Solving the Probability of 7 Games Played when i = 4 and Proving that the Probability is Maximized when p = 1/2

 Probability is a fundamental concept in mathematics that helps us understand the likelihood of an event occurring. In this blog post, we will solve the probability of a total of 7 games played when i = 4 and prove that the probability is maximized when p = 1/2.

Suppose two teams, A and B, play a series of games until one of them wins four games. Each team has a probability p of winning any individual game, and the outcomes of the games are independent. If i = 4, find the probability that a total of 7 games are played.

To solve this problem, we need to consider all possible game sequences that could lead to a total of 7 games played. We can use the combinatorial method to count the number of ways to win four games out of seven games played.

The number of ways to win four games out of seven games played can be calculated using the binomial coefficient formula:

C(7,4) = 7! / (4! * 3!) = 35

Where C(7,4) represents the number of ways to choose four games out of seven games played.

The probability of winning four games out of seven games played can be calculated using the binomial probability formula:

P(X=4) = C(7,4) * p^4 * (1-p)^3

Where X is the random variable representing the number of games won out of seven games played, p is the probability of winning any individual game, and (1-p) is the probability of losing any individual game.

Substituting i = 4 and simplifying, we get:

P(X=4) = 35 * p^4 * (1-p)^3

To find the probability that a total of 7 games are played, we need to consider all possible game sequences that could lead to a total of 7 games played. There are two possible scenarios:

Scenario 1: The first team wins four games and the second team wins three games. The number of ways to achieve this outcome is given by:

C(7,4) * p^4 * (1-p)^3 * p^3 * (1-p)^4

Simplifying, we get:

C(7,4) * p^7 * (1-p)^7

Scenario 2: The second team wins four games and the first team wins three games. The number of ways to achieve this outcome is given by:

C(7,4) * p^3 * (1-p)^4 * p^4 * (1-p)^3

Simplifying, we get:

C(7,4) * p^7 * (1-p)^7

Therefore, the probability that a total of 7 games are played is:

P(X=7) = C(7,4) * p^7 * (1-p)^7 + C(7,4) * p^7 * (1-p)^7

Simplifying, we get:

P(X=7) = 2 * C(7,4) * p^7 * (1-p)^7

To prove that the probability is maximized when p = 1/2, we can take the derivative of the probability function with respect to p and set it equal to zero:

d/dp [2 * C(7,4) * p^7 * (1-p)^7] = 0