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.