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The arrangement of the numbers around the circumference of a standard dart board is as shown below

  20 1 18 4 13 6 10 15 2 17 3 19 7 16 8 11 14 9 12 5

Oddly enough, no one seems to know for sure how this particular arrangement was selected. ... it's clear that the numbers are ordered to mix the large and small together, and possibly to separate numerically close values as far as possible (e.g., 20 is far from 19), no one seems to know of any simple criterion that uniquely singles out this particular arrangement as the best possible in any quantitative sense.

http://www.mathpages.com/home/kmath025.htm

Question

This appears to be an unsolved problem. How did the inventor of the standard dartboard come up with the order of the numbers in such a way as to minimise scores that are produced by inaccurate throws?

Can anyone see a pattern or was it just trial and error?

Given that computers weren't available then (pre 1900), can anyone suggest a pencil and paper method that produces a near-optimum result (and specifically this result) in a reasonable time?

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    $\begingroup$ I assume it'd be easy to do something like this simply by randomly picking large numbers, arranging them, and situating the smaller numbers to create the pattern that Okx describes. $\endgroup$ – 2xedo Aug 22 '15 at 20:38
  • $\begingroup$ My bet: A coincidence. It was a guess and nothing more :) $\endgroup$ – warspyking Aug 22 '15 at 22:15
  • $\begingroup$ complex Math was possible before computers, logarithms for example using log books, technology is faster but does not replace mathematical concepts. Whatever can be done with technology can be done by hand as well, it just may take months or years rather than seconds $\endgroup$ – Mousey Aug 28 '15 at 23:18
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The numbering system on a standard dartboard is designed in such a way as to reduce ‘lucky shots’ and reduce the element of chance. The numbers are placed in an order to encourage accuracy and punish inaccuracy. The placing of low scoring numbers either side of large numbers e.g. 1 and 5 either side of 20, 3 and 2 either side of 17, 4 and 1 either side of 18, will punish poor throwing. If you shoot for the 20 segment, the penalty for lack of accuracy is to land in either a 1 or a 5. That is basically it.

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  • $\begingroup$ Yes. I'm really asking if we think this can be achieved by trial and error - without a computer. If so, it could take a very long time and yet, the article seems to suggest, the result is near-optimum. $\endgroup$ – chasly from UK Aug 22 '15 at 14:16
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This is more an observation of the pattern than a method to get it, but if we assume a shooter has a spread of one space, meaning for example if aiming at 20, there is an equal chance of hitting 20,5,or 1, then we get these expected values for each target.

20 1 18 4 13 6 10 15 2 17 3 19 7 16 8 11 14 9 12 5
8.6 13 7.6 11.6 7.6 9.6 10.3 9 11.3 7.3 13 9.6 14 10.3 11.6 11 11.3 11.6 8.6 12.3

The expected values range from 7.3 to 14, a pretty big spread. But if we order targets by expected value, we get

17 13 18 20 12 15 6 19 10 16 11 2 14 4 9 8 5 1 3 7

This is semi-close to being ordered. Basically, if you evenly hit the target you're aiming at or one of it's neighbors, the best places to shoot for are actually 1,3 and 7, while the worst are 17, 13, and 18. There are still a couple inconsistencies, such as 14 being so high on the list, but this gives a general framework.

Other observations

Even spread is impossible: Consider 20. With value $a$ on its left and $b$ on its right, the expected value is $(20+a+b)/3$. Now consider spot $a$. 20 is one neighbor, call the other neighbor $c$. So if we have $(20+a+c)/3 = (20+a+b)/3 => c = b$ which is impossible, because there are no repeating values.

Smallest possible spread: If we order the scores 20,1,19,2... I think we get the smallest difference in expected values, from 17 = 8 to 10 = 13.66

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