No, you are wrong and hmexx is right. By using the algorithm that hmexx says (use the result of your own coin flip to make the guess of the other's coin flip)

No, you are wrong and hmexx is right. By using the algorithm that hmexx says (use the result of your own coin flip to make the guess of the other's coin flip) what this is really doing is reducing the problem to "What is the probability that two coin flips are the same?" The probability that two coin flips are the same is 50% (HH or TT vs HT or TH). The algorithm could also be "Use your own coin flip and then guess the opposite of your own result" and the probabilities would be the same, ie. 50%. Give the fact that you have a 50% chance of losing $2 and a 50% chance of winning $1, this is not a game that C should play, since the expected value is -$0.50.