The Ellsberg paradox is boring because it can have no possible effect on utility or the log score of your beliefs (note that this means it should not be conflated with the mother with the candy). This was, in fact, the main criticism of the previous post. Given that, the fact that it can’t lead to being dutch booked is trivial. I would rather this post was a lot shorter and had fewer generalizations.
I’m not sure what you mean. If it’s because the situation was too symetrical, I think I adressed that.
For example, you could add or remove a couple of red balls. I still choose red over green, and green-or-blue over red-or-blue. I think the fact that it still can’t lead to being dutch booked is going to be a surprise to many LW readers.
I agree, I don’t think there is any way to dutch-book someone for being wrong but consistent with the laws of probability (that is, still assigning 1⁄3 probabilities to r,g,b even when that’s wrong). They simply lose money on average. But this is an extra fact, unrelated to the triviality that is not being able to dutch-book someone based on an arbitrary choice between two equivalent options. Once they start paying for equivalent options, then they get money-pumped.
Once they start paying for equivalent options, then they get money-pumped.
Okay. Suppose there is an urn with 31 red balls, and 60 balls that are either green or blue. I choose to bet on red over green, and green-or-blue over red-or-blue. These are no longer equivalent options, and this is definitely not consistent with the laws of probability. Agreed?
(My prior probability interval is P(red) = 31⁄91 exactly, P(green) = (1/2 +- 1⁄6)(60/91), P(blue) = (1/2 -+ 1⁄6)(60/91).)
It sounds like you expected (and continue to expect!) to be able to money-pump me.
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.
If the bet pays $273 if I drew a red ball, I’d buy or sell that bet for $93. For green, I’d buy that bet for $60 and sell it for $120. For red-or-green, I would buy that for $153 and sell it for $213. Same for blue and red-or-blue. For green-or-blue, I’d buy or sell that for $180.
(Appendix A has an exact specification, and you may wish to (re-)read the boot dialogue.)
[ADDED: sorry, I missed “let’s drop the asymmetry” .. then, if the bet pays $9 on red, buy or sell for $3; green, buy $2 sell $4; red-or-green, buy $5 sell $7; blue, red-or-blue same, green-or-blue, buy or sell $6. Assuming risk neutrality for $, etc etc no purchase necessary must be over 18 void in Quebec.]
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 think the fact that it still can’t lead to being dutch booked is going to be a surprise to many LW readers.
I would not make this prediction. I would think anyone who would understand that claim without having to look up “dutch book” should find that obvious.
The Ellsberg paradox is boring because it can have no possible effect on utility or the log score of your beliefs (note that this means it should not be conflated with the mother with the candy). This was, in fact, the main criticism of the previous post. Given that, the fact that it can’t lead to being dutch booked is trivial. I would rather this post was a lot shorter and had fewer generalizations.
I’m not sure what you mean. If it’s because the situation was too symetrical, I think I adressed that.
For example, you could add or remove a couple of red balls. I still choose red over green, and green-or-blue over red-or-blue. I think the fact that it still can’t lead to being dutch booked is going to be a surprise to many LW readers.
I agree, I don’t think there is any way to dutch-book someone for being wrong but consistent with the laws of probability (that is, still assigning 1⁄3 probabilities to r,g,b even when that’s wrong). They simply lose money on average. But this is an extra fact, unrelated to the triviality that is not being able to dutch-book someone based on an arbitrary choice between two equivalent options. Once they start paying for equivalent options, then they get money-pumped.
Okay. Suppose there is an urn with 31 red balls, and 60 balls that are either green or blue. I choose to bet on red over green, and green-or-blue over red-or-blue. These are no longer equivalent options, and this is definitely not consistent with the laws of probability. Agreed?
(My prior probability interval is P(red) = 31⁄91 exactly, P(green) = (1/2 +- 1⁄6)(60/91), P(blue) = (1/2 -+ 1⁄6)(60/91).)
It sounds like you expected (and continue to expect!) to be able to money-pump me.
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.
If the bet pays $273 if I drew a red ball, I’d buy or sell that bet for $93. For green, I’d buy that bet for $60 and sell it for $120. For red-or-green, I would buy that for $153 and sell it for $213. Same for blue and red-or-blue. For green-or-blue, I’d buy or sell that for $180.
(Appendix A has an exact specification, and you may wish to (re-)read the boot dialogue.)
[ADDED: sorry, I missed “let’s drop the asymmetry” .. then, if the bet pays $9 on red, buy or sell for $3; green, buy $2 sell $4; red-or-green, buy $5 sell $7; blue, red-or-blue same, green-or-blue, buy or sell $6. Assuming risk neutrality for $, etc etc no purchase necessary must be over 18 void in Quebec.]
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 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?
I would not make this prediction. I would think anyone who would understand that claim without having to look up “dutch book” should find that obvious.