Saturday, February 5, 2011

To Thwart a Terrorist

In "Rational Choice", we have been learning about probability, one of the elements of which is dependence. It is important to know whether a second event is dependent upon the outcome of the first.

Rolling 6s again and again is for not especially unlikely.
For example, suppose I rolled a die. What are the chances of rolling a 6? If the die is fair, the answer is 1/6. I roll the die, and I get a 6. Now I am going to roll the die a second time. What are the chances of getting another 6?

Some might want to say that the chances are pretty low of getting another 6. After all, I just got one, and it's a fair die, so it's probably due for a different one of its equally-probable numbers. Besides, the chances of rolling two sixes in a row are pretty slim, right?
Wrong. The first roll has nothing to do with the second. The die isn't keeping track. It's just as likely that it rolls two 6's as it is that it rolls a 6 and a 5, or a 6 and a 4. The die has no interest in switching things up.

So these trials are independent in this example. Cool.

Once the card is removed, the odds change.
Other examples, of course, can have dependence. Suppose, for example, I have a deck of cards. What are the chances of drawing an Ace from the deck at random? There are 4 aces, 52 cards, so the chances are 4/52 (about 8%). Let's say I draw an Ace, and put it off to the side. Now, what are the chances of drawing a second Ace?

Well, if you're learning from the dice example, you'll be inclined to say, "It's the same chances. The cards don't care what you drew last time." But that's not true. The cards do care, because the Ace you drew last time was removed from the pile. Now there are only 3 Aces in the deck, and only 51 cards from which  to select. Your chances are 3/51, or roughly 6%. The odds went down.

Of course, you're only reading this for the fun part, as alluded to by the title. Fair enough. The textbook uses the following amusing example to illustrate this fact of probability:

Suppose your friend needs to get on a plane, but he's afraid that a terrorist might bring a bomb on board. How can you persuade him that it is safe? You try telling him that the chances of a terrorist with a bomb being on his plane are very very slim, but that doesn't convince him.

"Alright," you say, "Would you agree that is is almost impossible that two people would take bombs onto your plane?"

"Sure."

"Then you should bring a bomb onto the plane. The risk that there would be another bomb will then become negligible."

The joke makes an excellent point. It's obvious to us that bringing our own bomb won't stop the terrorist. That's because the two events are independent. Sure, we think they're unlikely to co-occur, but they won't actually affect one another. So once you've rolled one six, or flipped three heads, or scratched off twenty-seven "Sorry, You Are Not a Winner" tickets, there is absolutely no reason to believe that your next result will be any different.

3 comments:

  1. A king has four prisoners. He will kill them tomorrow morning, but gives them a chance to live; the prisoners must come up with a way to outsmart the king.

    The way in which the prisoners die is:
    1) All the prisoners line up, front to back.
    2) Each prisoner gets a hat, white or black.
    3) Each prisoner must guess the color of his/her hat without looking up.
    4) The prisoner must announce his/her guess out loud.

    Black kills, white lives.

    The prisoners have one night to come up with a system to have as many survivors as possible.

    What is the best code/system for the prisoners to adopt in order to survive?

    (Hint: The least number of killed prisoners is 1.)

    ReplyDelete
  2. 1. No,
    2. I made a mistake, he must guess correctly to live,
    3. Yes.

    ReplyDelete
  3. Are they allowed to communicate once the hats are on? Talk, nudge, anything?

    ReplyDelete

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