[EMAIL PROTECTED] wrote:
>Interesting point. Yes, if the Ss do something other than a random guess
>the binomial model would be violated. The question then becomes what
>would they do if they are uncertain? I suspect that they would fall back
>on visual inspection...which piece appears to be different than the others
>(less green pepper, more browned, etc) Such information is probably
>relevant often enough that "guessing" would be well above 1/3.
>
>Using blindfolded Ss will deal with that problem, and gets us back to
>the question that Dennis is asking. I'm guessing that rather than going
>through some sort of a systematic process (e.g. binary decision for the
>first piece, progress to second piece only if first piece was judged
>"same".....) Ss will in fact do something more like guessing. Only they
>will condition their guesses such that if they picked slice A as different
>on the previous trial they will first consider slices B and C on the
>current trial (they will actually avoid selecting the same slice position
>on sequential trials). Furthermore they will try to equalize the number
>of position choices they make across the experiment so that they choose
>each of A, B, and C three times and one of those a fourth time.
>
>This leads to: trials are not independent.
>
>The question remains, are the trials not independent in any way that
>matters for the purposes of binomial probabilities? Off the top of my
>head, I'm not sure. I am aware of, through secondary sources, studies
>where any amount of guessing about random events (eg coin tosses) results
>in a lower number of correct outcomes than simply making the same choice
>every time....that is always choosing "tails" will be superior to making
>choices between "heads" and "tails" on different trials.
>
<<<snip>>>
Michael,
can you provide references for those studies which show that '...always
choosing "tails" will be superior to making choices between "heads" and "tails"
on different trials.' For a "fair" coin-tossing situation, I am at a loss as
to how the expected value of proportion of correct choices could be different
from .5 using either strategy.
Dan Nordlund
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