truck

What you are saying is that there is definitely only one truck with this number plate. What we want to calculate is the probability of this guy (let's call him Phil) seeing the same truck in two completely different locations within a given period of time. First thing to note is that you have to avoid the common fallacy of overstating the 'coincidence' by ignoring the fact that there are many unique trucks which are candidates for being seen twice in different locations. Suppose, for example, that in any one year there are 1000 different vehicles parked near this guy's house. What you have to calculate is the probability that ANY one of them might subsequently be seen by the guy in a different location. That is 1000 more times likely than the probability of one SPECIFIC vehicle being seen twice.

So we have to first work out the probability of a specific vehicle V being seen twice by Phil and then multiply that probability by 1000.

The probability of Phil seeing the specific vehicle in two completely different locations depends on

a)   the amount of travelling Phil does and
b)   the average amount of travelling an arbitrary vehicle might do.

If, for example, Phil NEVER visits any other city than his own then the probability of him seeing vehicle V in two cities is actually 0 (which is even less than 1 in 150 million of course!)

But let's suppose on average that Phil visits 10 different cities within the same year (or whatever period you want to restrict it to). Let's suppose that at any time there are 100,000 vehicles in each city and that on any visit let's suppose that Phil sees 1 in every 100 cars in the city (i.e. he sees 1000 vehicles).   Now we have to consider the probability that Phil sees vehicle V if it is in any one of those 10 cities at the same time as him. This is more or less 10 times the probability that Phil sees vehicle V in one specific other city C. To work out this probability we can assume something like the following:

a typical truck driver visits say 50 different cities every year staying on average 2 days at each (so he is 'away from home about 30% of the time)
if there are, say, 500 different cities in the country then there is a 10% chance he will visit city C some time during the year. There is about a 1 in 1500 chance he will be in city C on a particular day of the year.
there is about a 1 in 7500 chance that Phil will be in city C on the same specific day, so there is a 1 in 11,250,000 chance they will be in the same city on the same specific day. But as there are 365 days in a year you need to multiply this by 365 to get the probability that during any one year they will be in the same specific city on any one day. So the probability is about 1 in 35,000.
the probability that Phil sees V when they are in the same city is 1 in 100 so the probability that Phil sees the vehicle in the same specific city on any one day of the year is 1 in 350,000

Now, as I said, there are 10 specific cities where they could meet so we need to multiple the last probability by 10.

The result is a probability of 1 in 35,000

Finally there were 1000 candidate vehicles to choose from so we need to multiple the last probability by 1000.

That leaves a 1 in 35 chance of seeing the same vehicle in a different city.

It's still unlikely of course, but MUCH less unlikely than the 1 in 152,000,000 you came up with.

Obviously lots of assumptions and approximations but the point is that your initial estimate is several orders of magnitude out.

This all helps to explain why so-called one in a million events happen all the time. It's because there are millions of events that can happen!

Seeing the same truck