Best of Seven

Two teams, A and B, play a "Best of Seven" tournament. The first team to win four games wins the tournament. If the probability that team A will win any game is p, and the probability team B will win is (1-p),
  1. What is the probability team A (or B) will win the entire tournament?
  2. What is the expected number of games in the tournament? (If p is 0, or if p is 1, the expected number of games would be 4. When p is between 0 and 1, the expected number should rise toward 7.)

Source: Original, but I imagine it's elsewhere.

Solutions were received from Jayavel Sounderpandian, Jozef Hanenberg, Jimmy Chng Gim Hong, Joseph DeVincentis, Lou Cairoli, Nick McGrath, Philippe Fondanaiche, Sudipta Das, Paul Botham, and Alan O'Donnell. The solution were of course quite algebraic. A few examples follow:

From Philippe Fondanaiche: 

Question 1 
P(A) = Pr[ A wins ] = p^4*[1 + 4*q + 10*q^2 + 20*q^3 ] with q=1-p 
P(B) = Pr[ B wins ] = q^4*[1 + 4*p + 10*p^2 + 20*p^3 ] with q=1-p 
We can check that P(A) + P(B) = 1 

Question 2 
X is the length of the tournament with X=4,5,6,7 
Pr[ X=4 ] = p^4 + q^4 
Pr[ X=5 ] = 4*p*q*(p^3 + q^3) 
Pr[ X=6 ] = 10*p^2*q^2*(p^2 + q^2) 
Pr[ X=7 ] = 20*p^3*q^3 

The expected number of games in the tournament is defined by: 
E(X) = sum{ x*Pr[ X= x] } for x=4,5,6,7 

Therefore E(X) = 4*(1 + p + p^2 + p^3 -13*p^4 +15*p5 - 5*p^6 ) 
Let p = 1/2 + z with z within the interval [-1/2 , 1/2] 
So E(X) = f(z) = 93/16 -47/4*z^2 + 23*z^4 -20*z^6 
It is easy to demonstrate through the analysis of f(z)  that E(X) is minimum and equal to 4 for p=0 or z=-1/2 and E(X) is maximum and equal to 93/16 for p=1/2 or z=0.
From Joseph Devincentis: 
1. The chance that A wins is the sum of the following:

p^4            (p wins in 4 games)
p^4(1-p)(4)    (p wins in 5 games, losing any one of the first 4)
p^4(1-p)^2(10) (p wins in 6 games, losing any two of the first 5)
p^4(1-p)^3(20) (p wins in 7 games, losing any three of the first 6)

This is 35 p^4 - 84 p^5 + 70 p^6 - 20 p^7.

2. The expected number of games can be computed by taking the above
probabilities for p winning in 4,5,6, or 7 games and multiplying by the
number of games, adding them up, and adding the same thing with 1-p
substituted for p to account for the cases where B wins.

For A winning, you get: p^4 (4 + 20(1-p) + 60(1-p)^2 + 140(1-p)^3)
= 224 p^4 - 560 p^5 + 480 p^6 - 140 p^7

For B winning, you get (1-p)^4 (4 + 20p + 60p^2 + 140p^3)
= (1 - 4p + 6p^2 - 4p^3 + p^4) (4 + 20p + 60p^2 + 140p^3)
= 4 + 4p + 4p^2 + 4p^3 - 276p^4 + 620p^5 - 500p^6 + 140p^7

The total expected number of games is thus:
4 + 4p + 4p^2 + 4p^3 - 52p^4 + 60p^5 - 20p^6
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