is $p(B \space wins)=1/2$.

a.

The probability that the series lasts for 4 games is calculated as follows.

In order for the series to end after four games, one team must win the first four games in a row.

Now,

$p(A\space wins)=1/2*1/2*1/2*1/2=(1/2)^4=1/16$

$p(B\space wins)=1/2*1/2*1/2*1/2=(1/2)^4=1/16$

If the series lasts four games then either Team A or Team B wins and the probability is given by,

$p(series\space lasts \space4 \space games)=p(A\space wins)+p(B\space wins)=1/16+1/16=2/16=0.125$

Therefore, the probability that the series lasts four games is 0.125

b.

To find the probability that the series last five games, we proceed as follows.

In order for the series to end in 5 games, one team must win exactly 3 games out of the first four games not following any order and then win the fifth game.

Given that the fifth game has already be won, there remains 4 games of which 3 games must be won by either teams hence there are $\binom{4}{3}=4ways$ to determine the winnings.

Probability that team A wins is given by,

$p(A\space wins)=4*(1/2)^5=4/32$

And for team B

$p(B \space wins)=4*(1/2)^5=4/32$

Thus, probability that the series lasts 5 games is 4/32+4/32=8/32=1/4=0.25.

$p(series\space lasts\space 5\space games)=0.25$

c.

For the series to end in 6 games, one team must win exactly 3 out of the first 5 games in any order and then win the sixth game. The number of ways to determine the number of winnings for both teams in the first 5 games is given by, $\binom{5}{3}=10 ways$.

So the probability that team A and B wins respectively is,

$p(A \space wins)=10*(1/2)^6=10*1/64=10/64=5/32$

$p(B \space wins)=10*(1/2)^6=10/64=5/32$

Therefore,

$p(series \space lasts\space 6\space games )=p(A\space wins)+p(b\space wins)=5/32+5/16=10/32=0.3125$

Probability that the series lasts 6 games is 0.3125.

d.

For the series to last 7 games, one team must win exactly 3 out of the first 6 games(in any order) and then win the seventh game. To determine the winnings in the first 6 games, it can be done in $\binom{6}{3}=20ways.$ The probability that either team A or B wins respectively is given by,

$p(A\space wins)=20*(1/2)^7=20/128$

$p(B\space wins)=20*(1/2)^7=20/128$

$p(series\space lasts \space 7\space games)=p(A\space wins)+p(B\space wins)=2*(20/128)=0.3125$

Therefore, probability that the series lasts 7 games is 0.3125.