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