The inspection paradox

When it comes to probabilities, our intuitions are not reliable, as I have written about before (see here and here). On so many occasions, I have thought that the result to a problem was so obvious as to not be worth thinking about more deeply, only to find myself proven wrong. And the new solution seems also so obvious that you wonder why you ever believed the earlier wrong answer.

Another example of this comes from the ‘inspection paradox’. It works like this. Suppose you know that the bus or train on a route that you take regularly does not always conform to the schedule. But while its arrival times are unreliable, it comes on average every ten minutes. If one goes to the bus stop at random times, one would think that the average wait time would be five minutes. But anyone who has actually experienced this has the feeling that on average the wait time is longer than that. We tend to put that down to being misled by psychology, by the fact that time drags when we are waiting for something, making it seem longer than it actually is

But it turns out that psychology is not to blame, that the wait time is in fact longer than half the average time between bus arrivals and this is what the inspection paradox is about.

To take a concrete example, suppose that the time interval between buses on a route is two minutes followed by 18 minutes, then two minutes, then 18 minutes, and so on, so that the average time interval is ten minutes. If one marks those on a time line, then one can easily see that if one arrives at random (equivalent to picking a random spot on the line or ‘inspecting’ it), then one is nine times more likely to arrive during the 18 minute interval than the two minute one. Hence on average, the wait time will be longer than five minutes.

One can make it more concrete. The probability of arriving during the two-minute interval is 2/20 and will have an average wait time of one minute, while the probability of arriving during the 18 minute interval is 18/20 and the average wait time will be nine minutes. So the expected wait time overall is (2/20)x1+(18/ 20)x9, which works out to 8.2 minutes.

I learned about this from Amir Aczel who goes into it in more detail and provides a general proof of the result, and also uses it to explain why it is that countries with a lot of immigrants have higher life expectancies than those with less, other things being equal of course.


  1. eigenperson says

    There are a lot of examples of this sort of thing. My favorite is the fact that if you flip a coin until you get three heads in a row, the expected number of flips you will have to make is much more than 8 (it’s actually 14). It should be possible to win a lot of money by betting on this with the right kind of sucker.

  2. Rob Grigjanis says

    I remember being astounded, as a boy, to learn that the probability of two people in a group of 23 having the same birthday is 50%.

  3. says

    I think quite a bit in statistics is counter intuitive. In my experience, you just need to give up on what makes intuitive sense and trust the numbers.

  4. markdowd says

    Monty Hall was actually really easy for me to understand, as well as the general solution for X doors with 1 car, X-1 goats, and a chance to switch to any other door after each time Monty shows a goat.* It’s actually easier when you take it to the extreme of 1,000 doors. You have an extremely unlikely chance of getting it right, so stick with your original choice while you let Monty show you all the bad choices that are left in the pile, finally switching at the end to the “good” choice. The interesting thing about the general Monty Hall is that your chances of winning actually increase as the number of doors go up.

    The more difficult one for me was the related “Monty Fall” problem. Look it up and see if you can understand that one.

    General solution is, pick a door, and don’t switch. At the end, when two doors are left, switch.

  5. Peter B says

    How about this one?
    Task1: roll a die counting the number of rolls until you get a six. Record that count.
    Task2: perform task1 several thousand times.
    Question: which count will occur the most often.

  6. felicis says

    Nice one, Peter B!

    I think part of what makes things ‘counterintuitive’, even for people with a basic grasp of probability and statistics, is that our most common models are either the uniform or normal distribution – but exponential (and Poisson) distributions are both very common and behave very differently. Another issue comes up with it being easy to work out small discrete problems, but the solution then doesn’t scale as you might think (the ‘most numbers are small’ problem…

  7. flex says

    I think non-normal distributions can really throw people off. Not only as felicis above states with exponential distributions, but many problems show bi-modal (or greater) distributions.

    One of the problems I’ve encountered as an engineer is meeting Cpk requirements. This is basically a requirement that the sigma of the distribution be a certain number of multiples of sigma from the specification limit. Typically the ideal Cpk is 3 or greater, meaning that the specification limits are at least 3 sigma from the median.

    When dealing with a dimension from a 2-cavity mold, it may well be that combining all the data into one lump, your sigma is so large that you can’t meet the Cpk requirements. However, treating each cavity individually, you will find that the sigma for each cavity is quite small, but the distribution is bi-modal. Two normal distributions separated by a valley.

    Assuming normality of the totality of the data, you fail the Cpk requirement. Separating the data into two normal distributions based on the cavity of the tool, and you meet the Cpk requirement.

    The specifications have not changed, the data has not changed, the tools have not changed. The only change is the assumption of normality. It’s always good to review your assumptions.

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