De Gruyter Advent Calendar | Answer Door 2
Solution according to Dodgson (1893)
Dodgson assumes that the first counter in the bag is equally likely to be white or black.
In his solution, Dodgson first points out a common misconception. Because a white counter has been placed in the bag and a white counter removed from it, it could be assumed that the original state has been restored and that the probability of the counter left in the bag being white once a counter has been removed is therefore still ½.
According to Dodgson, this assumption is incorrect.
He describes the correct solution as follows:
Once the white counter has been inserted into the bag, the bag has, with equal probability, either two white counters or one white counter and one black counter.
In the first case, the probability of withdrawing a white counter is 1, in the second case ½.
The probability of the counter in the bag being white is therefore double the probability of it being black. The probability of the counter still in the bag being white is therefore 2/3.
The correct solution is therefore c).