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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.

Axoren
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1 Answers1

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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.

masiewpao
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    In this case, the bias is across rounds rather than within them. Very astute! – Axoren Jul 03 '21 at 16:00
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    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). – Bernhard Barker Jul 04 '21 at 06:50
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    @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 – masiewpao Jul 04 '21 at 10:24
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    @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). – masiewpao Jul 04 '21 at 10:29
  • @BernhardBarker I shouldn't have said at 'no' point in my previous comment, rather at no point when there are sufficiently many rounds of RPS left to play. I think sufficiently many rounds here means 4, as this is when collusion deviates from random to yield a higher expectation. Therefore if at any point before this player 1 or 2 deviates, they hurt their own expectation, because of the reason in the answer – masiewpao Jul 04 '21 at 10:38
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    @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.) – Darren Cook Jul 06 '21 at 20:08
  • What happens if the only way to end the game is for player 3 to chose, when he lost the last round ? – Tom-Tom Jul 06 '21 at 22:43
  • @Tom-Tom I am not sure I understand – masiewpao Jul 07 '21 at 10:39
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    @DarrenCook That is a very interesting point about showing first, the only issue is that gets into the realm of changing the rules of the game. Colluding strategies should still take place in the rules of the original game, i.e. showing at the same time (at least for our analysis, not in the real world!) – masiewpao Jul 07 '21 at 10:44
  • Just read your answer again - I think £2.50 is a typo, and it should be £7.50? (The pot is £15 for each game.) – Darren Cook Jul 07 '21 at 10:50
  • @DarrenCook Heuristically, if player 1 wins the first round, they make £10 (they put £5 into the pot themselves), they then lose £5 on the next round. The expectation/profit is then $£2.50$ per round. – masiewpao Jul 07 '21 at 11:08
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    @masiewpao Sorry, yes. I was thinking "make" as gross earnings, not net profit. – Darren Cook Jul 07 '21 at 14:51
  • @DarrenCook It's a fair point! I've known folk who analyse poker strategy for example in the way you suggest; essentially focusing on gross earnings. I don't think it makes a difference so long as we remember what we've done, but typically for games involving some element of cost per round, I prefer to include that in the expectation calculation, as usually that's what we're interested in at the end of the day. – masiewpao Jul 07 '21 at 15:11
  • @masiewpao. I am asking about a possible additional rule : whenever Player 3 loses a round (so Player 1 or Player 2 won that round), Player 3 can decide to end the whole game. That way Player 1 and Player 2 do not know when the game ends. If there is collusion between them, Player 3 will surely end the game quickly and diminish their gains. – Tom-Tom Jul 07 '21 at 18:12
  • @Tom-Tom Ah I see! This is very interesting. I actually believe in this case the game is inherently unfair to P1,P2. If we can assume long (indefinite) play, we can optimise in terms of expectation, which is essentially how I quantify an effective colluding strategy in the OP case. (caveat being I ignored risk of ruin effects) – masiewpao Jul 07 '21 at 23:41
  • @Tom-Tom (cont) Assuming only P3 can terminate (meaning P1,P2 must play indefinitely), I think variance effects come into play. In the long run, expectation amongst the players should be zero, (or very close to it), since as you surmise the colluding strategy will be ineffective. However, I think since we can expect arbitrarily long sequences of positive outcomes to occur in the long run, I THINK we can expect that in any finite game, that P3 will at some point be ahead. They can choose to terminate here. – masiewpao Jul 07 '21 at 23:47