A bag contains one counter, known to be either white or black. A white counter is put in, the bag shaken, and a counter drawn out, which proves to be white. What is now the chance of drawing a white counter?
At first sight, it would appear that, as the state of the bag, after the operation, is necessarily identical with its state before it, the chance is just what it then was, viz. ½. This, however, is an error.
The chances, before the addition, that the bag contains (a) 1 white (b) 1 black, are (a) ½ (b) ½. Hence the chances after the addition, that it contains (a) 2 white (b) 1 white, 1 black, are the same, viz. (a) ½ (b) ½. Now the probabilities, which these 2 states give to the observed event, of drawing a white counter, are (a) certainty (b) ½. Hence the chances, after drawing the white counter, that the bag, before drawing, contained (a) 2 white, (b) 1 white, 1 black, are proportional to (a) ½ · 1 (b) ½ · ½; i.e. (a) ½ (b) ¼; i.e. (a) 2 (b) 1. Hence the chances are (a) ⅔ (b) ⅓. Hence, after the removal of a white counter, the chances, that the bag now contains (a) 1 white (b) 1 black, are for (a) ⅔ and for (b) ⅓.
Thus the chance, of now drawing a white counter, is ⅔.