Crazy Hall Problem

The probability of you picking a goat the first time is $\frac{99999}{100000}$ , which is 99.999%. Then, all other goat doors are revealed. The probability of you then switching to get the car is (after having picked a goat) is 100%.

The probability of you picking the car the first time is $\frac{1}{100000}$ , which is 0.00001%. Then, 99998 out of the 99999 goats are revealed. Hence, the probability of you switching to get the car (if your first choice was a car) is 0%.

Hence, if you are definitely switching to the other door after picking an initial door, the only way to get the car is to first pick a goat, which is 99.999% likely. Otherwise, you could pick the car and you wouldn't get the car by switching, but this is only 0.00001% likely.

Hence it is 0.00001% likely to not get the car, and 99.999% likely to get the car.

This is a play on a famous stats question of three doors, two with goats behind them and one with a car. It's simply an expanded version of that problem.

The trick is to note that the probability does not change if more doors are opened. In the original, there is a $\frac{2}{3}$ chance that you will pick a goat. If the other goat is revealed, no matter which goat you picked, the car will be in the third door, causing a $\frac{2}{3}$ chance that switching doors will give you a car.

The same principle applies here. With 100,000 doors to choose from, there is a $\frac{99999}{100000}$ chance that you will pick a goat. When every other goat is revealed, there is now a $\frac{99999}{100000}$ chance that the final door is the car.