# endoself comments on A summary of Savage's foundations for probability and utility. - Less Wrong

33 22 May 2011 07:56PM

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Comment author: 29 October 2011 05:07:45AM 0 points [-]

1. In the hypothetical world where all probabilities that you were asked to bet on were known, would you be a Bayesian?

2. How stable is your preference for Knightian risk over uncertainty? In other words, how much more would winning on green have to be worth for you to prefer it to red (feel free to interpret this as much as is necessary to make it precise)?

Comment author: 04 November 2011 01:23:27AM 1 point [-]

I'm not really clear on the first question. But since the second question asks how much something is worth, I take it the first question is asking about a utility function. Do I behave as if I were maximising expected utility, ie. obey the VNM postulates as far as known probabilities go? A yes answer then makes the second question go something like this: given a bet on red whose payoff has utility 1, and a bet on green whose payoff has utility N, what is the critical N where I am indifferent between the two?

For every N>1, there are decision procedures for which the answer to the first is yes, the answer to the second is N, and which displays the Ellsberg-paradoxical behaviour. Ellsberg himself had proposed one. I did have a thought on how one of these could be well illustrated in not too technical terms, and maybe it would be appropriate to post it here, but I'd have to get around to writing it up. In the meantime I can also illustrate interactively: 1) yes, 2) you can give me an N>1 and I'll go with it.

Comment author: 07 November 2011 02:52:29AM *  0 points [-]

Okay. Let N = 2 for simplicity and let \$ denote utilons like you would use for decisions involving just risk and no uncertainty.

P(Red) = 1/3, so you are indifferent between \$-1 unconditionally and (\$-3 if Red, \$0 otherwise). You are also indifferent between \$-3 iff Red and \$-3N (= \$-6) iff Green (or equivalently Blue). By transitivity, you are therefore indifferent between \$-1 unconditionally and \$-6 iff Green. Also, you are obviously indifferent between \$4 unconditionally and \$6 iff (die ≥ 3).

I would think that you would allow a `pure risk' bet to be added to an uncorrelated uncertainty bet - correct me if that is wrong. In that case, you would be indifferent between \$3 unconditionally and \$6 iff (die ≥ 3) - \$6 iff Green, but you would not be indifferent between \$3 unconditionally and \$6 iff (Green ∨ Blue) - \$6 iff Green, which is the same as \$6 iff Blue, which you value at \$1.

This seems like a strange set of preferences to have, especially since both (die ≥ 3) and (Green ∨ Blue) are both pure risk, but it could be correct.

Comment author: 08 November 2011 03:14:17AM 1 point [-]

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?

Comment author: 09 November 2011 09:46:27PM *  0 points [-]

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.

Comment author: 10 November 2011 07:17:37PM 1 point [-]

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.

Comment author: 11 November 2011 04:41:25AM *  0 points [-]

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.

Comment author: 11 November 2011 06:14:46AM *  0 points [-]

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.

Comment author: 11 November 2011 05:05:10PM *  0 points [-]

Wouldn't you then prefer \$0 to \$1 iff (Green ∧ Heads) - \$1 iff (Green ∧ Tails)?

Comment author: 11 November 2011 08:14:05PM 0 points [-]

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.

Comment author: 07 November 2011 03:33:24AM *  1 point [-]

You are also indifferent between \$-3 iff Red and \$-3N (= \$-6) iff Green (or equivalently Blue).

Isn't this -1 and -4, not -1 and -1? I think you want -3/N = -1.5.

Comment author: 07 November 2011 01:37:01PM *  0 points [-]

I'm not quite sure what you're first sentence is referring to, but fool prefers risk to uncertainty. From his post:

given a bet on red whose payoff has utility 1, and a bet on green whose payoff has utility N, what is the critical N where I am indifferent between the two?

. . .

you can give me an N>1 and I'll go with it.