three persons toss coin once each and predict the next guy in a, b, c, a order. Ensure at least one right. i read about a puzzle where two guys are in two separate chambers and they each toss once and get one chance to predict the other guy's result. There is a system by which they can ensure that one of them always predicts right.
One of them always predicts matching his own result. The other always predicts the opposite of his own result.
How can i find out if such a system can be there in a 3 or more person setup ?
three persons in three separate rooms each toss a coin once and predicts the toss result of the next guy in the cyclic order a, b, c, a, ( to avoid the two person case) . can they develop a system to get at least one prediction right between them ?
Please pardon my bad wording and correct it if you will.
SORRY! I needlessly complicated the matter using the word 'choice'. hope it is better now.
 A: Let's call the three people A, B, C, and suppose A guesses B's result, B guesses C's, and C guesses A. If each guesses the other's result is the same as his own, at least one must be right. For if A and B are both wrong, then B had a different result than A, and C had a different result than B, so A and C had the same result, so C is right. 
A: Each player has $4$ possible strategies: heads, tails, same and opposite. (Randomizing doesn’t help, since we’re aiming for certainty.) If at least one player uses one of the constant strategies (heads or tails), there is always a result where no one guessed right. Thus all three players must use one of the strategies that depend on their result (same or opposite). One guess is certain to be correct exactly if the number of players who use opposite is even. Thus, in addition to the strategy profile in Gerry Myerson’s answer, there are three further strategy profiles in which one player uses same and two use opposite; and these are the only four strategy profiles that solve the problem.
