but the only really arbitrary seeming port is the choice about how to order the limits.
Deciding that the probabaility of overtaking needs to be > 0 might also count here.
Ah, right, good point. I missed that. 0.5 does seem like a more reasonable choice, so that we know the ordering is as fine-grained as possible.
The tit-for-tat strategy unravels all the way back to the beginning, and we’re back at total defection.
Is even that true? Yes, defecting the last n rounds dominates defecting the last n-1 rounds, but defecting the last 5 vs tit-for-tat all the way the equilibrium can be either depending on the initial population. So for all I know there might be a stable equilibrium involving only strategies that cooperate until close to the end.
I’m not sure exactly what setup you’re imagining. If we’re just thinking about Nash equiliblria or correlated equilibria, there’s no “initial population”. If we’re doing something like evolving strategies for the n-round iterated game, then things will still unravel, but it can unravel pretty slowly. A genetic algorithm would have to develop the ability to count rounds, and develop a special exception for defecting in the final round, and then later develop a special exception to defect in the second to last round, etc. So it’s possible that a genetic algorithm would get stuck and effectively never unravel. If so, though, that’s due to limitations in its search.
(And of course this is very dependent on initial population; if it starts out with a lot of defection, it might never develop tit-for-tat in the first place.)
But this tells us that according to my definition above, the petersburg bet is at least (again, because thats only one way) as good an any finite fixed price, which if it isn’t paradoxical means that its more valuable than any fixed price.
Ah, I see, so this approach differs a lot from Ole Peters’ (which suggests logarithmic utility for St Petersburg, just like for Kelly). He studies iterated St Petersburg, though (but w/o fractional betting—just an option to participate in the lotto, at the set price, or not).
OTOH, if we use a cutoff of 1⁄2 rather than 0, the story might be different; there might be a finite price after which it’s not worth it. Which would be interesting. But probably not, I think.
Defecting one round earlier dominates pure tit-for-tat, but defecting five rounds earlier doesn’t dominate pure tit-for-tat. Pure tit-for-tat is better against pure tit-for-tat. So there might be a nash equilibrium containing only strategies that play tit-for-tat until the last few rounds.
Ah, I see, so this approach differs a lot from Ole Peters’
I looked at his paper on the petersburg paradox and I think he gets the correct result for the iterated game. He doesn’t do fractional betting, but he has a variable for players wealth—implicitly, price/wealth is a betting fraction (and since payoffs are fixed, price is implicitly offered odds). Also, and this is quite confusing, even though in the beginning it sounds like he wants to repeat the game with the price fixed but wealth changing over time, his actual calculation assumes the wealth (or distribution over growth rates) is the same each time. He talks about this at the bottom of page 11 and argues that its fine because of commutativity. I’m not sure if that commutativity argument works out, but it means the part before is effectively calculating the growth rate of a betting fraction. And if theres no death in the game, then the highest growth rate does indeed optimize my criterion.
Conceptually though there are differences: Peters totally rejects ensemble averaging. This works in infinite games with no chance of death, because then one player will with certainty experience events at frequencies reflecting the true odds—so it works in ordinary kelly, and it works in this petersburg-bet-in-a-kelly, but it wouldn’t work on the versions with ending chance.
(Also what I said about buying multiples in the last comment was confused—that would be different from one bigger bet.)
OTOH, if we use a cutoff of 1⁄2 rather than 0, the story might be different; there might be a finite price after which it’s not worth it. Which would be interesting. But probably not, I think.
Probably not, no. And provably not for the triple payoff version, so it wouldnt avoid the paradox anyway.
Defecting one round earlier dominates pure tit-for-tat, but defecting five rounds earlier doesn’t dominate pure tit-for-tat. Pure tit-for-tat is better against pure tit-for-tat. So there might be a nash equilibrium containing only strategies that play tit-for-tat until the last few rounds.
Defecting in the last x rounds is dominated by defecting in the last x+1, so there is no pure-strategy equilibrium which involves cooperating in any rounds. But perhaps you mean there could be a mixed strategy equilibrium which involves switching to defection some time near the end, with some randomization.
Clearly such a strategy must involve defecting in the final round, since there is no incentive to cooperate.
