# Achtung! Use maths to smash the German tank problem – and your rival

## Leaky data is a double agent in intelligence analysis

Reducing security risks from open source software

**Big Data's Big 5** You've employed Benford's Law to out fraudsters hidden in seemingly random numbers. Now what do you do if you need answers but some of your data is missing? Welcome to the German tank problem, the second in *The Reg*'s guide to crafty techniques from the world of mathematics that can help you quickly solve niggling data problems.

The German Tank problem (and its solution) can allow you to estimate data that you don’t have in order to gain information that you want. A more formal statement of this process is: “The estimation of the maximum of a discrete uniform distribution by sampling (without replacement).” The less formal title, though, is snappier and gives a huge clue as to the provenance.

During the Second World War, the British became very interested in the number of tanks the Germans were producing. Intelligence sources suggested the production was about 1,000 to 1,500 tanks per month. In practice it was nowhere near that high and the wily British boffins found another much more accurate way of estimating production.

They got the squaddies on the battlefield to examine as many captured/disabled tanks as they could and collect all the serial numbers they could find. (In truth this must have been highly exasperating for the soldiers, who probably felt that knocking the tank out in the first place was good enough)

Let’s assume for a moment that tank serial numbers are simple. They all have numbers taken from the same series; the series starts at “1” and increments by one for each new tank.

If we find a tank with the serial number 24 then the Germans have produced at least 24 tanks. However we can be a bit smarter than that. Suppose we see the following numbers: 23, 16, 24, 17, 11, 6, 7, 9 and 12.

That’s nine numbers between seven and 24. If there were actually 4,000 tanks out there, it would seem highly unlikely that our sample would just happen to come from the first 24 produced. It is much more likely that there are only about 30-40 tanks in all out there. (To put that slightly more formally, it is likely that the entire population from which we have drawn our sample is around 30-40.)

And we can be even smarter. Suppose we examine five tanks (we’ll call the number of examined tanks E, so E=5) with serial numbers of: 432, 3245, 2187, 435, 1542.

The biggest serial number (which we’ll call B) is 3,245, and the smallest (s) is 432. We could reasonably expect as many ‘hidden’ numbers – serial numbers we guess must be out there but for which we have no data – above 3,245 as there are below 432.

In our nomenclature, n is our estimate of the number of tanks.

To put that graphically:

And to put it mathematically:

we hypothesised that (n-B) is roughly equal to (s-1), so if:

n-B = s-1

then we can say that: n=B+(s-1)

which we can solve for n by slotting in the actual numbers.

In our case B=3,245, s=432, n=number of tanks produced, so:

n= 3,245 + (432-1) = 3,676

This is merely one formula for estimating the maximum of a discrete uniform distribution by sampling (without replacement). We can also use:

n = (1+1/E)*B

E, if you remember, is the number of tanks examined, which was five, so:

n = (1 +1/5)*3,245 = 3,894

This particular formula was proposed (PDF) by Roger Johnson of Carleton College in Northfield, Minnesota in the journal Teaching Statistics, Volume 16, Number 2, Summer 1994

You might also wish to try a Bayesian approach. I have no particular axe to grind here. I don’t mind how you do it; the important point is that you can.

Now, as it happens, there are several over-simplifications in my original description, both in the reality of WWII and also if we were trying to solve the same kind of problem today.

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