Question

Two teams play a series of games until one of the teams wins n games. In every game, both teams have equal chances of winning and there are no draws. Compute the expected number of the games played when (a) n = 2; (b) n = 3. (To keep track of what you are doing, it can be easier to use different letters for the probabilities of win for the two teams).

Answer 1

a)let X are number of games played

P(X=2)=P(first team wins both games)+P(second team wins both games)

=2*P(first team wins both games)=2*(1/2)*(1/2)=1/2

P(X=3)=2*P(first team wins 1st and second team second and them first team)+2*P(second team first and then next two games by first team)

=2*(1/2)*(1/2)*(1/2)+2*(1/2)*(1/2)*(1/2)=1/2

hence expected number of games =xP(x)=2*(1/2)+3*(1/2)=2.5

b)

P(X=3)=2*P(all 3 games won by first team)=2*(1/2)*(1/2)*(1/2)=1/4

P(X=4)=2*P(out of fist 3 games 2 won by first team and then 4th game won by first team)

=2* =2*3*(1/2)⁴ =3/8

P(X=5)=2*P(out of fist 4 games 2 won by first team and then 4th game won by first team)

=2*=2*6*(1/2)⁵ =3/8

hence expected games to be played =xP(x)=3*(1/4)+4*(3/8)+5*(3/8)=4.125