Can you help with this statistics problem?

Question

The following question was posed many years ago in a newspaper column by Marilyn Vos Savant,
"Suppose you're on a game show, and you are given a choice of three doors. Behind one door is a car; behind the others, goats. You pick a door--say, No. 1--and the host, who knows what's behind the doors, opens another door--say, No. 3-- which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"

I know that yes, you should switch because the probability is 2/3 that it is in the other door. (If you don't agree, check out http://www.marilynvossavant.com/articles/gameshow.html for the full explination.)

How do I answer the questions at the end knowing that the probability is 2/3?

How many times must you simulate or play the game in order to be ninety-five percent confident of predicting the probability of winning the car: 1) if the contestant has decided to switch? 2) if the contestant has decided not to switch?

Answer 1

Vos Savant was wrong. There's no statistical reason to switch. This has been disproven already, when there are two doors left and you have one of the two, you have the same 1/2 chance of getting the car whether you switch or not.

The choice between two doors does not remember the previous choice of three doors. her million doors example destroys her case. The odds on a given decision are based on the number of options for THAT particular decision, not the original number of options as Vos Savant repeatedly, incorrectly, assumes.

Answer 2

I don't need to perform any simulations because I took a statistics course and understand the principles behind the answer.
I did simulate the situation in MS EXCEL 16000 times and discovered that by the time 139 simulations had been completed the ratio of the count of wins by switching to total attempts was never more than 5% different than the anticipated 2/3 chance of winning.