by hollis
I think I found another way to calculate this:Let X be player 1's chance of winning
Let Y be side A's chance of winning
p(A wins | p1 is playing A) = X*Y + X*(1-Y) + (1-X)*Y - (1-X)*(1-Y)
p(B wins | p1 is playing B) = X*(1-Y) + X*Y + (1-X)*(1-Y) - (1-X)*Y
Case 1: X = (1/2), Y = (1/2)
p(A wins | p1 is playing A) = 0.5*0.5 + 0.5 * (1-0.5) + (1-0.5) * 0.5 - (1-0.5)*(1-0.5)
p(A wins | p1 is playing A) = 0.25 + 0.25 + 0.25 - 0.25
p(A wins | p1 is playing A) = 0.50
Case 2: X = (2/3), y = (2/3)
p(A wins | X is playing A) = (2/3) * (2/3) + (2/3) * (1-2/3) + (1-2/3)*2/3 - (1-2/3)*(1-2/3)
p(A wins | p1 is playing A) = 4/9 + 2/9 + 2/9 - 1/9
p(A wins | p1 is playing A) = 7/9
p(A wins | p1 is playing A) = ~0.78
p(B wins | p1 is playing B) = (2/3) * (1-2/3) + (2/3) * (2/3) + (1-2/3)*(1-2/3) - (1-2/3)*(2/3)
p(B wins | p1 is playing B) = 2/9 + 4/9 + 1/9 - 2/9
p(B wins | p1 is playing B) = 5/9
p(B wins | p1 is playing B) = ~0.56
p(p1 wins) = (p(A wins | p1 is playing A) + p(B wins | p1 is playing B))/2
p(p1 wins) = (7/9 + 5/9) / 2
p(p1 wins) = (12/9) / 2
p(p1 wins) = 6/9
p(p1 wins) = 2/3
Our methods come to *slightly* different values for p(A wins | p1 is playing A) and p(B wins | p1 is playing B), so I am a little worried. However, it's not a big enough discrepancy that I am going to hold back on data analysis any longer!
Woohoo, stats sure does beat work!