We have three people to consider: A, B and C. Let's call the probability that A guesses correctly A1, and the probability that he guesses incorrectly A2. Similarly, we can define B1, B2, C1 and C2. Then the constraint E(wrong guesses) ≥ E(correct guesses) becomes A2 + B2 + C2 ≥ A1 + B1 + C1. We are interested in how large we can make p(correct guess an no wrong guess), which I'll call p. In the notation above, p = (1 - A2)(1-B2)(1 - C2) - (1 - A1 - A2)(1 - B1 - B2)(1 - C1 - C2).
Of course, each of the As, Bs and Cs should lie between 0 and 1, and we must have that each of A1 + A2, B1 + B2 and C1 + C2 is at most 1. I'll call these the bounding conditions.
We want to show that p ≤ 1/2, which we can now do by the circuitous method of supposing that we can
choose A1, A2, B1, B2, C1 and C2 meeting the constraint above but with p > 1/2, and showing that that leads to an absurd conclusion. So suppose we have such values for the As, Bs and Cs.
Lets first of all think about what happens if we change A1 and A2 a bit. We don't want to break the constraint, so whatever we add on to A1, we should also add on to A2. So what happens if we replace A1 by A1 + d, and A2 by A2 + d? Well, the value of p changes by an amount proportional to d, namely d*[-(1-B2)(1-C2) + 2(1 - B1 - B2)(1 - C1 - C2)]. So by either taking d to be positive (if -(1-B2)(1-C2) + 2(1 - B1 - B2)(1 - C1 - C2) is positive) or negative (otherwise), we can modify the values of A1 and A2 without lowering the value of p. In fact, we can keep on changing the values of A1 and A2 in the appropriate direction until we can go no further because of the bounding conditions. This happens when either at least one of A1 and A2 is 0 or else A1 + A2 = 1. So, we may as well suppose that one of these things happen: if not, we can make it happen without breaking the constraint or reducing the value of p.
We can quickly rule out the case A1 + A2 = 1. From the constraint, we get 1 - A2 = A1 ≤ A2 + B2 + C2, and so 2(1-A2) + (1 - B2) + (1 - C2) ≤ 3. By the AM-GM inequality
applied to 2(1-A2), 1 - B2 and 1 - C2 we now get 2*(1-A2)*(1-B2)*(1 - C2) ≤ 1, and so p = (1 - A2)(1 - B2)(1 - C2) ≤ 1/2. That's not possible, as we're assuming p > 1/2.
So either A1 = 0 or A2 = 0. Similarly we can assume that one of B1 and B2 is 0 and that at least one of C1 and C2 is zero (note for pedants: we can make all three of these assumptions at once, since the changes we made to force either A1 or A2 to be 0 didn't affect the values of the Bs or the Cs at all).
We can't have that all three of A1, B1 and C1 are 0, since in that case p would be 0. So let's suppose that A1 > 0, which forces A2 = 0. Then, using the constraint, B2 + C2 ≥ A1 > 0, so B2 and C2 can't both be zero. Let's suppose that B2 > 0, which forces B1 = 0. We now have 2 cases to consider: the case where C1 = 0, and the case where C2 = 0.
First, let's examine the case where C1 = 0. Then we get A1 ≤ B2 + C2, so that A1 + (1 - B2) + (1 - C2) ≤ 2. In this case, p = (1 - B2)(1-C2) - (1 - A1)(1 - B2)(1 - C2) = A1(1 - B2)(1 - C2) ≤ 8/27 < 1/2, by applying the AM-GM inequality to A1, (1 - B2) and (1 - C2). Once again, this case is eliminated since we are assuming p > 1/2.
Next, lets examine the case where C2 = 0. Then we get A1 + C1 ≤ B2, so that (1 - B2) + (A1 + C1) ≤ 1. In this case, p = (1 - B2) - (1 - A1)(1 - B2)(1 - C1) = (1 - B2)[1 - (1 - A1)(1 - C1)] = (1 - B2)(A1 + C1 - A1*C1) ≤ (1 - B2)(A1 + C1) ≤ 1/4 < 1/2, by applying the AM-GM inequality to 1-B2 and A1 + C1. This final case is therefore also eliminated since we are assuming p > 1/2.
We have now eliminated all possible cases, which is absurd. So our assumption must have been false. That is, we always have p ≤1/2. This, together with the arguments quoted above, shows that the heroes have no strategy which is guaranteed to succeed more than half the time.