Problem 72 - A First Course in Probability

Question

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the teams is stronger than the other and wins each game with probability \(.6,\) independently of the outcomes of the other games. Find the probability, for \(i=4,5,6,7,\) that the stronger team wins the series in exactly \(i\) games. Compare the probability that the stronger team wins with the probability that it would win a 2-outof-3 series.

Step-by-Step Solution

Answer

The stronger team has a higher probability of winning a best-of-7 series (0.984384) compared to winning a 2-out-of-3 series (0.84). The probabilities that the stronger team wins the series in exactly 4, 5, 6, or 7 games are 0.1296, 0.3456, 0.31104, and 0.198144, respectively.

Step 1: Considering the case when stronger team wins in exactly 4 games

In order for the stronger team to win in exactly 4 games, it must win all the first 4 games. The probability of this happening is: \( P(X=4) = (0.6)^4 = 0.1296 \)

Step 2: Considering the case when stronger team wins in exactly 5 games

In order for the stronger team to win in exactly 5 games, it must win 3 of the first 4 games and then win the 5th game. The probability of this happening is: \( P(X=5) = \binom{4}{3} (0.6)^3 (0.4) (0.6) = 0.3456 \)

Step 3: Considering the case when stronger team wins in exactly 6 games

In order for the stronger team to win in exactly 6 games, it must win 3 of the first 5 games and then win the 6th game. The probability of this happening is: \( P(X=6) = \binom{5}{3} (0.6)^3 (0.4)^2 (0.6) = 0.31104 \)

Step 4: Considering the case when stronger team wins in exactly 7 games

In order for the stronger team to win in exactly 7 games, it must win 3 of the first 6 games and then win the 7th game. The probability of this happening is: \( P(X=7) = \binom{6}{3} (0.6)^3 (0.4)^3 (0.6) = 0.198144 \)

Step 5: Calculating the probability of winning the best-of-7 series

To find the overall probability that the stronger team wins the best-of-7 series, we sum the probabilities of winning in exactly 4, 5, 6, and 7 games: \( P(\text{winning best-of-7}) = P(X=4) + P(X=5) + P(X=6) + P(X=7) = 0.1296 + 0.3456 + 0.31104 + 0.198144 = 0.984384 \)

Step 6: Calculating the probability of winning a 2-out-of-3 series

In order to compare the scenario of winning a best-of-7 series to a 2-out-of-3 series, we need to calculate the probability of winning a 2-out-of-3 series: Considering the cases when the stronger team wins in exactly 2 games and in exactly 3 games: \( P(X=2) = (0.6)^2 = 0.36 \) \( P(X=3) = \binom{2}{1}(0.6)^2 (0.4) = 0.48 \) \( P(\text{winning 2-out-of-3}) = P(X=2) + P(X=3) = 0.36 + 0.48 = 0.84 \)

Step 7: Comparing both probabilites

Now that we have calculated the probability that the stronger team wins a best-of-7 series and a 2-out-of-3 series, we can compare these values: \( P(\text{winning best-of-7}) = 0.984384 \) \( P(\text{winning 2-out-of-3}) = 0.84 \) So, it can be observed that the stronger team has a higher probability of winning a best-of-7 series (0.984384) compared to winning a 2-out-of-3 series (0.84).

Key Concepts

Binomial Probability Distribution

When we talk about the probability of winning a series, especially in scenarios with independent trials like athletic games, we often make use of a particular kind of probability distribution known as the binomial probability distribution. This type of distribution helps us to determine the probability of obtaining a fixed number of successful outcomes in a specific number of trials.

Under this distribution, the number of games won by the stronger team, given fixed probabilities for winning individual games, fits perfectly. Here's why: first, we have a sequence of independent events (each game being won or lost), and each event has only two possible outcomes (win or loss). Just as importantly, the probability of winning remains consistent across games.

Moreover, this distribution makes calculating the likelihood of winning an exact number of games straightforward. For instance, if the team needs to win 4 out of 7 games, we use the binomial formula which combines the probability of winning individual games and the number of ways in which these wins can be arranged.

Combinatorics in Probability

Combinatorics plays a critical role when it comes to calculating probabilities in a series of events. It's essentially the mathematics of counting without having to count each arrangement individually. This becomes essential in our exercise problem where we calculate the probability of winning in exactly 5, 6, or 7 games.

To compute these probabilities, we use a concept called the binomial coefficient, commonly written as \( \binom{n}{k} \), representing the number of ways to choose \( k \) successes out of \( n \) trials. This concept is pivotal because it allows us to account for all possible win combinations that lead to winning the series in a certain number of games. For example, if the team wins in exactly 6 games, there are \( \binom{5}{3} \) ways to distribute these 3 wins among the first 5 games, followed by a win in the 6th game.

Independent Events in Probability

We consider the wins and losses of the games to be independent events in probability. Essentially, the outcome of one game does not influence the outcome of another. This characteristic is foundational when determining the probabilities using the binomial distribution.

In our exercise, it's stated that each game is won with a probability of 0.6, independently of the other games. This independence is crucial for the calculations, as it allows us to raise the probability to the power of the number of wins needed and multiply this by the probability of losses raised to the power of the number of losses.

Understanding the concept of independent events is not only important for correctly applying the binomial formula but also for grasping why certain probabilities change (or not) in different scenarios. For instance, whether the series is best-of-7 or a 2-out-of-3, the fact that events are independent underlies the entire probability calculation process.