>>14087273wrong you fucking retard
prior to picking a door the probability of picking the door with the car behind it is always 1/3
let there be three doors A through C
assume you pick A
this then leaves monty with two options: he can either open door B or C, the probability of which is 1/2
let us assume that the car is behind door C; now monty must open door B, the probability of which is 1
in either case you multiply the probabilities to find the normalizing constant, in this case 1/2
therefore the odds for you sticking with your original pick, door A, given monty opened door B, are:
(1/3*1/2)/1/2 = 1/3
the odds if you switch to door C, given monty opened door B:
(1/3*1)/1/2 = 2/3
suck my dick
