Problem 22 - A First Course in Probability

Question

Suppose that two teams play a series of games that ends when one of them has won \(i\) games. Suppose that each game played is, independently, won by team \(A\) with probability \(p .\) Find the expected number of games that are played when (a) \(i=2\) and (b) \(i=3 .\) Also, show in both cases that this number is maximized when \(p=\frac{1}{2}\).

Step-by-Step Solution

Answer

For i=2, the expected number of games played (E) is given by \(E = 3(p^2q + q^2p) + 4(p^2q^2 * 2)\). By differentiating this expression with respect to p and equating it to 0, we find that the expected number of games is maximized when \(p = \frac{1}{2}\). For i=3, the expected number of games played (E) is given by \(E = 5(p^3q^2 + q^3p^2) + 6(p^3q^3 * 3)\). By differentiating this expression with respect to p and equating it to 0, we find that the expected number of games is maximized when \(p = \frac{1}{2}\). In both cases (i=2 and i=3), the expected number of games played is maximized when p=1/2, indicating that an equal probability of winning for both teams leads to the longest series of games.

Step 1: Probabilities of teams A and B winning a match and ending the series

Let's denote the probability of team A winning a game by p and the probability of team B winning a game as q = 1 - p since the outcomes of each game are mutually exclusive.

Step 2: a) Expected number of games when i=2

We proceed by finding the probability distributions for the possible number of games played for i = 2:#tag_list# 3 games: The series ends with a 2-0 or a 0-2 score. The probability of reaching this outcome is p^2 * q + q^2 * p, since there are 2 possibilities where the games can end (either in favor of team A or team B). 4 games: The series ends with a 2-1 or a 1-2 score. The probability of reaching this outcome is p^2 * q^2 * 2, since the game could end with either an extra win for team A or an extra win for team B.

Step 3: Expected value calculation for i=2

Now, we can find the expected number of games played (E) when i = 2 using the probabilities calculated above: \(E = 3(p^2q + q^2p) + 4(p^2q^2 * 2)\) To find the value of p that maximizes the number of games played, we differentiate the above expression with respect to p and equate it to 0: \( \frac{dE}{dp} = 0 \) Solving this equation, we get: \( p = \frac{1}{2} \)

Step 4: b) Expected number of games when i=3

We proceed by finding the probability distributions for the possible number of games played for i = 3:#tag_list# 5 games: The series ends with a 3-0 or a 0-3 score. The probability of reaching this outcome is p^3 * q^2 + q^3 * p^2, since there are 2 possibilities where the games can end (either in favor of team A or team B). 6 games: The series ends with a 3-1 or a 1-3 score. The probability of reaching this outcome is p^3 * q^3 * 3, since the game could end with 3 different combinations of wins and losses for both teams.

Step 5: Expected value calculation for i=3

Now, we can find the expected number of games played (E) when i = 3 using the probabilities calculated above: \(E = 5(p^3q^2 + q^3p^2) + 6(p^3q^3 * 3)\) To find the value of p that maximizes the number of games played, we differentiate the above expression with respect to p and equate it to 0: \( \frac{dE}{dp} = 0 \) Solving this equation, we get: \( p = \frac{1}{2} \) In both cases, when i=2 and i=3, the expected number of games played is maximized when p = 1/2, indicating that an equal probability of winning for both teams leads to the longest series of games.

Key Concepts

Understanding Probability Theory

Probability theory is a branch of mathematics concerned with analyzing random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non-deterministic events or measured quantities that may either happen or not happen.

For instance, when two teams play a series of games, the outcome of each game can be considered a random event. If we say that team A wins with probability 'p', this is an expression of the likelihood of that event occurring. Similarly, the probability 'q' for team B equals 1-p, since both outcomes are mutually exclusive and all probabilities must sum up to 1. This balances out the possible outcomes for each and every game they play.

Probability theory helps us quantify the uncertainty and make predictions about the average behavior over many repetitions of the game. This concept is crucial in calculating expected values, which predict the number of games likely to be played, and understanding distributions, such as the binomial distribution used to model the number of successes in a series of independent experiments.

Expected Value Calculation in Action

The expected value, or mathematical expectation, combines probabilities and impacts to give a long-term average outcome of random events. It's like a weighted average for random variables, where each possible outcome is weighted by its probability of occurrence.

Here's how it works in the context of our games example: If we take all possible outcomes of the series of games, and multiply each by the probability of its occurrence, and then sum all these products together, we get the expected value. It tells us the average number of games that will be played in the long run.

This methodology was applied in the solution. For example, when team A needs to win 'i' games, the expected number of games 'E' is calculated using the respective probabilities of outcomes. The exercise illustrates how the formula for 'E' changes depending on the number of games needed to win the series. It also demonstrates that the most games are expected to be played when the two teams are equally matched (i.e., when p equals 0.5), showcasing the relationship between probabilities and the expected value.

Binomial Distribution's Role

The binomial distribution is a probability distribution that summarizes the likelihood that a value will take on one of two independent values under a given number of observations or trials. It is an appropriate model when we're dealing with two possible outcomes in each trial, such as a win or loss in a game.

In the example involving teams A and B, the binomial distribution answers questions like, 'What is the probability that team A will win 'i' games out of 'n' played?' This is based on the fixed probability 'p' of team A winning a single game. As the series of games progresses, we are essentially looking at different outcomes that follow a binomial distribution.

The exercise provided applies the principles of binomial distribution to calculate the expected number of games. By understanding that each game played is an independent event with two outcomes—win or loss—it's evident why the solution utilizes binomial probabilities to derive expectations over a range of possible game scenarios.