Against a real partner — not a copy of yourself — it’s better to leave some wiggle room rather than simply rejecting offers less than 50% in case your partner has a slightly different notion of fairness from yours. For example, you could reject increasingly lower offers with increasing probability such that their expected utility is maximized at 50%.
If the first player knows the second player’s distribution, then their optimum strategy is always a single point, the one where
(1-offer) * P(offer accepted)
is maximized. You can do this by setting P(50%) = 1 and P(x) < 1 / 2(1-x) for all x < 50%. Choosing a distribution only just under these limits maximizes player 2′s payoff for irrational player 1′s, while providing incentive for smarter player 1′s to always choose 50%.
In general, it never makes sense for acceptance probability to decrease for larger amounts offered, and so the reject probability is a cumulative distribution function for a threshold value. Hence any viable strategy is equivalent to drawing a threshold value from some distribution. So in principle, both players are precommitting to a single number in each round drawn from some distribution.
Nonetheless, the game does not become symmetric. Even when both players enter the game with a precommitted split drawn from a distribution, the first player has the disadvantage that they cannot win more than the amount they commit to, while the second player will receive a larger payout for any proposal above their committed level. So for any distribution other than “always 50%”, the first player should propose unfair splits slightly more often than the second player rejects them.
However, in settings where the players are known to choose precommitted splits from a distribution, one player or the other can always do better by moving their cumulative distribution closer to “always 50%”. This is the only stable equilibrium. (Edit: I messed up the assumptions in the maths, and this is completely wrong)
As seen above, a population of player 2′s with known precommitment strategy can induce player 1 to offer 50% all the time. But this still isn’t a stable equilibrium! Player 2′s can likewise choose a rejection function that incentivizes any offer short of 100%. This can be seen as slightly skewed version of Prisoner’s Dilemma, where either side choosing a distribution that incentivizes greater than 50% pay-off to themselves is defecting, and one that incentivizes 50% is cooperating.
Against a real partner — not a copy of yourself — it’s better to leave some wiggle room rather than simply rejecting offers less than 50% in case your partner has a slightly different notion of fairness from yours. For example, you could reject increasingly lower offers with increasing probability such that their expected utility is maximized at 50%.
If the first player knows the second player’s distribution, then their optimum strategy is always a single point, the one where
(1-offer) * P(offer accepted)
is maximized. You can do this by setting P(50%) = 1 and P(x) < 1 / 2(1-x) for all x < 50%. Choosing a distribution only just under these limits maximizes player 2′s payoff for irrational player 1′s, while providing incentive for smarter player 1′s to always choose 50%.
In general, it never makes sense for acceptance probability to decrease for larger amounts offered, and so the reject probability is a cumulative distribution function for a threshold value. Hence any viable strategy is equivalent to drawing a threshold value from some distribution. So in principle, both players are precommitting to a single number in each round drawn from some distribution.
Nonetheless, the game does not become symmetric. Even when both players enter the game with a precommitted split drawn from a distribution, the first player has the disadvantage that they cannot win more than the amount they commit to, while the second player will receive a larger payout for any proposal above their committed level. So for any distribution other than “always 50%”, the first player should propose unfair splits slightly more often than the second player rejects them.
However, in settings where the players are known to choose precommitted splits from a distribution, one player or the other can always do better by moving their cumulative distribution closer to “always 50%”. This is the only stable equilibrium.(Edit: I messed up the assumptions in the maths, and this is completely wrong)As seen above, a population of player 2′s with known precommitment strategy can induce player 1 to offer 50% all the time. But this still isn’t a stable equilibrium! Player 2′s can likewise choose a rejection function that incentivizes any offer short of 100%. This can be seen as slightly skewed version of Prisoner’s Dilemma, where either side choosing a distribution that incentivizes greater than 50% pay-off to themselves is defecting, and one that incentivizes 50% is cooperating.
That seems reasonable yes.