Here’s a little brainteaser to think about if the Super Bowl ends up being a bit dull. Imagine that you are standing at the baggage carousel at an airport waiting for your bag. A percentage x of the bags from your flight have already appeared on the carousel and yours is not among them. How large does x have to be before there is a probability greater than one half that your bag has been lost by the airline?
Of course, we need to make a few assumptions before we can attempt a proper mathematical analysis of the situation.
First, you can assume that the airline loses two percent of all the bags that it handles. According to this 2006 article from the USA Today, this figure is roughly accurate for the least reliable U.S. airlines. The industry average is closer to .63 percent, but that number is ugly and harder to work with.
We will also assume that your bag is as likely to occupy any place in line as any other. I think it also simplifies the computations if you assume there are exactly 100 bags. If you number the bags from one to 100 with one being the first bag to appear and one hundred being the last, then the number assigned to your bag is chosen randomly from the numbers between one and one hundred.
Using Bayes’ Theorem I’ve managed to cook up my own little formula for the probability that your bag has been lost given that x percent of the bags have appeared without yours being among them. My formula gives the correct answers for the extreme cases. That is, if all of the bags have appeared then my formula says there is probability one that your bag is not there. And when none of the bags have yet appeared my formula reutrns a probability of 1/50, or two percent, exactly as it should be.
Still, the answer it gives me to the specific question I posed above seems a bit hard to believe. So I figured I would turn it over to all of you. Let me know what you come up with!