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I have an electronic weighing-machine, which I believe to be internally very accurate. It will weigh up to $100$ kg, but not activate below $10$ kg. The digital display reports to one decimal place. The problem is that I don't know whether the reading is rounded (with worst error $\pm 50$ g) or truncated (with worst error $-100$ g and expected bias $-50$ g). I have a great quantity of books and papers that can be stacked on the machine to make any weight within its limits, but nothing of accurately known weight.

I guess that any solution must be statistical. A good solution would minimize the number of weighings, given a tolerance probability of a false indication. (Assume a 50/50 prior distribution for rounding/truncation; for illustration, a targeted probability could be $0.1\%$.)

John Bentin
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  • Do you know that these are the only possibilities and that the machine is otherwise accurate? (E.g. do you also have to account for the possibility of some Gaussian error before either the rounding or the truncation?) – Qiaochu Yuan Jun 26 '13 at 22:07
  • @QiaochuYuan: Here, "very accurate" means "treat pre-display errors as negligible". I make these assumptions for definiteness and simplicity, as a first approximation. They idealize reality, but are not contradicted by my experience of packing bags for air travel or checking melons bought in the market. – John Bentin Jun 27 '13 at 09:58

4 Answers4

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Hint: Suppose you weigh a batch of individual objects that are a little over $10$ kg and record their weights. Now stack them together. What do you expect the result to be in the two cases? If you do this a few times, you will know.

Ross Millikan
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Assuming the scales are perfectly accurate then:

In the case of truncation, if you choose two random objects with recorded weights p and q, then the weight of them both together will be recorded as either p + q or p + q + 100 g, with equal probabilities.

In the case of rounding, if you choose two random objects with recorded weights p and q, then the weight of them both together will be recorded as either p + q - 100 g or p + q, with equal probabilities.

So as soon as you see a single +100g or -100g result you will have absolute certainty which is the case, but until then you won't be able to distinguish between them. So 10 such comparisons will give a probability of less than 0.1% that you still do not know which is the case.

MikeFee
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  • Corrected solution. – MikeFee Jun 26 '13 at 22:18
  • If the machine rounds, and I weigh objects of actual mass $11.03$ and $11.04$ kg separately, the machine will report $11$ kg for each: a total of $22$ kg. The combined weight will be reported as $22.1$ kg. – John Bentin Jun 27 '13 at 07:43
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Assuming uniformly and independantly distributed masses (modulo $100$g), the probability that the display for $A+B$ equals the sum of the individual displays is $\frac12$ with rounding as well as with truncating. However, when the weight of $A+B$ does not equal the sum, the error is always in the same direction with truncation, but is greater and less with equal probability when rounding. Thus perform enough such comparisons of sums and (random) parts weighings to distinguish statistically between these two options. (Formally, a single case of the sum apperaring too heavy excludes truncation, but you may want to add more tests to accustom for error probabilities). Nine instances of the sum being lighter (so in about 18 comparisons / 54 weighings should be enough for your targeted error probability.

Hagen von Eitzen
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A deterministic solution seems possible. Make up a bundle such that the addition of a little extra weight is enough to increase the reading to the next decimal digit. Repeat with another bundle and little extra weight. Now weigh the two bundles together. Then the addition of the little extra weights will make no difference to this reading if the machine rounds, but will increase the reading to the next decimal digit if the machine truncates. Here, weights of below $20$ g (approximately the weight of four A4 sheets of photocopier paper) are small enough to count as "little".

John Bentin
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