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fool comments on A summary of Savage's foundations for probability and utility. - Less Wrong
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Comments (88)
That's right.
I take it what is strange is that I could be indifferent between A and B, but not indifferent between A+C and B+C.
For a simpler example let's add a fair coin (and again let N=2). I think $1 iff Green is as good as $1 iff (Heads and Red), but $1 iff (Green or Blue) is better than $1 iff ((Heads and Red) or Blue). (All payoffs are the same, so we can actually forget the utility function.) So again: A is as good as B, but A+C is better than B+C. Is this the same strangeness?
Not quite.
I think that the situation that you described in less strange then the one that I described. In yours, you are combining two 'unknown probabilities' to produce 'known probabilities'.
I find my situation stranger because the only difference between a choice that you are indifferent about and one that you do have a preference about is the substitution of (Green ∨ Blue) for (die ≥ 3). Both of these have clear probabilities and are equivalent in almost any situation. To put this another way, you would be indifferent between $3 unconditionally and $6 iff (Green ∨ Blue) - $6 iff Green if the two bets on coloured balls were taken to refer to different draws from the (same) urn. This looks a lot like risk aversion, and mentally feels like risk aversion to me, but it is not risk aversion since you would not make these bets if all probabilities were known to be 1/3.
Ohh, I see. Well done! Yes, I lose.
If I had a do-over on my last answer, I would not agree that $-6 iff Green is worth $-1. It's $-3.
But, given that I can't seem to get it straight, I have to admit I haven't given LW readers much reason to believe that I do know what I'm talking about here, and at least one good reason to believe that I don't.
In case anyone's still humouring me, if an event has unknown probability, so does its negation; I prefer a bet on Red to a bet on Green, but I also prefer a bet against Red to a bet against Green. This is actually the same thing as combining two unknown probabilities to produce a known one: both Green and (not Green) are unknown, but (Green or not Green) is known to be 100%.
$-6 iff Green is actually identical to $-6 + $6 iff (not Green). (not Green) is identical to (Red or Blue), and Red is a known probability of 1/3. $6 iff Blue is as good as $6 iff Green, which, for N=2, is worth $1. $-6 iff Green is actually worth $-3, rather than $-1.
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.