I understand that. But you said the court tested the witness's reliability under the same conditions as those to which the witness testified. So the mistake is more significant in the real experiment (the testimony) than in a 50:50 distribution. I get that. But it's also more significant, and by the same margin, in the control experiment (the court's evaluation of the witness). Given that the distribution of mistakes is skewed in identical ways in both the experiment and the control experiment, the percentage of correct identifications should be the same in both, too.
how's this, then: the court observed the witness as 100 taxis drove by, with an 85-15 distribution, and the witness identified each color with 80% accuracy. therefore, 80% of 85 = 68 & 80% of 15 = 12 cars were identified correctly, for a total of 68 + 12 = 80 cars as expected. now, if the witness were to see 15 blue cars in a row, 12 would be identified as blue, but if he sees 85 green cars in a row, then 85 - 68 = 17 would also be (mis)identified as blue. the 80% accuracy still holds, but because one color is so dominant over the other, the misidentification of the dominant color outweighs the correct identification of the infrequent color.