Blog: Plutonium

There was a 1 in 10 chance that you chose the right door initially. But since 8 doors have been essentially removed from play, shouldn't it be a 50% chance now? I don't see how there could be a 9/10 chance that it is behind the remaining door since 8 of those doors have been removed from the equation.

The chance that you chose the right door initially, is still only 1 in 10. The chance that it is behind one of the other 9 doors is 9 in 10. Since 8 of those doors have been opened, that chance has transferred to the 9th door.

Think about it in terms of set theory. There is the set of doors you have chosen, and the set of doors you haven't chosen. There is only 1 door in the set of doors you have chosen, and 1 in 10 chance that the car is behind it. There are 9 doors in the set of doors you haven't chosen, and a 9 in 10 chance that the car is behind one of them. If 8 of those doors are opened, then there is still a 9 in 10 chance that the car is behind the set of doors you haven't chosen. Only if the 9th door is opened to reveal either a car or a goat, does the probability change to 0 or 1.

If people still aren't convinced, there is a test you can do, although it requires two people.

Ask a friend to lay out 10 cards face down. One of the cards must be the Ace of Spades, and your friend must know which one it is.

You choose a card. Your friend then turns up eight of the cards, leaving the card you have chosen, and one other card, face down. They must know the location of the Ace of Spades, to avoid turning this up, if it is not the card you have chosen.

Now you are asked if you want to stick with your original card, or switch to the one remaining card.

Repeat this exercise several times, and see how often you get the Ace of Spaces when you (a) stick, or (b) swap.