Ah, RPS -- possibly the greatest game ever. Or possibly not. Anyway, I may write more about it later, but for now I'm interested in finding a way to use RPS to select a winner from among three people with equal probability.
Playing two-out-of-three between one pair and then pitting the winner against the odd man out won't work, because it gives the odd man a 50% chance of winning while the other two have only a 25% chance each. Requiring the odd man to win two in a row against the winner of the pair reduces his odds to 25% and raises the odds of the other two to 37.5%. Is there a reasonable way to get the odds to 33% for each player, or is it impossible to do because of the nature of the game?
I'm shocked you haven't received a response yet. I'd use a round robin format...Player A/Player B, BC and AC. Play until someone wins both matches in a round.
How about the following:
Players A, B, and C all choose either R, P, or S simultaneously. If they all choose the same thing, then it's a draw and they try again. If they all choose three different items (i.e., one R, one P, one S), it's also a draw.
If two players pick one item and one player picks a second, they see whether the player who picks the unique item would win over the other two. If he does, then he's the winner. If the player who picks the unique item would lose to the other two, then he's eliminated and the other two play a conventional game of RPS to decide between them.
Examples:
A/B/C choose R/R/R. Draw.
A/B/C choose R/S/P. Draw.
A/B/C choose S/S/R. C wins (R beats S.)
A/B/C choose S/S/P. Eliminate C. (S beats P.) A and B play again.
It might take a few rounds to determine the winner since 1/3 of the initial rounds will end up in a Draw, but that's already the case with 2-person RPS. This algorithm should be fair to all three players.
I didnt read the previous response, Im sure it is right considering how much is written...
but just do a round robin... player A plays both B and C.
player B plays both A and C. etc...
play until one of the three players has won against the other two. Just keep doing it. Thats all ive got, I think it works pretty good.
Hm, all of these are tournament style solutions, which means they're open ended and could go on forever. I like PH's best -- it's the most elegant, but it also requires some rule changes (basically creating a three-player variant).
I think any 3-player version of RPS that's reasonably faithful to the rules is going to be open-ended since that's already inherent in the 2-player version, which could also theoretically go on forever.
If you want something simple, quick, and fair that's guaranteed to end in one round, then how about a 3-player version of Odds-and-Evens:
Players A, B, and C each simultaneously stick out either 0, 1, or 2 fingers. Then count the total number of fingers between all players, and take the remainder modulo 3.
If (Total mod 3) = 0, then A is the winner.
If (Total mod 3) = 1, then B is the winner.
If (Total mod 3) = 2, then C is the winner.
This generalizes very easily to N players.
(straightens tie, checks hair, ready to declaim!)
I lived in Japan for 14 years, and watched them choose one person out of a group of any number of people (when a non-serious decision) with this game. They used the method described by Paul Hsieh.