@Ashrem reread the problem. It entirely hinges on Monty revealing that 1 of the 2 doors not picked does not contain the prize. That leaves the player with a 50/50 situation. Even if there was a 'good' prize and a 'not so good' prize the player will still have a 50/50 choice. It does not matter what they chose the first time and the probability does not change.
I'm fully aware of the detail of the exercise. You are likely making the most common error of looking at the event as a singularity.
If 10,000 people pick from A, B, and C, where there is only a prize in C, ~3,333 are going to end up with the prize. At that point in time, If you eliminate either A or B (depending what the person selected), what you are actually doing is selecting ~3,333 people who have a prize, and ~6,666 people who don't have a prize. (it kind of sounds impossible until you consider it is being done
one person at a time). So when he eliminates one wrong choice, at that moment 2/3 people have the other wrong choice. If everyone of the 10,000 picks the other door, 2/3 are picking right and 1/3 is picking wrong.
The individual's odds at that moment are the same, but the aggregate answer is that there's a 66% chance of improving your position.