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In an undergraduate seminar on game theory I attended, it was mentioned in an answer to a question posed to the presenter that, when computing a payoff matrix, the headings in the rows and columns aren't individual actions, but are rather entire strategies; in other words it's as if you pretty much decide what you do in all circumstances at the beginning of the game. This is because when evaluating strategies nobody cares when you decide, so might as well act as if you had them all planned out in advance. So in that spirit, I'm going to use the following principle:
An Agent should chose the strategy that it predicts gives the greatest outcome, weighted by probability of that outcome etcetc...
to approach every one of these problems.
On Omega's coin flip: Omega has given you the function you have to apply to your strategy, you just apply it, and the result's bigger if you answer "yes". Although, realistically, there's no way that Omega has provided nearly enough information for you to trust him, but whatever, that's the premise.
On Parfit's hitchhiker: Again, Ekman has access to your strategy. Just pick one that does benefit him, since those are the only ones that have you not dying at the outcome. If you don't have 100$, find something else you could give him.
On the Democratic Pie: Well, your problem has no strong Nash equilibrium. No solution is going to be stable. I don't really know how this works when you have more than two players, (undergraduate, remember) but I suggest looking into not using a pure strategy. If each voter votes randomly, but choses the probability of his votes appropriately, things work out a little better. You can then compute which semi-random strategy gets you the highest expected size of your slice, etcetc.... Find a book, I don't know how this works. (If I wanted to solve this problem on my own, I would trying to do it with 8 coconuts first, rather than a Continuum of cake.) (This also spells Doom for the AIs respecting whatever Constitution is given then. Not just Doom, Somewhat Unpredictable Doom.)
On the Prisoner's Dilema's Infinite Regress: I don't know.
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In an undergraduate seminar on game theory I attended, it was mentioned in an answer to a question posed to the presenter that, when computing a payoff matrix, the headings in the rows and columns aren't individual actions, but are rather entire strategies; in other words it's as if you pretty much decide what you do in all circumstances at the beginning of the game. This is because when evaluating strategies nobody cares when you decide, so might as well act as if you had them all planned out in advance. So in that spirit, I'm going to use the following principle:
to approach every one of these problems.
On Omega's coin flip: Omega has given you the function you have to apply to your strategy, you just apply it, and the result's bigger if you answer "yes". Although, realistically, there's no way that Omega has provided nearly enough information for you to trust him, but whatever, that's the premise.
On Parfit's hitchhiker: Again, Ekman has access to your strategy. Just pick one that does benefit him, since those are the only ones that have you not dying at the outcome. If you don't have 100$, find something else you could give him.
On the Democratic Pie: Well, your problem has no strong Nash equilibrium. No solution is going to be stable. I don't really know how this works when you have more than two players, (undergraduate, remember) but I suggest looking into not using a pure strategy. If each voter votes randomly, but choses the probability of his votes appropriately, things work out a little better. You can then compute which semi-random strategy gets you the highest expected size of your slice, etcetc.... Find a book, I don't know how this works. (If I wanted to solve this problem on my own, I would trying to do it with 8 coconuts first, rather than a Continuum of cake.) (This also spells Doom for the AIs respecting whatever Constitution is given then. Not just Doom, Somewhat Unpredictable Doom.)
On the Prisoner's Dilema's Infinite Regress: I don't know.
Further elaboration on the cake problem's discrete case:
Suppose there are two slices of cake, and three people who can chose how these will be distributed, by majority vote. Nobody votes so that they alone get both slices, since they can't get a majority that way. So everybody just votes to get one slice for themselves, and randomly decides who gets the other slice. There can be ties, but you're getting an expected 2/3 of a slice whenever a vote is finally not a tie.
To get the continuous case:
It's tricky, but find a way to extend the previous reasoning to n slices and m players, and then take the limit as n goes to infinity. The voting sessions do get longer and longer before consensus is reached, but even when consensus is forever away, you should be able to calculate your expectation of each outcome...