Problem 78 - A First Course in Probability

Question

\(A\) and \(B\) play a series of games. Each game is independently won by \(A\) with probability \(p\) and by \(B\) with probability \(1-p .\) They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series. (a) Find the probability that a total of 4 games are played. (b) Find the probability that \(A\) is the winner of the series.

Step-by-Step Solution

Answer

The probability of a total of 4 games played is \(2p^2(1-p)\), and the probability that A is the winner of the series is \(3p^2(1-p)\).

Step 1: Case 1: AABB

The probability of this outcome is given by p^2 (1-p)^2, as we have two games won by A, and two by B.

Step 2: Case 2: BBAA

The probability of this outcome is given by p^2 (1-p)^2 just like in Case 1, as we have two games won by A, and two by B.

Step 3: Case 3: AABA

In this outcome, A wins the first and third games, and B wins the second game. The probability of this case is p^2(1-p).

Step 4: Case 4: BAAB

Similar to Case 3, A wins two games, and B wins one game. The probability of this case is p^2(1-p).

Step 5: Total Probability for 4 games

Adding the probabilities of all four cases, we get the total probability for 4 games: \(2p^2(1-p)^2 + 2p^2(1-p) = 2p^2(1-p)(1 + (1-p)) = 2p^2(1-p) \) (b) Probability that A is the winner of the series

Step 6: Case 1: A wins in 3 games

A must win the first two games in a row, and then B can win the third game. The probability of this case is: \(p^2(1-p)\).

Step 7: Case 2: A wins in 4+ games

We have already calculated the probability of a total of 4 games played: \(2p^2(1-p)\). Among these, A is the winner in cases AABA and BAAB. The probability of A winning in 4 games is: \(2p^2(1-p)\)

Step 8: Total Probability for A to win the series

Adding the probabilities of both cases, we get the total probability of A winning the series: \(p^2(1-p) + 2p^2(1-p) = 3p^2(1-p)\)

Key Concepts

Independent Events

In probability theory, independent events are events that do not influence each other. This means the outcome of one event does not affect the outcome of another. For example, when you flip a coin multiple times, each flip does not affect the others.
In terms of our game scenario, each game played by players A and B is independent. The result of one game does not impact the result of the next game. This independence allows us to calculate the probability of various sequences of wins and losses without needing to consider previous game outcomes.

• If A wins a game with probability \( p \), this probability remains constant every time they play, regardless of past wins or losses.
• Independent events simplify computations by allowing probabilities to be multiplied across events, which is crucial in determining sequence probabilities, such as AABA or AABB.

Winning Series

A winning series requires a player to win more games than their opponent by a specified margin. In our scenario, the series ends when a player's total wins exceed the other's by two. This requirement ensures that one player decisively wins the series rather than just leading momentarily.
Consider the series between players A and B:

• The series will continue until, for instance, A has won two more games than B or vice versa.
• This means that one of the players has consistently pulled ahead, demonstrating dominance in the games.
• The condition impacts the strategy as players aim to secure the necessary margin to win.

Probability Calculation

Calculating the probability of various game outcomes relies on understanding the various sequences of wins and losses. Probability calculations require assessing all feasible sequences of independent game results and summing their probabilities.
The calculations often involve:

• Identifying all potential sequences that lead to a certain outcome, such as winning in exactly 4 games.
• Using the probabilities \( p \) and \( 1-p \) for players A and B to win, respectively, to compute the probability of each sequence.
• Combining probabilities from different sequences that yield the same final outcome.

For instance, using our steps:- Cases AABB and BBAA have probabilities calculated as \( p^2(1-p)^2 \).- Cases AABA and BAAB are calculated as \( p^2(1-p) \).- These are then summed to find the total probability, highlighting the importance of considering all possible sequences to get the accurate probability of an event.

Game Outcomes

Game outcomes in this context refer to the different scenarios of ending the series. We focused on scenarios where a total of four games are played and where A wins the series.

• Outcomes in four games involve both players winning different games in sequences like AABB or AABA. Some players win exactly two games before others can catch up, which allows the series to continue.
• Player A is the series winner if they reach the required win margin first. This could happen in less than or more than four games.
• The step-by-step approach shows that by calculating possible game sequences and their probabilities, one can predict outcomes like A winning in three or four games with confidence.

Understanding these outcomes helps in grasping how probability can predict the likelihood of different endings to a series, based on known probabilistic foundations.