Yes. Break it down into the possible cases, assign probability for the case, and value for the case. Multiply each probability by its value and sum these up.
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The fact that there is no communication ensures that whether A is right and whether B is right are completely indpendent.
Conceptually, predicting a coin flip is like making a statement "1" or "0" which will then be XOR'd with a 1 or 0 from a securely, randomly, uniformly generated OTP of which no copies exist. In other words, the plaintext is immediately lost forever and you just have the ciphertext.
As a result of this, we must truly consider that A being right is a 50/50 proposition. It is also indpendent of B being right.
Thus we have the following four cases:
A right, B right - Result value * Percent chance = EV
0, 0 = +1 * 0.25 = +0.25
0, 1 = +1 * 0.25 = +0.25
1, 0 = +1 * 0.25 = +0.25
1, 1 = -2 * 0.25 = -0.5
------------------------------- sum of above:
+0.25. Therefore, as long as each team member is independently predicting their own coin (which means that their prediciton will be xor'd by a random bit) C should play this game long-term.
(The values of +1 is because each round starts with the team giving C +1. If at the end of the roudn C must return 3 then this is +1 -3 = -2 for the round.)
Now here is another interesting question. What if under the same conditions A and B both try to predict C's coin toss, of which there is only one? Should C now play? Here is the answer is: "No", because A can predict heads, B can predict heads, and then it looks like this:
A right, B right - Result value * Percent chance = EV
0, 0 = +1 * 50% = 0.5
0, 1 = +1 * 0% = 0 } not possible
1, 0 = +1 * 0% = 0 }
1, 1 = -2 * 50% = -1
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-0.5
In this case, C should not play. This is because in this case the events are not truly independent, there is a way to break the 25% 25% 25% 25% into 50% and 50% - namely by picking the same prediction together.