Probability of carrying an autosomal recessive trait given both parents are carriers
Monty Hall does not apply here because there is no decision point past where information is given: the information is already known that the mother is normal phenotype, and the probability for each child of the mother's parents is independent; in the Monty Hall problem, it is known that exactly 1 of the doors contains the prize. Because this sum is known (that only 1 door has the prize), if you know the contents of one door it changes the probability of the others. That is, the occurrence of the prize/goat is sampled without replacement: the thing behind each door is dependent on what is behind the other doors.

2/3 is the correct probability for the normal phenotype mother to be a carrier. Therefore the probability for a child to be homozygous recessive is (2/3) * (1/2) * (1/2) = 1/6.