Recently I've taken to recording how many throws it takes me to complete a game of Round the World in darts. The game is quite simple: you must score a 1 before moving on to the 2, once the 2 is completed you can move on 3 and so on around the board before finishing on the red bullseye or the green ring around it. The minimum number of throws required for this feat is 21.
I decided to measure the efficiency of my score by dividing it into 21 and expressing this as a fraction. Thus: $$ \text{efficiency }=\frac{21}{\text{score}} \times 100$$Figure 1 shows a graph of the resultant efficiencies for scores ranging from 105 to 21.
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Figure 1: permalink |
Figure 2 shows a table of selected scores and their associated efficiencies (rounded to the nearest whole number).
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Figure 2: permalink |
As can be seen, it becomes increasingly difficult to achieve an efficiency close to 100%. For example, 22 scores 95% and 21 scores 91% but all other scores are under 90%. I'm recording these results in a newly created AirTable database. See Figure 3.
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Figure 3 |
I'm keeping track of the number of throws using a counter on my iPhone. See Figure 4.
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Figure 4 |
There is a defect of sorts in this way of measuring efficiency because it supposes that 21 steps required to finish are all equal. Indeed from 1 to 20 they are but the final bullseye and green ring have a combined area that is considerably smaller than the numbered sectors. One might complete steps 1 to 20 with 20 throws and then spend ten more throws before hitting the central area of the dartboard. The final score of 30 with result efficiency of 70% doesn't fully reward the extraordinary skill required to progress from 1 to 20 in only 20 throws.
The bull's-eye has an outerbull area (also know as the single bull, which scores 25) and an inner bull (also known as a double bull's-eye, which scores 50). The circular scoring area of the standard dartboard has a diameter of 34" and the bull's-eye has a diameter of 3". So this means that each numbered sector has an area of 45.04 square inches and the bull's-eye has an area of 7.069 square inches which is thus more than six times smaller. Hitting the bull's-eye in a single throw ought to be rewarded more than hitting one of the numbered sectors in a single throw.
Hitting the bull's-eye is equivalent to hitting six numbered sectors in succession. The numbers should range from 1 --> 20 and then from 21 --> 26 but 26 is difficult to work with. Let's go with 21 --> 25 so that the numbered sectors count for 80% and the bull's-eye 20%. If somebody hits the numbers 1 to 20 in twenty throws, they are assured of an 80% score. A formula then involves two statistics, \(x\) and \(y\) where the former represents the throws taken to traverse 1 to 20 and the latter represents the throws needed to hit the bull's-eye. The formula thus becomes:$$\text{efficiency }=\frac{20}{x} \times 80 + \frac{1}{y} \times 20$$This is a fairer estimate of efficiency that doesn't unduly penalise somebody for having difficulty hitting the bull's-eye. I've changed my AirTable database to reflect these changes.
ADDENDUM: March 6th 2024
There's a major problem with this final efficiency formula that I came up with and I only noticed it today. Recently, on February 26th, I achieved an efficiency of 62% after scoring 41 in the 1 to 20 section and 1 in the bull's-eye:$$ \text{efficiency } =\frac{20}{40} \times 80 + \frac{1}{1} \times 20 \approx 62.0 \%$$Today I needed 41 for the 1 to 20 section but needed two throws to get the bull's-eye. However, I was shocked to see that my efficiency was 10% less as the result of the calculation:$$ \text{efficiency } =\frac{20}{41} \times 80 + \frac{1}{2} \times 20 \approx 52.0 \%$$This is clearly not reasonable but the problem only emerged as the consistency of my dart throwing improved.
For the time being, I'll revert to my original formula:$$ \text{efficiency }=\frac{21}{\text{score}} \times 100$$This then produces more reasonable results:$$ \text{efficiency }=\frac{21}{41} \times 100 \approx 51.2 \%$$$$ \text{efficiency }=\frac{21}{43} \times 100 \approx 48.8 \%$$This formula is far from perfect but it will have to do for the time being until I come up with something better.
For the time being, I'll revert to my original formula:$$ \text{efficiency }=\frac{21}{\text{score}} \times 100$$This then produces more reasonable results:$$ \text{efficiency }=\frac{21}{41} \times 100 \approx 51.2 \%$$$$ \text{efficiency }=\frac{21}{43} \times 100 \approx 48.8 \%$$This formula is far from perfect but it will have to do for the time being until I come up with something better.
