[mcprwklzpq]

This was not what bothered me. I needed to see how this puzzle does not contradict my intuition that if i flip a coin to choose between two last doors i do get 1/2 chance of winning whatever are the chances for each door. So with 1000 doors: (1/1000)/2 + (999/1000)/2 = 1/2 chance of winning. And in general (A+B)/2 = 1/2 if A+B=1.

[AmericanChopper]

If you flip a coin at the decision stage, then you would have a 50/50 chance of winning. But that approach would mean you’re disregarding information that you can use to increase your chances of success above 50/50. When you make your initial choice you have a 1/1000 chance of guessing correctly, meaning there is a 999/1000 chance that the prize is behind another door. By eliminating 998 doors from contention, you’re not changing those odds. There is still a 999/1000 chance for the prize to be behind a door other than the one you initially picked. Only now, instead of having 999 doors to choose from, you only have one. So by switching your choice, you increase your chance of success from being whatever it was initially, to being (1 - whatever it was initially). Which will always be greater than 0.5.

This is materially the same as being given a choice between opening one door randomly, or being able to open every other door. Because the conditions of opening doors prior to the final choice is that the door you initially picked can’t be opened at this phase, and neither can the winning door.