Is Rock Paper Scissors with two players but three contestants a fair game?

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.

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.