Two hundred and seventy, against fourteen hundred

Five gearboxes, or fifty

Serial numbers run 1 to N and you have captured k of them at random. The green ticks are what you saw. The highest of them, m, is your floor: the fleet is at least that big, and almost certainly bigger, because the top of the range is exactly the part you are least likely to have sampled.

How much bigger? Your k captures cut the run from 1 to m into k gaps, and those gaps hide the m − k serials you did not see — m/k − 1 of them per gap. There is one more gap above m, and no reason for it to hide fewer. Add it: N̂ = m + m/k − 1.

·

The shortfall is exactly one gap

That was an intuition. Here is the fact it turns into. Over every way of drawing k serials from N — not a simulation, all of them, counted exactly — the highest one you see averages k(N+1)/(k+1), so the raw maximum falls short by (N−k)/(k+1). That number is the count of serials hiding above your maximum. The correction adds (m−k)/k, the count hiding in an average gap below it — and the average of that, over every possible capture, is (N−k)/(k+1) as well. Not close to it. It. The correction does not approximately undo the bias; it undoes it, and the two exact fractions on the right of the table below are the same fraction.

Using all of your data made it worse

There is an obvious rival. The serials you saw should average about halfway up the range, so 2 × mean − 1 is an estimate too — and it is also exactly unbiased, and it uses every number you captured rather than throwing away all but one. It is the worse estimator, and not by a little: its variance is (k + 2)/3 times the other's, whatever N is. At ten serials it is four times as wide; at fifty, seventeen.

Both distributions below are exact — every possible capture, weighted, with no sampling anywhere on this page except the picture at the top. The maximum is a sufficient statistic: once you know it, the rest of your serials say nothing more about how many were built, and an estimator that listens to them anyway is listening to noise.

How many serials is enough

The spread of the good estimator is √( (N−k)(N+1) / (k(k+2)) ), which falls off roughly like N/k — so accuracy is bought with serial numbers, linearly, and five of them are worth very little. The Allied figures were not built on five. Every tank carried serials on its chassis, engine, gearbox and road wheels, each running in its own sequence, and captured and destroyed vehicles fed hundreds of numbers a month into the estimate. That is the row that matters below, and it is why the answer came out at a couple of per cent rather than a factor of five.

What it was for

The figures usually quoted come from Ruggles and Brodie's account of the work, published in 1947 once the German production records had been captured and the two kinds of estimate could finally be marked against them. This page cannot check these three columns — they are history, not arithmetic, and the only thing it can honestly claim about them is that the method in the middle column is the one drawn above.

The interesting column is the third, and the interesting comparison is not statistics beat spies. It is that the two methods failed differently. The intelligence estimates are wrong by a factor of four to eight, always in the same direction, because they were assembled from reports by people with reasons to be impressed. The serial-number estimates are wrong by a few per cent in both directions, because they were assembled from a quantity nobody had a motive about: how far apart the numbers were.

What this page checks about itself

Every distribution on this page is exact. The spread of the maximum comes from counting the captures with each maximum, and the spread of the mean from a count of how many captures have each total — both in whole numbers, held in BigInt, and reduced to a fraction only at the end. Nothing here is simulated, so nothing here has a random seed to be lucky with.