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Sergio Tenenbaum
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Manuscript Workshop Announcement: A. J. Julius, “Reconstruction” The University of Toronto Centre for Ethics will be hosting a workshop on A. J. Julius’s manuscript “Reconstruction” on Friday, June 6, from 11 to 6. See below the fold for more details.... Continue reading
Posted May 15, 2014 at PEA Soup
Maxims and MRIs: Kantian Ethics and Empirical Psychology A two-day workshop to be held at the University of Toronto, Centre for Ethics, Toronto, ON, Canada May 9-10, 2014 In recent years, many moral philosophers have drawn inspiration from the exciting... Continue reading
Posted Apr 6, 2014 at PEA Soup
Hi Doug: No, I think that if you allow preference sets with no upper bound, you will have other cases in which you need to allow that it would be rational to choose counterpreferentially. At any rate, I find it very intuitive that if ST stops at a point that is not too far in the series she acts rationally (and if she goes to the end she acts irrationally). And if accepting (R3) commits you to denying this, it seems to me a high cost (but, of course, you might have independent reasons to deny the consistency of the ST scenario as I am describing it).
Hi Doug: I guess I don't understand why would we say that intransitive preferences are rational if we then conclude that an agent that chooses on the basis of such preferences is irrational no matter what she chooses (is this what you are proposing?). If your preferences put you in a position that whatever you do is irrational, then I would say that they are not the preferences of a rational agent. Generally when people say that some sets of intransitive preferences are rational, they mean that an agent with such a set of preferences could still choose rationally. I am also not sure why you think that the probabilistic case is more interesting. Of course, you could reproduce the same structure with probabilities, if the probabilities of moving up too far are never large enough to offset the gains of continuing to the next stage.
Hi Doug: Here are the things that ST might intend at N. I assume that this is choice under certainty so the prospect is just the state of affairs in parenthesis. (1) Stop at N (Pain at level N &$X) (2) Continue at N then stop at N + 1 (Pain at level N +1 (indistinguishable from N), $X & 100,000) (3) Continue at N and N + 1, stop at N + 2 (Pain at level N + 2 (indistinguishable from N + 1) & $X + 200,000) . . . On Quinn's proposal, it is rational to choose (intend to perform) act (1) even though (2) is preferred to (1). This seems to be a straight violation of (R3). If the reply is that if ST really prefers (2), she should choose (2), then, by parity of reasoning, she should choose (3) as she prefers (3) over (2), and so forth. Given that the preferences are intransitive, for every act she intends to perform, there is an act that you prefer over the one you intended to perform. Your reply to Chrisoula seemed to assume that the choice must be between only two alternatives, but I don't know why we should restrict the choice set in this manner. I take it that "by following one's intransitive preferences", Chrisoula means that you choose according to preference at each choice node. One could have the view that ST has intransitive preferences but it is perfectly rational for her to go all the way to the last setting (and also to switch back to the first setting and no money if she is later given the option). Or one could have the view that intransitive preferences are not rational (or not even possible). But I agree with Chrisoula that if you reject both these views, it'll be hard to endorse (R3).
I don't think you need any such assumption. The ST can rationally believe that he is just as likely to stop at N + 1 as he is to stop at N, but this couldn't suffice to make it the case that it is rational to continue, because the argument will generalize, and would have as a consequence that it is rational to go to the last setting. I have troubles assessing objective probabilities in cases we are assuming that ST is rational. If ST is rational, and Quinn is right, the OP that he will stop at N is 1. If we now look at the counterfactual possibility that he wouldn't, then I would guess that insofar as ST is rational and Quinn is right, the OP he'll stop at N+1 is again 1 (since this is the closest to the original plan). But you don't even need it to be 1, it could be just arbitrarily large and either you conclude that it is rational to stop at N or that the only rational place to stop is the last setting. At any rate, many solutions to the ST puzzle conclude that ST must choose counterpreferentially, which would be in violation of R3. I think these are the only plausible solutions. In general, other solutions will be committed to there being a most preferred outcome in the series (at least for a rational ST), which rejects the initial setup of the puzzle and, to my mind, arbitrarily restricts what can count as a rational set of preferences. Of course, not everyone agree with me here. But it is at least not obvious that any plausible way of dealing with the ST puzzle and similar cases will be compatible with R3.
I forgot to say: "relative to this set of options" just meant to leave open the possibility that if somehow ST got unhooked and then faced the same scenario, it would be perfectly rational for her to choose to stop at a different setting.
Good question! Here is a possibility (but I haven't looked back at the Quinn, so I might be completely wrong). Couldn't we say that Quinn takes the lesson of the ST puzzle to be that you cannot read off that X is worse than Y from X is dispreferred to Y even when "worse than" is being read instrumentally (exactly because preferences can be rational but nontransitive, while "worse than" is transitive)? So let us say that ST decides to stop at N before the whole process begins. If this is the case, then, relative to this set of options ( I am understanding 'set of options' aggregatively here, which I don't think it is possible in your third interpretation), N + 1 is worse than N, even though N is dispreferred to N + 1. Had ST chosen to stop at N + 1, than N would be worse than N + 1. This depends on a "pick and stick to it" solution (which I believe Quinn favours), but I think it could be adapted to other types of solution.
My colleague Jonathan Weisberg has created an excellent free iPhone app. Among other things, the app allows you to run informal surveys, run intuition polls, etc. I was one of the beta testers and I can say that the interface... Continue reading
Posted Nov 20, 2012 at PEA Soup