On Tue, 2003-08-05 at 15:32, Joseph Halpern wrote: > The arguably more interesting version of Ellsberg's paradox has > balls of three different colors in the urn: 30 reds, and 60 that are > some combination of blue and yellow. A ball is drawn. > > In situation A, you get to choose between betting on red and betting on > yellow (you get $1 if you guess right and 0 otherwise). In situation B, > you get to choose between between on red+blue or betting on yellow+blue. > (If you bet on red+blue, you get a dollar if the ball drawn is either > red or blue). > > Most people choose to bet on red in the first case and blue+yellow in > the second. That's inconsistent with having a subjective probability on > the balls (no matter what your attitude is to risk). > > Note that, unlike your discussion below, there's only one decision. > There is no second decision. > > -- Joe
This paradox seems to centre on whether the urn is changed between bets. We bet on red for A because we assume the ratio chooser is minimizing the yellow count. For the same reason, if the situations are independent, then we'd rightly choose blue+yellow for B. If, however, if the urn is actually unchanged, then we should obviously choose red+blue for B (as we're guessing blue > yellow, thus yellow < red, thus blue + yellow < blue + red). Am I missing something? Does make you pause for a second or too though :) Tim.
