The
specific issue was small sample sizes and how they can result in skewed, and
hence unrepresentative statistical results.
Arty was struggling a little with the concept and asked the following
question: “Are burglars more likely to get burgled than non-burglars because
burglars are a small group within the total population?”
For
those who are mathematically inclined, a couple of answers follow. If you are not mathematically inclined, jump
to the Venn diagram below.
The
short answer to Arty’s question is:
No,
burglars are not more likely to get burgled than non-burglars because burglars are a
small group within the total population.
Descriptive, post-event, statistical distributions don’t alter pre-event
statistical probabilities of the event occurring.
The
long answer is:
No,
so long as a series of burglaries is completely random each burglar has the
same chance of being burgled as the average citizen, even though the relatively
small sample size of burglars will put their rates of being burgled at the
extremes – either lower than average or higher than average.
In
reality, it’s a little more complicated than that. Self-burglary is probably rare, so a real-world
burglar is very slightly less likely to be burgled than is the average citizen.
Think
of it this way, in a population in which there is only one burglar, the chance
of the burglar being burgled is zero, unless the burglar is incredibly stupid
or particularly forgetful. If there are
two burglars, then each burglar is half as likely to be burgled as the average
citizen, because an average citizen might be burgled by either of the burglars
while the burglars only have to worry about the other burglar (so long as
they are neither incredibly stupid nor particularly forgetful). The equations describing the probabilities
are:
Pr(Burgled
if Burglar) = Pr(Bd|Br) = B/(p-1)*(n-1)/n
Pr(Burgled
if not Burglar) = Pr(Bd|notBr) = B/(p-1)
where
B is the number of burglaries, p is the population size and n is the number of
burglars.
Pr(Bd|Br)
approaches Pr(Bd|notBr) as n increases.
According
to Home Office statistics, the probability of a household being burgled in any
given year is about 2.5%. If we assume
that the average burglar performs about five burglaries in a working week (I
know this is low, and that it acts to increase the calculation of burglars),
then 1 out of 10,000 citizens is a burglar.
In a population of p=63,000,000, there will be n=6,300 burglars carrying
out about B=25,000 in a year, so:
Pr(Bd|Br)
= B/(p-1)*(n-1)/n = 2.4996%
Pr(Bd|notBr)
= B/(p-1) = 2.5000%
So,
no, there is a 0.0004% lower probability of being burgled
if you are a burglar in the UK – if the distribution of crime was
random. Which it isn’t.
Burglars
tend to work close to home and they tend to live in high crime areas. This is partly because of the burglars,
partly because a depressed economy leads to crime, including burglary.
I
have to get a little creative here and appeal to Pareto. Let’s say that in a standard city of a
million inhabitants, 80% of the crime takes place amongst 20% of the
population. This means that there are
100 burglars in total, n=80 of whom live among the high-crime sub-population of
p=200,000 people, who are the victims of B=20,000 burglaries a year. So:
Pr(Bd|Br)
= B/(p-1)*(n-1)/n = 9.8750%
Pr(Bd|notBr)
= B/(p-1) = 10.0000%
And
n=20 of whom live among the low-crime sub-population of p=800,000 people, who
are the victims of B=5,000 burglaries a year.
So:
Pr(Bd|Br)
= B/(p-1)*(n-1)/n = 0.5938%
Pr(Bd|notBr)
= B/(p-1) = 0.6250%
Therefore,
within the city as a whole, the probability of being burgled is:
Overall
Pr(Bd|Br) = 9.8750%*0.8 + 0.5938%*0.2 = 8.01876%
Overall
Pr(Bd|notBr) = 10%*0.19992 + 0.6250%*0.79998 = 2.49992%
(The
0.19992 and 0.79998 figures arise after removing the burglars from the two
sub-populations.)
So,
yes, burglars will tend to get burgled more often than an average non-burglar from
the total population – but it has nothing to do with burglars being a small
subset of the total population per se. It is instead related to the facts that 1) a
burglar is more likely to live near other burglars and 2) that burglars tend to
work close to home.
Unfortunately,
I think all this mathematics would have totally fried Arty’s circuits, so let’s
use a sort of Venn diagram instead. Imagine that
you are a burglar in a population of 100, of which 10 are burglars.
Assume
that four burglaries take place.
Burglar
|
Total burglar
|
Non-burglar
|
||
Active
1/10
|
Inactive
9/10
|
|||
Burglary 1
|
0
|
1/99
|
1/99*9/10 = 1/110
= 0.909%
|
1/99
= 1.010%
|
Burglary 2
|
0
|
1/99
|
1/99*9/10 = 1/110
= 0.909%
|
1/99
= 1.010%
|
Burglary 3
|
0
|
1/99
|
1/99*9/10 = 1/110
= 0.909%
|
1/99
= 1.010%
|
Burglary 4
|
0
|
1/99
|
1/99*9/10 = 1/110
= 0.909%
|
1/99
= 1.010%
|
TOTAL
|
3.636%
|
4.040%
|
So,
just using statistics (and not criminal demographics), we see that so long as
you are bright enough as a burglar to not burgle yourself, you have a lower
probability of being burgled as a burglar (when burglaries are completely random).
The probability of being burgled as a burglar increases as the
population of burglars increases. This
can be seen easily when you think that if there is only one burglar (you), then
you won’t get burgled – ever. Add another burglar and you suddenly have a
chance of being burgled. In a population
consisting entirely of burglars, the probability of your being burgled is
totally unaffected by your being a burglar because everyone in the population is
a burglar.
Therefore,
as a burglar, your best option for increasing your home security is to remove
the number of burglars from the population, which can do conveniently by
reporting other burglars to the police – I’d suggest doing that immediately,
before they report you. If you are not
bright enough to avoid burgling yourself, you might want to report yourself to
the police.
However,
Arty was not specifically worried about his home security. He was thinking about the problem of small
sample sizes. Let’s say the distribution
of burglaries looked like this:
Probability
of being burgled (general)
|
4/100 = 4%
|
Probability
of being burgled (non-burglar)
|
3/90 = 3.33%
|
Probability
of being burgled (burglar)
|
1/10 = 10%
|
However,
a very slightly difference scenario is this:
Probability
of being burgled (general)
|
4/100 = 4%
|
Probability
of being burgled (non-burglar)
|
4/90 = 4.44%
|
Probability
of being burgled (burglar)
|
0/10 = 0%
|
Neither
the 10% figure nor the 0% is a realistic approximation of the probability of
being burgled as a burglar (at 3.636%). A
sample of four burglaries is simply too small, and part of the reason that
sample is too small is because the population being considered (100
inhabitants) is too small. To get a
proper idea of probabilities, you need a significantly larger sample size.
--------------------------------
While this article is pretty much targeted at one person, if I’ve encouraged any burglars out there to report their comrades (or themselves) to the local constabulary, then I think I’ve done my good deed for the day.
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