In a game of Rock Paper Scissors with two players, there are $3$ outcomes every round each with multiplicity $3$.

  • Player 1 can win.
  • Player 2 can win.
  • Both players can draw.

If we were to assign a winning result to a third "Player" when Player 1 and 2 draw, does this become a fair 3-player game?

  • Player 1 wins.
  • Player 2 wins.
  • Both players draw, so Player 3 wins.

Player 1 and 2 have no incentive to cooperate to reduce Player 3's chances because the only way to do so is for one to forgo their own chance at winning.

Player 3 cannot negatively or positively reduce the chances of either Player 1 or Player 2 because they make no moves in this game.

This looks to create a Nash Equilibrium and a fair game.

Am I missing something? I'm worried that the fact that Player 3 does not participate at all may be causing some bias I'm overlooking.

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    $\begingroup$ I wonder if there have been investigations published in the literature on the use and bias of spectator agents in general. $\endgroup$
    – TheSimpliFire
    Jul 3 '21 at 16:02
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    $\begingroup$ I would advise you to not accept an answer so early, as it discourages other people from answering. $\endgroup$
    – Joe
    Jul 3 '21 at 16:12

By allowing collusion, it is possible for player 1 and 2 to engage in a strategy that benefits each individually, while meaning player 3 never wins. For example, player 1 and 2 agree to win and lose in an alternating fashion.

This gives a 50% win rate for both players, up from a 33% win rate in the non-colluding, random case. This colluding strategy is better for players 1 and 2, and essentially steal 'equity' from player 3.

For example, if the game is such that every player puts £5 into a pot, then on average player 1 and 2 make £2.50 a game by colluding, vs £0 in the random 'fair' case. Player 3 obviously loses £5 a game in the colluding case.

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    $\begingroup$ In this case, the bias is across rounds rather than within them. Very astute! $\endgroup$
    – Axoren
    Jul 3 '21 at 16:00
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    $\begingroup$ Collusion makes for an interesting case, as player 2 (designated to lose in the last round) will, if playing optimally, try to win that round instead. Player 1, knowing this, will try to counter player 2 (which player 2 will know) and both will probably just end up playing randomly on the last round. Player 1 will then try to win the 2nd-last round (assuming they are designated to lose in that round). Player 2 may then try to win the 3rd-last round. It probably requires a bit of analysis to figure out for how many rounds collusion would work (if it works at all). $\endgroup$ Jul 4 '21 at 6:50
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    $\begingroup$ @BernhardBarker I agree, I actually think my answer is not very precise. However I did think about the specific situation you mention, and I was worried about this 'regress' back to the random strategy since if one player knows the other is trying to steal, they will have to revert back to a random strategy $\endgroup$
    – masiewpao
    Jul 4 '21 at 10:24
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    $\begingroup$ @BernhardBarker (cont) However for sufficiently long runs of playing RPS, I believe at no point does it benefit either player to deviate. The reason being that the moment one player deviates, the other player will respond by returning to a random strategy. But the player who originally deviates knows this, and hence shouldn't deviate if there are sufficiently many rounds left (as then deviation has a worse expectation). $\endgroup$
    – masiewpao
    Jul 4 '21 at 10:29
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    $\begingroup$ @BernhardBarker If P1 or P2 doesn't trust the other, it creates the chance that P3 wins more than they do. They can avoid this by also adding to the agreement that the player whose turn it is to lose always shows first. The vulnerable player cannot now be taken advantage of. (I suppose P3 might object that this is against the spirit of the game... but as we've seen, the best P3 can hope for is to break even, so he should run away before the first round.) $\endgroup$ Jul 6 '21 at 20:08

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