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 probability of the stronger team winning in exactly 4, 5, 6, or 7 games can be found using the binomial probability formula for each case (i.e., exactly i-1 successes out of the first i-1 games). The probabilities for winning in exactly 4, 5, 6, or 7 games are 0.1296, 0.20736, 0.20736, and 0.165888, respectively. The total probability of winning the best-of-7 series is 0.710208. In comparison, the probability of winning a best-of-3 series is 0.648. Thus, the stronger team has a higher probability of winning a best-of-7 series compared to a best-of-3 series.

Step 1: Calculate the Probability of Winning in Exactly 4 Games

For the stronger team to win in exactly 4 games, they must win all 4 games. We can use the binomial probability formula: \(P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\) In this case, n = 4 (total number of games), k = 4 (number of games won), and p = 0.6 (probability of winning each game). Substituting these values: \(P(X = 4) = \binom{4}{4}(0.6)^4(1-0.6)^{4-4} = 1(0.6)^4(0.4)^0 = (0.6)^4 = 0.1296\)

Step 2: Calculate the Probability of Winning in Exactly 5 Games

For the stronger team to win in exactly 5 games, they must win 3 of the first 4 games and then win the last game. We can use the binomial probability formula again: \(P(X = 3) = \binom{4}{3}(0.6)^3(1-0.6)^{4-3} = 4(0.6)^3(0.4)^1 = 4(0.216)(0.4) = 0.3456\) Since they need to win the last game as well, we multiply this probability by the probability of winning the last game (0.6): \(0.3456 * 0.6 = 0.20736\)

Step 3: Calculate the Probability of Winning in Exactly 6 Games

For the stronger team to win in exactly 6 games, they must win 3 of the first 5 games and then win the last game. We can use the binomial probability formula again: \(P(X = 3) = \binom{5}{3}(0.6)^3(1-0.6)^{5-3} = 10(0.6)^3(0.4)^2 = 10(0.216)(0.16) = 0.3456\) Since they need to win the last game as well, we multiply this probability by the probability of winning the last game (0.6): \(0.3456 * 0.6 = 0.20736\)

Step 4: Calculate the Probability of Winning in Exactly 7 Games

For the stronger team to win in exactly 7 games, they must win 3 of the first 6 games and then win the last game. We can use the binomial probability formula again: \(P(X = 3) = \binom{6}{3}(0.6)^3(1-0.6)^{6-3} = 20(0.6)^3(0.4)^3 = 20(0.216)(0.064) = 0.27648\) Since they need to win the last game as well, we multiply this probability by the probability of winning the last game (0.6): \(0.27648 * 0.6 = 0.165888\)

Step 5: Calculate the Total Probability of Winning the Series

To find the total probability of winning the series, we add the probabilities from steps 1-4: \(0.1296 + 0.20736 + 0.20736 + 0.165888 = 0.710208\)

Step 6: Compare with the Probability of Winning a Best-of-3 Series

In a best-of-3 series, the stronger team must win 2 of the 3 games. Using the binomial probability formula: \(P(X = 2) = \binom{3}{2}(0.6)^2(1-0.6)^{3-2} = 3(0.6)^2(0.4)^1 = 3(0.36)(0.4) = 0.432\) However, we need to add the probability that the stronger team wins all 3 games (total = 3): \(P(X = 3) = \binom{3}{3}(0.6)^3(1-0.6)^{3-3} = 1(0.6)^3(0.4)^0 = (0.6)^3 = 0.216\) Adding the probabilities for winning 2 games and 3 games: \(0.432 + 0.216 = 0.648\)

Step 7: Conclusion

The probability that the stronger team wins a best-of-7 series is 0.710208, while the probability that they win a best-of-3 series is 0.648. Thus, the stronger team has a higher probability of winning a best-of-7 series compared to a best-of-3 series.

Key Concepts

Binomial Probability Formula

The Binomial Probability Formula is a fundamental concept in probability theory. It is used to calculate the probability of a given number of successes in a fixed number of independent trials, with the same chance of success on each trial. This formula can be expressed as \(P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\), where:

• \(n\) is the total number of trials,
• \(k\) is the number of successful trials,
• \(p\) is the probability of success on each trial, and
• \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) successes from \(n\) trials.

This formula is widely used in various fields to make decisions based on probabilistic outcomes. In the context of the athletic teams' series, it helps in computing the likelihood that the stronger team wins a specific number of games in a series.

Series of Games

A series of games, such as best-of-7 or best-of-3, often serves as a deciding factor in tournaments. The win condition for a series of games involves achieving a certain number of victories. This could range from winning 4 out of 7 games to 2 out of 3 games, depending on the series type.
The strategy employed involves analyzing each possible scenario where the stronger team could win the series. For example, winning the series in exactly 4 games means winning straight 4 games without losing. Similarly, analyzing for 5, 6, and 7 games involves calculating the chances for different combinations of wins and losses that still lead to a series win.

Probabilistic Comparisons

Probabilistic comparisons involve evaluating different scenarios to determine which outcome is more likely. In sports or any scenario involving a series of events, comparing probabilities can help determine the most advantageous approach.
Consider the athletic teams' series: Not only does calculating the probability of winning in 4, 5, 6, or 7 games help forecast the outcome, but comparing the odds with a shorter series, like best-of-3, offers insights. For instance, the stronger team has a probability of 0.710208 of winning a best-of-7 series, whereas the probability for a best-of-3 series is 0.648. The comparison shows a higher chance of a series victory in the longer series. This form of analysis enables better strategic decisions when planning and predicting outcomes in competitions.

Independence of Events

In probability theory, the independence of events is a key assumption that significantly simplifies analysis. Two events are considered independent if the outcome of one does not affect the outcome of the other. When it comes to a series of games, each game is treated as an independent event, assuming past wins or losses do not impact future games.
In the given example, the probability of the stronger team winning any single game remains consistent (0.6), regardless of previous results. This assumption is crucial because it underpins the use of the binomial probability formula, ensuring calculations for each potential series outcome remain accurate. Understanding this concept helps in recognizing when this type of analysis is applicable and when external factors might need accounting, possibly rendering the events dependent.