Question

Two teams A and B play a series of games until one team wins three games. We assume that the games are played independently and that the probability that A wins any game is p. What is the probability that the series lasts exactly four games?

Answer 1

The probability that the series lasts exactly four games is
`3p(1-p)(p^(2) + (1 - p)^(2))`

Explanation:

For each game, there are only two possible outcomes. Either team A wins, or team A loses. Games are played independently. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

We also need to know a small concept of independent events.

Independent events:

If two events, A and B, are independent, we have that:

`P(A \cap B) = P(A)*P(B)`

What is the probability that the series lasts exactly four games?

This happens if A wins in 4 games of B wins in 4 games.

Probability of A winning in exactly four games:

In the first two games, A must win 2 of them. Also, A must win the fourth game. So, two independent events:

Event A: A wins two of the first three games.

Event B: A wins the fourth game.

P(A):

A wins any game with probability p. 3 games, so n = 3. We have to find P(A) = P(X = 2).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(A) = P(X = 2) = C_(3,2).p^(2).(1-p)^(1) = 3p^(2)(1-p)`

P(B):

The probability that A wins any game is p, so P(B) = p.

Probability that A wins in 4:

A and B are independent, so:

`P(A4) = P(A)*P(B) = 3p^(2)(1-p)*p = 3p^(3)(1-p)`

Probability of B winning in exactly four games:

In the first three games, A must win one and B must win 2. The fourth game must be won by 2. So

Event A: A wins one of the first three.

Event B: B wins the fourth game.

P(A)

P(X = 1).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(A) = P(X = 1) = C_(3,1).p^(1).(1-p)^(2) = 3p(1-p)^(2)`

P(B)

B wins each game with probability 1 - p, do P(B) = 1 - p.

Probability that B wins in 4:

A and B are independent, so:

`P(B4) = P(A)*P(B) = 3p(1-p)^(2)*(1-p) = 3p(1-p)^(3)`

Probability that the series lasts exactly four games:

`p = P(A4) + P(B4) = 3p^(3)(1-p) + 3p(1-p)^(3) = 3p(1-p)(p^(2) + (1 - p)^(2))`

The probability that the series lasts exactly four games is
`3p(1-p)(p^(2) + (1 - p)^(2))`