OK.. after seeing the results, I was wrong too...
Try this one though (taken from http://www.jimloy.com/puzz/monty.htm):
Martin Gardner's version, published in October 1959, involved three condemned prisoners, one of whom will be pardoned at random. One prisoner cons the warden into naming one of the other prisoners (other than the prisoner who is asking this of the warden) who will not be pardoned. Do this prisoner's (the one talking to the warden) chances of being pardoned then go up to 50%? This is identical to the Monty Hall trap, and this prisoner's chances are still 1/3, but the probability that the third prisoner will be pardoned have gone up to 2/3.
The correct term for a thousand million is a milliard.
However, this has never actually caught on for reasons that are beyond me.
The "American" system seems to have become standard.
Not only is it easier for people who are less mathematically inclined to use,
it also makes profits sound bigger. ;-)
Although they are usually referred to as the American and British numeration systems,
both were actually invented by the French.
I, for one, believe that it should only advance to the higher "term" when it becomes
necessary to repeat a particular term twice in a row.
100,000 = One hundred thousand
1,000,000 = One thousand thousand = One million
1,000,000,000 = One thousand million
1,000,000,000,000 = One million million = One Billion
10 ^ 100 = Ten thousand quintillion
What Martian says is 100% correct.
I've seen an example where there are 100 doors with 1 car and 99 goats.
Now, you choose a door, then the host reveals 98 doors which do not contain the car.
This leaves 2 doors: the door you chose and 1 other.
The probablity of initially choosing the door with the car behind it is 1 in 100.
On its own the above statement is true. However, we know for a FACT that
the host will always narrow the doors down to 2, while always avoiding the door with the car behind.
With this added information the statement now becomes false. Why?
Because we will always end up with 2 doors, 1 with the car behind it, and 1 with a goat behind it.
Your initial choice is irrelevant.
Thus, the probability of initially choosing the door with the car behind it,
with the knowledge that the host will reveal 98 incorrect doors, is 1 in 2!
The irony of the question in the original post is any fool will say 50/50 for their first answer... and be correct!
It is to be noted that this only works because of the rules of the game.
Were it random what doors were revealed, it would be an entirely different situation.
In this case, the car could be revealed at any time and every time a car is not revealed increases the chance that the door you initially chose contains the car.
Anyway, the trick is not to dwell on it too much, or you may find yourself looking for answers which do not exist!