Think of it like this:

What is the probability that two coins come up with the same face value? 0.5

If we use the guessing scheme where each player guesses his own guess as the result of the other player's coin, then players A and B win if their coins are the same. Their coins are the same with probability 0.5.

Thus, Expectation[game] = Expectation[Game|A & B Win] * P(A & B win) + Expectation[Game|A & B Lose] * P(A & B Lose) = 1 * .5 + -2 * .5 = -1, so C should not play.

If the players guessed randomly then the expectation of the game would be as you say it is.

If you want to say someone is wrong, it would be better to provide an explanation of exactly why they're wrong rather than point everyone elsewhere and make then figure out the hole in the logic.

Point out the hole directly.

_but_ they can communicate a strategy ahead of time in step 1.

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.

Spoiler. The below contains an exact response

-------------------------- 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

-------------------

-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.

--------------------------

The problem with this analysis is that with the "choose the same as your flip strategy", the case where A is right and B is wrong do not occur. Likewise when A is wrong and B is right does not occur. The results of each coin flip are independent, however the strategy produces a non-independent result.

  A, B, GuessA, GuessB, Winner, AmountC, Chance, EV
  H, H, H,      H,      AB,     -3,      0.25,   -0.75
  H, T, H,      T,      C,      +1,      0.25,   +0.25
  T, H, T,      H,      C ,     +1,      0.25,   +0.25
  T, T, T,      T,      AB,     -3,      0.25,   -0.75

Sum(EV)=-1 per round So C should not play

C loses 2, not 3, when he loses (he gets one from AB at the start of each round). C's EV should therefore be -0.5

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.

Your analysis is incorrect because A and B are allowed to decide on a strategy ahead of time - and there are two simple and similar strategies that allow A and B to guess correctly, together, half the time.

The first is described by: 'If I get heads, I will guess you got heads, otherwise I will guess you got tails' The second: 'If I get heads, I will guess you got tails, otherwise I will guess you got heads'

These are functionally identical, in that in half of the cases they get the same result (first strategy wins these) and in the other half they get opposite results (second strategy wins these) - they get 2$ for winning and give 1$ when they lose, since they win half the time under either strategy C should not play this game.

Play with different strategies:

  import random
 
  # Modify strategy here
  def turn():
      toss = random.randint(0, 1)
      guess = random.randint(0, 1)
      # Uncomment to make A/B win
      # guess = toss
      return toss, guess

  random.seed()

  score_ab, score_c = 0, 0

  for i in range(1, 1001):
      score_ab -= 1
      score_c  += 1
      toss_a, guess_a = turn()
      toss_b, guess_b = turn()

      if toss_a == guess_b and toss_b == guess_a:
        score_ab += 3
        score_c  -= 3

      print i, 'ab:', score_ab, 'c:', score_c

  if score_ab > score_c:
      print "Players A, B win"
  else:
      print "Player C wins"

But they _do_ get to communicate beforehand and could collude to always vote the value of the coin they tossed as their guess for the other coin toss?

Isn't that step 1 where they can agree on a strategy (of which randomly, independently guessing is just one possible strategy)?

Are you the author of the original problem? Perhaps the English prose could be tighter to be more correct with the solution you have in mind.