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June 24th, 2011

Equal Weight and Asymmetric Uncertainty

Not for the first time, I’m unsure what the Equal Weight View says about a case. Here it is.

Jack and Jill and Microsoft Excel

Jack and Jill both have the same evidence; they know that a paricular table has 83 rows and 97 columns. They both try to figure out how many cells are in the table. They know that the number of cells quals the number of rows times the number of columns. Jill concludes that it has 8051 cells. Jack can’t conclude anything about the number of cells. He thinks it might be 8051, but it might be something else. When they are each told about the other’s conclusions (or lack thereof), what should their final conclusions be?

A position half-way between Jack’s view and Jill’s view is presumably something like a view that the table probably has 8051 cells, but that it is a serious possibility that the table has a differen number of cells.

But surely Jill shouldn’t move her views in that way, should she? If she should hold firm here, is this just a counterexample to (some versions of) the Equal Weight View?

Posted by Brian Weatherson in Uncategorized

5 Comments »

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5 Responses to “Equal Weight and Asymmetric Uncertainty”

  1. Sophie Horowitz says:

    It seems to me that on some ways of filling out the case, Jill should lower her confidence that the answer is 8051. This is precisely because Jack’s uncertainty gives Jill some higher-order evidence to the effect that the problem was harder than she thought it was, and that it’s therefore more likely that she made a mistake. The more likely it is that she’s made a mistake, the less confident she should be in her answer.

    Consider what will happen if she doesn’t lower her confidence. She will be totally confident that the answer is 8051. But since Jack is her peer, it seems that Jill should still take his uncertainty as evidence that she messed up. Of course, it’s a math problem, so if she messed up, her answer is wrong. And she knows this. So she should think something like this: “The answer is 8051, but there’s a good chance the answer isn’t 8051.” That’s incoherent, and also seems a little crazy. So it looks like the EWV is getting the right answer here.

    But this is an interesting kind of case for other reasons. Jill’s epistemic state, if she lowers her confidence, is still strange. Let’s say she’s now .7 confident that the answer is 8051, and .3 confident that it’s something else. If the answer were something else, she should have gotten that answer to begin with. So if the answer were something else, she should be .7 confident in that other answer, and much less than .7 confident in 8051. Let’s suppose she knows all of this. Now, what should she think about the answer she got? She has .3 credence that the answer isn’t 8051, in which case her credence in 8051 should be less than .7. And she has .7 credence that the answer is 8051, in which case her .7 confidence is rational. So she should take it to be somewhat likely that her current credence in 8051 is too high, and not likely at all that her credence in 8051 is too low. This doesn’t necessarily seem like a counterexample to the EWV, but it’s an awkward conclusion nonetheless.

  2. Fritz Warfield says:

    I agree that the case is underspecified in important ways.

    Is this a case of peer disagreement?
    If so, why is it that Jill can do simple multiplication but Jack can’t? (as you say, Jack “can’t conclude anything about the number of cells”). I guess you didn’t specify that Jill got her answer by doing simple multiplication so I’m reading that in. If it was important that she not have arrived at her answer this way then I’ve missed something. Quite possibly I have missed something.

    But if I haven’t then this doesn’t look like a case of peer disagreement. And if it’s not a case of peer disagreement, then why would the equal weight view (a difficult to understand and very implausible view on the understanding I do manage to have of it) have to say anything at all about it?

  3. Brian Weatherson says:

    I was assuming Jack and Jill are peers, but I don’t se why that raises any doubts about how to apply EWV. After all, peerhood is a dispositional state. If Jack and Jill have the same prior probability of getting it right, they are peers. Now there are plenty of things that I can do some, but not all, of the times I attempt to do them. There’s no mystery at all about the possibility that on one occasion where I, and someone equally good (in general) at that task, attempt the task one of us succeeds and the other does not. If one’s level of ability at a task means that one succeeds at the task on some but not all occasions, then you’d expect asymmetric outcomes like this on a regular basis.

    If peerhood rules out that kind of possibility, there basically aren’t any peers anywhere in the world ever.!

  4. Fritz Warfield says:

    I’m actually sympathetic to the idea that peerhood is extremely rare — when peerhood is defined in the Tom Kelly way this seems straightforward (my former student Nate King has defended this ably in print).
    The sufficient condition you offer Brian, isn’t sufficient for peerhood on Tom’s way of defining peer. But maybe that means so much the worse for that way of defining peers.
    If we accept your sufficient condition (for some reason in my circle that condition is usually called “Elga’s condition”) we surely have to allow proponents of the EWV to still reject cases in which one “peer” stumbles so badly. If my peer notices that on this occasion, for whatever reason, that my basic math skills aren’t working and hers are, she needn’t give any ground to my view. This would treat the case like a performance error case that every view on peerhood acknowledges is a case in which there is no pressure to revise in light of a peer’s disagreement.

  5. Jeff says:

    So they have a prior belief that they’re peers. Let’s also assume that Jill has prior strong evidence that this is a relatively simple math problem, and that she is quite reliable at assessing and solving relatively simple math problems. That seems necessary to feed the intuition that we have that Jill shouldn’t update significantly toward Jack’s uncertainty.

    Given that explicitly made assumption, Jill should update her uncertainty just slightly; what she should update rather more is her probability that Jack is actually her peer. If we don’t hold fixed the strength of belief in Jack’s peerhood, I don’t see any problem with this example. But maybe an Equal Weight proponent would disagree. So much worse for the Equal Weight View.

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