The Mathematics of Number Plate Spotting

There are (at least) two variations on the game of Number Plate Spotting: Consecutive Number Plate Spotting (CNPS) and Non-Consecutive Number Plate Spotting (NCNPS). CNPS, popularised by the comedian and broadcaster Richard Herring, involves spotting all numbers in consecutive order, i.e. you must spot a 1 before you can start looking for a 2, and so on. NCNPS involves simply ticking off each number once it has been spotted, in whichever order you happen to spot them.

With the introduction of the new system of number plates the game has become more complicated as some people allow the spotting of, e.g. BN02 FDC as a “2”, and OY59 DXG as a “59”, while purists argue that these numbers are in fact abbreviated or encoded versions of the year and month in which the number plate was issued, and so do not represent integers in the range 1 – 999. I am personally persuaded by the purist argument in this case.

Even the pre-Millennial version of the game had variants as there were still different formats to the number plate, specifically the antique ABC 123, followed by the
PQR 123 S format, and finally the W 123 XYZ. Again some might argue that only one format should be chosen at the outset and number plates of other formats discounted, though this does not affect the mathematics.

On Twitter the other day, and in response to a message read out on the Andrew Collins and Josie Long BBC 6Music radio programme, I stated that the non-consecutive version of the game was “approximately 1,000 times easier than true CNPS”. But as soon as I sent the message I began to question this figure. I have since done the maths.

Assume then that we are only counting number plates whose format contains an integer in the range 1 – 999.

For CNPS each number must be spotted before the next. Assuming that there is an even distribution of numbers then on average it will take 999 distinct number plates before the given number is spotted. As there are 999 numbers to be spotted this means that the average game of CNPS will take 999 x 999 number plates, which is 998,001 in total.

The maths of NCNPS is a little more complicated*. Each number has to be spotted but in any order, hence:

The first number plate will contain a number not previously spotted, so the probability of it containing a number we are looking for is 1.

The probability that the second number plate contains a number that we have not already spotted is 998/999, which means that, on average, we will have to view 999/998 number plates before we can spot a new number.

Once we have the second number, the probability that the next number plate contains a number that we haven’t already spotted is 997/999, which means that, on average, we will have to view 999/997 number plates before we can spot the third number.

And so on, until the 999th number, for which we will, on average, have to view 999/1 number plates before we spot the final number we are looking for.

This means that, on average, the total number of number plates we need to view in order to spot all 999 numbers is

1 + 999/998 + 999/997 + 999/996 + … 999/1

This is a harmonic series and there is a formula** for calculating the total:

n x [ln(n) + γ]

where γ is the Euler constant (here approximated to 0.57722) and n is the total number of items to be spotted, in this case 999.

Hence we have

999 x [ln(999) + 0.57722] = 7,476.49 (2 d.p.)

Finally we can divide 998,001 by 7,476.49 which gives

133.49 (2 d.p.)

Which means that, purely in terms of the average number of number plates needed to be viewed to complete each version of the game, and discounting the modern format of number plate entirely, NCNPS is about 133.5 times easier than CNPS.

* I am of course here indebted to Simon Whitehouse’s excellent article on the topic of Panini football stickers and the coupon collector problem.

** A formula which I initially got wrong, and remains an approximation, but it’s close enough to illustrate the point I think.


  1. Joshua says:

    Of course, with NCNPS comes the less measurable difficulty of remembering which numbers you have and haven’t spotted. Under CNPS, you only need to remember the last number spotted.

  2. Ryan Cullen says:

    You are also assuming that you have no knowledge of the next plate, however a good CNPSer would have notes reminding them where upcoming plates are regularly located (or even a website which had a twitter account to send DMs after plates were found, and yes I know how sad that was).

  3. hoppo says:

    “Assuming that there is an even distribution of numbers then on average it will take 999 distinct number plates before the given number is spotted.”

    Not true. This assumes that you will only spot a given number after you’ve seen every other number in random order first. You would have to be pretty unlucky for this to happen?

    In reality, if we assume that every number has equal frequency then it’s equally likely that it will be the next number you see, or the 999th, or any number in between.

    So on average, you will have to look at about 500 different numbers to spot the next one you need.

    Of course, skilled players such as Mr Herring keep detailed notes of where upcoming numbers might be lurking (by ‘prespotting’ neighbour’ cars, etc.) so in actual fact the true average number of numberplates you would need to look at to advance your score would probably be considerably less than 500.

  4. Warren says:

    You should also try to factor in that cnps and ncnps is becoming harder in a non-linear fashion due to the removal of older cars from the pool of vehicles.

    I’ve been stuck on 250 for a year now.

  5. joyfeed says:

    It seems that I have oversimplified the harmonic series formula, most notably omitting the Euler constant, so I am working on that now. Regarding the points raised above, however:

    hoppo: the assumption is that each number plate is independent of the next so could contain the same number again or any number from 1 – 999, and that every number plate viewed has an equal chance of containing the number we are looking for, i.e. a 1/999 chance, and so the average number of plates we would have to see before we spotted the specific number we were looking for would be 999. Your point about the skilled Herring-style operator is quite right though, and in practice non-random elements such as memory, notebooks and tip-offs from friends would all come into play to reduce the total we would need to view before a successful spot.

    Ryan: agreed, see point above.

    Warren: Absolutely – I spotted 202 only this week after I think about a year of looking. The chances of any of us completing this idiotic and pointless task are descending exponentially down the probability toilet.

  6. The maths looks pretty spot on, so I wrote a python script to model the problem:

    The output I got was:
    Mean of 100 goes of CNPS upto 1000: 999246.22 cars
    Mean of 100 goes of NCNPS upto 1000: 7407.75 cars

    Which agree with your numbers pretty closely!

  7. Lee says:

    Lightweights, the Northern Ireland version runs 1 – 9999.

  8. Emile says:

    I see your Northern Ireland 9,999 and raise you Republic of Ireland 1 – 99,999

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