Question

Suppose that two teams are playing a series of games, each of which is independently won by team A with probability p and by team B with probability 1−p. The winner of the series is the first team to win four games. Find the expected number of games that are played, and evaluate this quantity when p = 0.5.

Answer 1

Answer:

4 games played = 1/56

5 games played = 5/56

6 games played = 15/56

7 games played = 35/56

Step-by-step explanation:

The probability of each time winning is 0.5

So there are a couple of ways the series could go.

1. Team A could win all first 4 matches. We can depict this as A-A-A-A
2. Team A could win 4, while having lost 1.  Lets depict this a A-B-A-A-A
3. Team A could win 4, while having lost 2. A-B-B-A-A-A
4. Team A could win 4, while having lost 3. A-B-B-B-A-A-A

These 4 possibilities could be repeated with B winning as well. It should be noted that these are the only ways for the series to end. We find the number of permutations of each possibility above to find their probability.

I advise you to study 'how to permute identical objects' for this.

These permutations are stated below:

1. 4!/4! = 1

2. $\frac{5!}{4! * 1!}$ = 5

3 $\frac{6!}{4! * 2!}$ = 15

4 $\frac{7!}{4! * 3!}$ = 35

These are they ways A or B could win. The total is 1+5+15+35 = 56. The answer given thus reflects the possibilities.