Marilyn Savant's solution—that choosing a new door increases your odds from 1:3 to 1:2—doesn't seem very controversial to me. In a single instance it probably doesn't make a great difference, but I think if you played this game a number of times, the decision to change doors would yield better results.
This problem divides people into lateral thinkers (who choose to change doors) and vertical thinkers (who stick with their first choice). Lateral thinkers view the problem in time and in terms of two discrete sets of initial conditions and reason: "Under the first set of initial conditions my first choice yields a 1:3 probability, under the new set of initial conditions a change yields a 1:2". Vertical thinkers view the problem in space and say, "There was a 1:3 chance that my first choice was correct. Now there is a 1:2 choice that my first choice was correct—after all, either choice will be 1 of two doors!"
Personally, I would change my first choice.
Consider instead a bifurcation in time, which gives you two sets of doors. Set A consists of three doors (two bags of dogfood and one girl). Set B consists of two doors (one bag of dogfood and one girl). Which set of doors yields the greatest chance of getting the girl? Clearly, Set B. Making a decision under condition B yields a greater chance of getting the girl. If you allow this fact to have retroactive effect on the first choice (which was out of three doors) it is simply a case of choosing bewteen a 1 in 3 chance and a 1 in 2 chance.
Last edited by Hydra; 11/08/06 03:08 AM.
