one should never be required to strictly c-prefer a mixture of A and B to C whenever neither A nor B is c-preferred to C
There are candidate counterexamples to this claim. For example, imagine A is giving a benefit to Amy, and B and C each designate the same action of giving a benefit to Bobby. Then if you’re impartial, you won’t strictly c-prefer either of A or B to C, but you might strictly c-prefer the 50:50 mixture AB to C on the basis that it’s fairer to randomize who gets the benefit.
Also, imprecise consequentialism (plus Dissent, Unanimity, and Justification) has an even more counterintuitive implication than ‘you can be required to strictly c-prefer a mixture of AB to C even though neither A nor B is c-preferred to C.’ It implies:
You can be required to strictly c-prefer a mixture of A, B, and C to D, even though (i) neither A nor B is c-preferred to D, (ii) C is c-dispreferred to D, and (iii) the mixture has an arbitrarily high probability of resulting in C.
Here’s an example to illustrate:
I made the mixture have a 60% chance of C just to avoid the diagram being all bunched up. But the steeper you make the diagonals A and B, the higher you can push the probability of C and yet still have the mixture dominate D.
Extra stuff:
There are candidate counterexamples to this claim. For example, imagine A is giving a benefit to Amy, and B and C each designate the same action of giving a benefit to Bobby. Then if you’re impartial, you won’t strictly c-prefer either of A or B to C, but you might strictly c-prefer the 50:50 mixture AB to C on the basis that it’s fairer to randomize who gets the benefit.
Also, imprecise consequentialism (plus Dissent, Unanimity, and Justification) has an even more counterintuitive implication than ‘you can be required to strictly c-prefer a mixture of AB to C even though neither A nor B is c-preferred to C.’ It implies:
You can be required to strictly c-prefer a mixture of A, B, and C to D, even though (i) neither A nor B is c-preferred to D, (ii) C is c-dispreferred to D, and (iii) the mixture has an arbitrarily high probability of resulting in C.
Here’s an example to illustrate:
I made the mixture have a 60% chance of C just to avoid the diagram being all bunched up. But the steeper you make the diagonals A and B, the higher you can push the probability of C and yet still have the mixture dominate D.