Outreach in Computer Science
Bayesian solutions to real-world risk problems
Seeing the same truck
friend, who writes novels, posed the following to me:
"I'm stuck on something for this
thriller I'm writing. I
have this character who can roughly compute probabilities off the top
of his head. He's smart more than autistic, if you know what I mean.
guy sees a truck (he knows the plates) parked outside his
then sees the same truck show up again in a separate part of the
country later. What are the odds of it being a coincidence?
Say there are
150,000,000 private homes in America. And only one of these
trucks has the number plate.
Does that make the odds
150,000,000 to 1?
Or this is all completely
The problem is not very
stated and my friend’s suggested answer is an example of the
kind of false reasoning explained in detail in this
Here is the reply I gave:
The probability is definitely
in 150 million as you suggest. It is a little complicated to get to a
proper answer, but here goes:
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
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.
I said, it is probably useful to look at a simpler example like the one
here to really get your head around
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
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
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
- 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
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
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
Return to Main Page
Making Sense of Probability: Fallacies, Myths and Puzzles