P6 entails that there are (uncountably) infinitely many events. It is at least compatible with modern physics that the world is fundamentally discrete both spatially and temporally. The visible universe is bounded. So it may be that there are only finitely many possible configurations of the universe. It's a big number sure, but if it's finite, then Savage's theorem is irrelevant. It doesn't tell us anything about what to believe in our world. This is perhaps a silly point, and there's probably a nearby theorem that works for "appropriately large finite worlds", but still. I don't think you can just uncritically say "surely the world is thus and so".

If this is supposed to say something normative about how I should structure my beliefs, then the structural premises should be true of the world I have beliefs about.

Not running into the Allais paradox means that if you dump an undetermined ball into a pool of balls, you just add the bets together linearly. But, of course, you do that enough times and you just have the normal result.

So yeah, I'm pretty sure Allais paradox.

Ah, I see what you mean now. So, through no fault of your own, I have conspired to put the wrong boots in front of you. It's not about the probability depending on whether you're buying or selling the bet, it's about assigning an extra value to known proportions.

Of course, then you run in to the Allais paradox... although I forget whether there was a dutch book corresponding to the Allais paradox or not.

How do you choose the interval? I have not been able to see any method other than choosing something that sounds good (choosing the minimum and maximum conceivable would lead to silly Pascal's Wager - type things, and probably total paralysis.)

The discontinuity: Suppose you are asked to put a fair price f(N) on a bet that returns N if A occurs and 1 if it does not. The function f will have a sharp bend at 1, equivalent to a discontinuity in the derivative.

An alternative ambiguity aversion function, more complicated to define, would give a smooth bend.

I intentionally designed the bets so that your agent would take none of them individually, but that together they would be free money. If it has a correct belief, naturally a bet you won't take might look a little odd. But to an agent that honestly thinks P(green | buying) = 2/9, the green and blue bets will look just as odd.

And yes, your agent would take a bet about (green or blue). That is beside the point, since I merely first offered a bet about green, and then a bet about blue.

I mean that I could offer you $9 on green for 2.50, $9 on blue for 2.50, and $9 on red for 3.01, and you wouldn't take any of those bets, despite, in total, having a certainty of making 99 cents. This "type 2" dutch book argument (not really a dutch book, but it's showing a similar thing for the same reasons) is based on the principle that if you're passing up free money, you're doing something wrong :P

I mean that I could offer you $9 on green for 2.50, $9 on blue for 2.50, and $9 on red for 3.01, and you wouldn't take any of those bets, despite, in total, having a certainty of making 99 cents. This "type 2" dutch book argument (not really a dutch book, but it's showing a similar thing for the same reasons) is based on the principle that if you're passing up free money, you're doing something wrong :P

Yes. So basically you are biting a certain bullet that most of us are unwilling to bite, of not having a procedure to determine your decisions and just kind of choosing a number in the middle of your range of choices that seems reasonable.

You're also biting a bullet where you have a certain kind of discontinuity in your preferences with very small bets, I think.

Ah, I see. But now you'll get type 2 dutch booked - you'll pass up on certain money if someone offers you a winning bet that requires you to buy.

I'm confused what your notation means. Let's drop the asymmetry for now and just focus on the fact that you appear to be violating the laws of probability. Does your (1/2 +- 1/6) notation mean that if I would give you a dollar if you drew a green ball, you would be willing to pay 1/3 of a dollar for that bet (bet 1)? Ditto for red (bet 2)? But then if you paid me a dollar if the ball came up (green-or-red), you would be willing to accept 1/2 of a dollar for that bet (bet 3)?

In that case, the dutch book consists of bets like (bet 1) + (bet 2) + (bet 3): you pay me 1/3, you pay me 1/3, I pay you 1/2 (so you paid me 1/6th of a dollar total). Then if the ball's green I pay you a dollar, if it's red I pay you a dollar, and if it's (green-or-red) you pay me a dollar.