I guess you mean: you offer me a bet on green for $2.50 and a bet on blue for $2.50, and I’d refuse either. But I’d take both, which would be a bet on green-or-blue for $5. So no, no dutch book here either.
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 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.
You mean, I will be offered a bet on green, but I may or may not be offered a bet on blue? Then that’s not a Dutch book—what if I’m not offered the bet on blue?
For example: suppose you think a pair of boots is worth $30. Someone offers you a left boot for $14.50. You probably won’t find a right boot, so you refuse. The next day someone offers you a right boot for $14.50, but it’s too late to go back and buy the left. So you refuse. Did you just leave $1 on the table?
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.
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.
No, this doesn’t sound like the Allais paradox. The Allais paradox has all probabiliies given. The Ellsberg paradox is the one with the “undetermined balls”. Or maybe you have something else entirely in mind.
I guess you mean: you offer me a bet on green for $2.50 and a bet on blue for $2.50, and I’d refuse either. But I’d take both, which would be a bet on green-or-blue for $5. So no, no dutch book here either.
Or do you have something else in mind?
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 wouldn’t take any of them individually, but I would take green and blue together. Why would you take the red bet in this case?
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.
You mean, I will be offered a bet on green, but I may or may not be offered a bet on blue? Then that’s not a Dutch book—what if I’m not offered the bet on blue?
For example: suppose you think a pair of boots is worth $30. Someone offers you a left boot for $14.50. You probably won’t find a right boot, so you refuse. The next day someone offers you a right boot for $14.50, but it’s too late to go back and buy the left. So you refuse. Did you just leave $1 on the table?
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.
I do not run into the Allais paradox—and in general, when all probabilties are given, I satisfy the expected utility hypothesis.
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.
No, this doesn’t sound like the Allais paradox. The Allais paradox has all probabiliies given. The Ellsberg paradox is the one with the “undetermined balls”. Or maybe you have something else entirely in mind.
What I mean is possible preference reversal if you just have a probability of a gamble vs. a known gamble.
I wouldn’t take any of them individually (except red), but I’d take all of them together. Why is that not allowed?