Nice post! Here’s an illustrative example in which the distribution of p matters for expected utility.
Say you and your friend are deciding whether to meet up but there’s a risk that you have a nasty, transmissible disease. For each of you, there’s the same probability p that you have the disease. Assume that whether you have the disease is independent of whether your friend has it. You’re not sure if p has a beta(0.1,0.1) distribution or a beta(20,20) distribution, but you know that the expected value of p is 0.5.
If you meet up, you get +1 utility. If you meet up and one of you has the disease, you’ll transmit it to the other person, and you get −3 utility. (If you both have the disease, then there’s no counterfactual transmission, so meeting up is just worth +1.) If you don’t meet up, you get 0 utility.
It makes a difference which distribution p has. Here’s an intuitive explanation. In the first case, it’s really unlikely that one of you has it but not the other. Most likely, either (i) you both have it, so meeting up will do no additional harm or (ii) neither of you has it, so meeting up is harmless. In the second case, it’s relatively likely that one of you has the disease but not the other, so you’re more likely to end up with the bad outcome.
If you crunch the numbers, you can see that it’s worth meeting up in the first case, but not in the second. For this to be true, we have to assume conditional independence: that you and your friend having the disease are independent events, conditional on the probability of an arbitrary person having the disease being p. It doesn’t work if we assume unconditional independence but I think conditional independence makes more sense.
The calculation is a bit long-winded to write up here, but I’m happy to if anyone is interested in seeing/checking it. The gist is to write the probability of a state obtaining as the integral wrt p of the probability of that state obtaining, conditional on p, multiplied by the pdf of p (i.e. P(s1,s2)=∫P(s1,s2|p)f(p)dp). Separate the states via conditional independence (i.e. P(s1,s2|p)=P(s1|p)P(s2|p)) and plug in values (e.g. P(you have it|p)=p) and integrate. Here’s the calculation of the probability you both have it, assuming the beta(0.1,0.1) distribution. Then calculate the expected utility of meeting up as normal, with the utilities above and the probabilities calculated in this way. If I haven’t messed up, you should find that the expected utility is positive in the beta(0.1,0.1) case (i.e. better to meet up) and negative in the beta(20,20) case (i.e. better not to meet up).
Reflecting on this example and your x-risk questions, this highlights the fact that in the beta(0.1,0.1) case, we’re either very likely fine or really screwed, whereas in the beta(20,20) case, it’s similar to a fair coin toss. So it feels easier to me to get motivated to work on mitigating the second one. I don’t think that says much about which is higher priority to work on though because reducing the risk in the first case could be super valuable. The value of information narrowing uncertainty in the first case seems much higher though.
Nice example, I see where you’re going with that.
I share the intuition that the second case would be easier to get people motivated for, as it represents more of a confirmed loss.
However, as your example shows actually the first case could lead to an ‘in it together’ effect on co-ordination. Assuming the information is taken seriously. Which is hard as, in advance, this kind of situation could encourage a ‘roll the dice’ mentality.