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 i games. (a) If i 4, find the probability that a total of 7 games are played. (b) Find the expected number of games that are played when i 3.

Answer 1

a)

If i is to be 4 and we wish to find the probability of playing7 games overall, then there must have been 6 games in which eachteam won 3

P(3 in 6) = 6^C3 (p)³*(1-p)³ =20 (p³) (1 - 3p + 3p² - p³)

Now we need to use this and consider each of the cases whereteam A wins the 7th game and where Team B wins the 7th game.

P(7 games) =P(6 games and A wins the 7th) +P(6 games and Bwins the 7th)

                  =   p*(20p³)(1-3p+3p²-p³)+ (1-p)*(20p³)(1-3p+3p²-p³)

b)

If i = 3

Number of chances that both teams are win three games

AAA AABA AABBA

BBB ABAA ABBAA

BAAA BBAAA

BBAB BBAAB

BABB BBABB

ABBB BABAB

AABBB
  
ABABA

i.e if any of the team must win three games, then the game stops.

The probability that A wins is p

The probability that B wins is = 1-p = q

Therefore i = 3 the expected number of games played is

E(x) = 3[p*p*p + q*q*q + 4(p*p*q*p + p*q*p*p + q*p*p*p + q*q*p*q + q*p*q*q + p*q*q*q) + 5(p*p*q*q*p +p*q*q*p*p + q*q*p*p*p + q*q*p*p*q + q*p*p*q*q + p*p*q*q*q + q*p*q*p*q + p*q*p*q*p)]

E(T