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fool comments on The Ellsberg paradox and money pumps - Less Wrong
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Comments (72)
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.
I wouldn't take any of them individually (except red), but I'd take all of them together. Why is that not allowed?