In a World Series, two teams play each other in at least four and at most seven games. The first team to win four games is the winner of the World Series. Assuming that both teams are equally matched, what is the probability that a World Series will be one (a) in four games? (b) in five games? (c) in six games? (d) in seven games? Explain.
Be "equally matched" I will assume
P(win) = 1/2 = prob(loss)
winning in 4 games = (1/2)^4 = 1/16
in 5 games, has to lose once)
LWWWW
WLWWW
WWLWW
WWWLW ----- each of these has a prob of (1/2)^5
but there are 4 cases,
so prob(5games) = 4/32 = 1/8
6 games , 2 losses, 4 wins
number of ways = 6!/(2!4!) = 15 , but that includes the case of ending with a loss, which can't happen
so number of 6 games is 14
prob(6games) = 14/(1/2)^6 = 14/64 = 7/32
7 games, 4 wins, 3 losses
number of cases = 7!/(3!4!) = 35 , less the case of ending in L
we have 34 such cases
prob(7 games) = 34/(1/2)^7 = 34/128 = 17/64
To calculate the probabilities of different outcomes in a World Series, we need to understand the number of possible ways the series can play out.
In a best-of-seven series, there are a total of 4 potential outcomes:
1. One team wins in 4 games (4-0)
2. One team wins in 5 games (4-1)
3. One team wins in 6 games (4-2)
4. One team wins in 7 games (4-3)
Since both teams are equally matched, we assume the probability of either team winning a single game is 0.5 (50%).
Now let's calculate the probabilities for each scenario:
(a) Probability of World Series ending in 4 games (4-0):
In this case, the first team must win the first 4 games in a row.
The probability of this happening is: 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 or 6.25%
(b) Probability of World Series ending in 5 games (4-1):
In this scenario, the series could play out in either of these two ways:
- The first team wins games 1-4, and the second team wins game 5.
- The second team wins game 1, and the first team wins games 2-5.
The probability of either of these sequences happening is:
- (0.5 * 0.5 * 0.5 * 0.5 * 0.5) + (0.5 * 0.5 * 0.5 * 0.5 * 0.5) = 0.15625 or 15.63%
(c) Probability of World Series ending in 6 games (4-2):
Similarly, we have two possible sequences:
- The first team wins games 1-4, loses game 5, and wins game 6.
- The second team wins game 1, wins games 2-4, loses game 5, and loses game 6.
The probability of either of these sequences happening is:
- (0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5) + (0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5) = 0.09375 or 9.38%
(d) Probability of World Series ending in 7 games (4-3):
Finally, we have two potential sequences:
- The first team wins games 1-4, loses games 5 and 6, and wins game 7.
- The second team wins game 1, loses games 2-4, wins game 5, loses game 6, and wins game 7.
The probability of either of these sequences happening is:
- (0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5) + (0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5 * 0.5) = 0.03125 or 3.13%
Therefore, the probabilities are as follows:
(a) 6.25%
(b) 15.63%
(c) 9.38%
(d) 3.13%