From: Ray Vickson Subject: Re: The Hitchhiker Problem Date: Thu, 10 Aug 2000 15:33:39 -0400 Newsgroups: sci.math,rec.puzzles Summary: [missing] David Petry wrote: > Here's a neat problem which I thought of while hitchhiking. > At least I think it's neat. I also think it'd be neat to see > Marilyn vos Savant tackle the problem, but that's a whole > different subject. > > I'd be interested to know if this is a variant on a standard > puzzle. Hardly a "puzzle", but rather a deep probabilistic result in Renewal Theory. In fact, I have heard it said that many early probability theory papers have erroneous results because they relied on the common sense type of arguments you outlined, which happen to deliver wrong answers. Nowadays, we understand better what's going on. Let's change it from hitchhikers to lightbulbs (i.e., how long has the current lightbulb been in operation). This type of question belongs to the general topic of the "Inspection Paradox": if an average lightbulb burns for 1000 hours, and I walk into a room at a random time, how long does the current lightbulb have left to go? The answer depends on the probability distribution of lightbulb lifetimes, but you get a particularly simple answer in the case of exponentially-distributed lifetimes. The answer in this case is: still 1000 hours left to go. Interestingly, the average age of the current lightbulb is also 1000 hours. This means that the particular lightbulb that just happens to be installed in the room when I walk in has an average total life of 2000 hours (even though I bought it at a hardware store that sells bulbs having 1000-hour average lifes). I discuss this with students in my probability modelling classes, and some of them find it hard to believe; more precisely, they believe each single step in the argument, but find it hard to believe where they end up. I get them to do a simple simulation in EXCEL, and presto!--it's true. Whether this has any relevance to the hitchhikers problem is problematic, though. > > > It's a multiple choice question. > > Let's say that as an experiment, you pick up hitchhikers, > and every time you ask the hitchhiker how long he has been > waiting for a ride. You collect all the data, and find that on > average, the hitchhikers tell you they have been waiting 20 > minutes. > > The question is, how long does the average hitchhiker > wait for a ride? > > A) 40 minutes, because if you didn't pick up a given hitchhiker, > he would, on average, have to wait an additional 20 minutes > beyond the 20 minutes he's already waited. (In other words, > by picking him up, you're effectively cutting his expected waiting > time in half). > > B) 10 minutes, because the chances of coming across a > hitchhiker who has to wait a long time for a ride are much greater > than the chances of coming across a hitchhiker who gets picked > up quickly, thus skewing your data in favor of long waits, and a > simple calculation shows that the real figure is half of what your > data suggests. > > C) 20 minutes, because you are no different from anyone else who > picks up hitchhikers, so if everyone collected the same data, they > would all get the same result, which implies that on average, > hitchhikers wait 20 minutes. > > D) Impossible to tell from the data (please explain) > > E) Other (please explain) -- R. G. Vickson Department of Management Sciences University of Waterloo Waterloo, Ontario, CANADA ============================================================================== From: Ray Vickson Subject: Re: The Hitchhiker Problem Date: Fri, 11 Aug 2000 15:41:18 -0400 Newsgroups: sci.math,rec.puzzles Patrick Hamlyn wrote: > Ray Vickson wrote: > > >Let's change it from hitchhikers to lightbulbs (i.e., how long has the > >current lightbulb been in operation). This type of question belongs to > >the general topic of the "Inspection Paradox": if an average lightbulb > >burns for 1000 hours, and I walk into > >a room at a random time, how long does the current lightbulb have left > >to go? > >The answer depends on the probability distribution of lightbulb > >lifetimes, but you get > >a particularly simple answer in the case of exponentially-distributed > >lifetimes. The answer in this case is: still 1000 hours left to go. > >Interestingly, the average age of the > >current lightbulb is also 1000 hours. This means that the particular > >lightbulb that just > >happens to be installed in the room when I walk in has an average total > >life of 2000 hours (even though I bought it at a hardware store that > >sells bulbs having 1000-hour average lifes). I discuss this with > >students in my probability modelling classes, and some of them find it > >hard to believe; more precisely, they believe each single step in > >the argument, but find it hard to believe where they end up. I get them > >to do a simple > >simulation in EXCEL, and presto!--it's true. > > You might be able to make this more palatable to them by pointing out that the > selection process is biased toward longer-lived lamps - the shorter-lived ones > having a tendency to winnow themselves out of the gene pool before you get to > inspect them. I do all of this, using knotted-string analogies, videotape thought experiments, etc. A semi-intuitive explanation is not hard to give, and after a while they believe it. Of course, rigorous proof requires Blackwell's Renewal Theorem, which I don't state at all in the undergrad. version and only mention without proof in the graduate version. -- R. G. Vickson Department of Management Sciences University of Waterloo Waterloo, Ontario, CANADA