datdatruth

It doesn't, you're simply wrong. See my other comments.

This does not work because the text says "Players A and B build a team, they have one fair coin each" NOT one fair coin total.

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.

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