Blog: Plutonium

Posted by Plutonium on 28th July 2012 at 09:03 (4305 Views)
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JUL
28
2012
Monty Hall II
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Well, there have been some interesting comments on my last post, but what I think we need to do, is review the situation.

In the Monty Hall problem, most people's reasoning would be: there are two remaining closed doors, so there is a 50% chance that the car is behind one or the other. Therefore, the your chances of winning the car are not improved by swapping doors, they stay the same.

In reality, there is still only a 1/3 chance that you picked the right door to start with, and a 2/3 chance that the car is behind the one remaining closed door. So you double your chances by swapping doors.

Many people still find that they cannot accept this argument. However, it becomes more obvious if there are more doors.

For example, suppose there are 10 doors. You choose one. The compere opens 8 of the other doors, revealing goats, leaving only two doors still closed, the one you chose, and one other door.

(BTW in this scenario it is to my mind self-evident that the compere knows which door the car is behind, otherwise there is a 4/5 chance that he will inadvertently open the door that has the car.)

Now, there is still only a 1 in 10 chance that you chose the right door. It follows that there is a 9 in 10 chance that the car is behind the one remaining closed door.

In that scenario, do you stick or swap? It's your call.
Updated 28th July 2012 at 09:04 by Plutonium
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  1. Evil -
    JUL
    29
    2012
    Evil's Avatar
    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.
  2. Plutonium -
    JUL
    29
    2012
    Plutonium's Avatar
    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.
  3. Plutonium -
    JUL
    29
    2012
    Plutonium's Avatar
    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.

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