Originally posted by: pinion9
You are absolutely wrong with your example because it becomes a new game. With probability like this you cannot take into account past performance or tries; you must view it as a new game altogether. In your example with the box you have two remaining, so there is absolutely a 50/50 chance of getting the one empty box.
Each time the round ends you put your box back into the pile, but you MUST choose the same box each time. Yes, overall you have a 1/10 chance of picking the box with money. However, with only two boxes left, at that very moment, there is a 50/50 chance of you having the money. There are two boxes, one has money. 50/50 chance.
As for taking the door, you are wrong. You open 1 door of the 3 and now you have 2 doors left. You pick 100 random people to open one of the remaining doors and the distribution will be about 1/2, not 2/3.
If you flip a coin nine times and get heads each time, what is the probablity of getting a heads on the next flip? --- Exactly 50% because past performance matters none and you play the game at that very instant.