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endoself comments on A summary of Savage's foundations for probability and utility. - Less Wrong
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Hmm. Now we have that $6 iff Green is worth $1 and $-6 iff Green is worth $-3, but $6-6 = $0 iff Green is not equivalent to $1-3 = $-2.
In particular, if you have $6 conditional on Green, you will trade that to me for $1. Then, we agree that if Green occurs, I will give you $6 and you will give me $6, since this adds up to no change. However, then I agree to waive your having to pay me the $6 back if you give me $3. You now have your original $6 iff Green back, but are down an unconditional $2, an indisputable net loss.
Also, this made me realize that I could have just added an unconditional $6 in my previous example rather than complicating things by making the $6 first conditional on (die ≥ 3) and then on (Green ∨ Blue). That would be much clearer.
I pay you $1 for the waiver, not $3, so I am down $0.
In state A, I have $6 iff Green, that is worth $1.
In state B, I have no bet, that is worth $0.
In state C, I have $-6 iff Green, that is worth $-3.
To go from A to B I would want $1. I will go from B to B for free. To go from B to A I would pay $1. State C does not occur in this example.
Wouldn't you then prefer $0 to $1 iff (Green ∧ Heads) - $1 iff (Green ∧ Tails)?
Indifferent. This is a known bet.
Earlier I said $-6 iff Green is identical to $-6 + $6 iff (not Green), then I decomposed (not Green) into (Red or Blue).
Similarly, I say this example is identical to $-1 + $2 iff (Green and Heads) + $1 iff (not Green), then I decompose (not Green) into (Red or (Blue and Heads) or (Blue and Tails)).
$1 iff ((Green and Heads) or (Blue and Heads)) is a known bet. So is $1 iff ((Green and Heads) or (Blue and Tails)). There are no leftover unknowns.
Look at it another way.
Consider $6 iff (Green ∧ Heads) - $6 iff (Green ∧ Tails) + $4 iff Tails. This bet is equivalent to $0 + $2 = $2, so you would be willing to pay $2 for this bet.
If the coin comes out heads, the bet will become $6 iff Green, with a value of $1. If the coin comes out tails, the bet will become $4 - $6 iff Green = $4 - $3 = $1. Therefore, assuming that the outcome of the coin is revealed first, you will, with certainty, regret having payed any amount over $1 for this bet. This is not a rational decision procedure.
How about:
Consider $6 iff ((Green and Heads) or (Blue and Tails)). This is a known bet (1/3) so worth $2. But if the coin is flipped first, and comes up Heads, it becomes $6 iff Green, and if it comes up tails, it becomes $6 iff Blue, in either case worth $1. And that's silly.
Is that the same as your objection?
Yes, that is equivalent.
Very well done! I concede. Now that I see it, this is actually quite general.
My point wasn't just that I had a decision procedure, but an explanation for it. And it seems that, no matter what, I would have to explain
A) Why ((Green and Heads) or (Blue and Tails)) is not a known bet, equiprobable with Red, or
B) Why I change my mind about the urn after a coin flip.
Earlier, some others suggested non-causal/magical explanations. These are still intact. If the coin is subject to the Force, then (A), and if not, then (B). I rejected that sort of thing. I thought I had an intuitive non-magical explanation. But, it doesn't explain (B). So, FAIL.