Mishihari Lord said:
The answer to the dragon question seemed odd to me so I looked up some references and reviewed the proofs and I'm pretty sure the ones I saw were wrong.
Let there exist X doors. One door is the correct choice.
The chance of picking the correct door is 1/X. The chance of picking the incorrect door, therefore, is 1-1/X.
After you have chosen a door, the set of X doors may be viewed as two subsets: "your door," and "all other doors." The chance that subset "your door" is correct is 1/X. The chance that subset "all other doors" is correct is 1-1/X.
For any X greater than 2, the chance that subset "all other doors" is correct is greater than the chance that subset "your door" is correct.
By way of example, let X = 100.
When you randomly choose a door, there is a 1 / 100 chance that you picked correctly (subset "your door"). There is a 99 / 100 chance that you picked incorrectly (i.e., 99% chance subset "all other doors" is correct).
When the non-winning doors are removed, you still have the two subsets: "your door" vs. "all other doors." "All other doors" still has a 99 / 100 chance to be correct.
In other words, switching works because you are more likely to be wrong on your first choice than right on your first choice.