But then, similarly, it must involve defecting on the second-to-last round, etc.
So it should not have any probability of cooperating—at least, not in the game-states which have positive probability.
Right? I think my argument is pretty clear if we assume subgame-perfect equilibria (and so can apply backwards induction). Otherwise, it’s a bit fuzzy, but it still seems to me like the equilibrium can’t have a positive probability of cooperating on any turn, even if players would hypothetically play tit-for-tat according to their strategies.
(For example, one equilibrium is for players to play tit-for-tat, but with both players’ first moves being to defect.)
Ah, right, good point. I missed that. 0.5 does seem like a more reasonable choice, so that we know the ordering is as fine-grained as possible.
I’m not sure exactly what setup you’re imagining. If we’re just thinking about Nash equiliblria or correlated equilibria, there’s no “initial population”. If we’re doing something like evolving strategies for the n-round iterated game, then things will still unravel, but it can unravel pretty slowly. A genetic algorithm would have to develop the ability to count rounds, and develop a special exception for defecting in the final round, and then later develop a special exception to defect in the second to last round, etc. So it’s possible that a genetic algorithm would get stuck and effectively never unravel. If so, though, that’s due to limitations in its search.
(And of course this is very dependent on initial population; if it starts out with a lot of defection, it might never develop tit-for-tat in the first place.)
Ah, I see, so this approach differs a lot from Ole Peters’ (which suggests logarithmic utility for St Petersburg, just like for Kelly). He studies iterated St Petersburg, though (but w/o fractional betting—just an option to participate in the lotto, at the set price, or not).
OTOH, if we use a cutoff of 1⁄2 rather than 0, the story might be different; there might be a finite price after which it’s not worth it. Which would be interesting. But probably not, I think.
Defecting one round earlier dominates pure tit-for-tat, but defecting five rounds earlier doesn’t dominate pure tit-for-tat. Pure tit-for-tat is better against pure tit-for-tat. So there might be a nash equilibrium containing only strategies that play tit-for-tat until the last few rounds.
I looked at his paper on the petersburg paradox and I think he gets the correct result for the iterated game. He doesn’t do fractional betting, but he has a variable for players wealth—implicitly, price/wealth is a betting fraction (and since payoffs are fixed, price is implicitly offered odds). Also, and this is quite confusing, even though in the beginning it sounds like he wants to repeat the game with the price fixed but wealth changing over time, his actual calculation assumes the wealth (or distribution over growth rates) is the same each time. He talks about this at the bottom of page 11 and argues that its fine because of commutativity. I’m not sure if that commutativity argument works out, but it means the part before is effectively calculating the growth rate of a betting fraction. And if theres no death in the game, then the highest growth rate does indeed optimize my criterion.
Conceptually though there are differences: Peters totally rejects ensemble averaging. This works in infinite games with no chance of death, because then one player will with certainty experience events at frequencies reflecting the true odds—so it works in ordinary kelly, and it works in this petersburg-bet-in-a-kelly, but it wouldn’t work on the versions with ending chance.
(Also what I said about buying multiples in the last comment was confused—that would be different from one bigger bet.)
Probably not, no. And provably not for the triple payoff version, so it wouldnt avoid the paradox anyway.
Defecting in the last x rounds is dominated by defecting in the last x+1, so there is no pure-strategy equilibrium which involves cooperating in any rounds. But perhaps you mean there could be a mixed strategy equilibrium which involves switching to defection some time near the end, with some randomization.
Clearly such a strategy must involve defecting in the final round, since there is no incentive to cooperate.
But then, similarly, it must involve defecting on the second-to-last round, etc.
So it should not have any probability of cooperating—at least, not in the game-states which have positive probability.
Right? I think my argument is pretty clear if we assume subgame-perfect equilibria (and so can apply backwards induction). Otherwise, it’s a bit fuzzy, but it still seems to me like the equilibrium can’t have a positive probability of cooperating on any turn, even if players would hypothetically play tit-for-tat according to their strategies.
(For example, one equilibrium is for players to play tit-for-tat, but with both players’ first moves being to defect.)
Yeah you’re right. I just realized that what I had in mind originally already implicitly had superationality.