Burglars Burgling Burglars

A discussion came up recently between a couple of mathematically inclined people (let’s call them “neopolitan” and “the Frog”) who had read Daniel Kahneman’s excellent book, “Thinking, Fast and Slow”, and someone not quite so mathematically inclined (“Arty”) who is still working through the book.

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.

If burglaries are conducted totally at random, what is the chance that you (as a burglar) will be burgled?  Each time a burglary happens there is a 1/10 chance that you will be the person committing the burglary, so you won’t be burgled, and a 9/10 chance that there will be chance of someone being burgled by someone else.  The probability involved depends on whether people get burgled twice.  This does happen so let’s assume that all burglaries are totally random and there is a very slight chance that you could get burgled four times in a row.  What are the probabilities?

	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:

The statistics derived from this ridiculously small sample would be:

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:

The statistics derived from this sample would be:

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.

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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.