I think the portfolio argument is an unnecessary detour though. There's a two-line proof by induction.
1. The payoff in the base case of (0,1) or (1,0) is 2.
2. If we are at (r,b), r >=b , have $X, and stake (r-b)/(r+b) on red, the payoff if we draw red and win is X * (1+(r-b)/(r+b)) * 2^(r+b-1) / (r+b-1 choose r-1) = X * 2^(r+b) * r / ((r+b) * (r+b-1 choose r-1)) = X * 2^(r+b) / (r+b choose r).
Similarly, if we draw black and lose, the payoff is X * (1-(r-b)/(r+b)) * 2^(r+b-1) / (r+b-1 choose r) = X * 2^(r+b) * b / ((r+b) * (r+b-1 choose r)) = X * 2^(r+b) / (r+b choose r). QED
I think the portfolio argument is an unnecessary detour though. There's a two-line proof by induction.
1. The payoff in the base case of (0,1) or (1,0) is 2.
2. If we are at (r,b), r >=b , have $X, and stake (r-b)/(r+b) on red, the payoff if we draw red and win is X * (1+(r-b)/(r+b)) * 2^(r+b-1) / (r+b-1 choose r-1) = X * 2^(r+b) * r / ((r+b) * (r+b-1 choose r-1)) = X * 2^(r+b) / (r+b choose r).
Similarly, if we draw black and lose, the payoff is X * (1-(r-b)/(r+b)) * 2^(r+b-1) / (r+b-1 choose r) = X * 2^(r+b) * b / ((r+b) * (r+b-1 choose r)) = X * 2^(r+b) / (r+b choose r). QED