Let’s assume that the odds you assign of the person telling the truth is greater than 1/3^^^^3. One thing that is clear is that if you faced that decision 3^^^^3 times, each decision independent from the others… then you should pay each time. When you aggregate independent decisions, it narrows your total variance, forcing you closer to an expected utility maximiser (see this post).
The odds I assign to the person telling the truth are themselves uncertain. I can’t assign odds accurately enough to know if it’s 3^^^^3, or a millionth of that or a billion times that.
Now, one typical reply to “the odds I assign are themselves uncertain” is “well, you can still compute an overall probability from that—if you think that the odds are X with probability P1 and the odds are Y with probability 1-P1, you should treat the whole thing as having odds of X P1 + Y (1-P1)”.
But that reply fails in this situation. In that case, if you face 3^^^^3 such muggings, then if the odds the first time are P1, the odds the second time are likely to be P1 too. In other words, if you’re uncertain about exactly what the odds are, the decisions aren’t independent and aggregating the decision doesn’t reduce the variance, so the above is correct only in a trivial sense.
Furthermore, the problem with most real-life situations that even vaguely resemble a Pascal’s mugging is that the probability that the guy is telling the truth depends on the size of the claimed genocide, in ways that have nothing to do with Kolomogorov complexity or anything like that. Precisely because a higher value is more likely to make a naive logician willing to be mugged, a higher value is better evidence for fraud.
The odds I assign to the person telling the truth are themselves uncertain. I can’t assign odds accurately enough to know if it’s 3^^^^3, or a millionth of that or a billion times that.
Now, one typical reply to “the odds I assign are themselves uncertain” is “well, you can still compute an overall probability from that—if you think that the odds are X with probability P1 and the odds are Y with probability 1-P1, you should treat the whole thing as having odds of X P1 + Y (1-P1)”.
But that reply fails in this situation. In that case, if you face 3^^^^3 such muggings, then if the odds the first time are P1, the odds the second time are likely to be P1 too. In other words, if you’re uncertain about exactly what the odds are, the decisions aren’t independent and aggregating the decision doesn’t reduce the variance, so the above is correct only in a trivial sense.
Furthermore, the problem with most real-life situations that even vaguely resemble a Pascal’s mugging is that the probability that the guy is telling the truth depends on the size of the claimed genocide, in ways that have nothing to do with Kolomogorov complexity or anything like that. Precisely because a higher value is more likely to make a naive logician willing to be mugged, a higher value is better evidence for fraud.