The Dartboard The Dartboard It's the playing area -- get to know it! The Darting environment consists of a couple of elements, which all center around the dartboard itself. The dartboard, the throwing area, and lighting are all important factors. They also differ between different darts styles. Hard-Tip: Regulation size: 17-3/4" diameter. That's the full board. 17-3/4" (call it 18") is the side of the full board to the edge. 13-1/4" is the diameter of the playable area. Soft-Tip: Regulation size: 15-1/2" diameter. 15" is the diameter of the playable area. Practice Boards Various practice boards are made. In steel-tip darts one common format is to reduce the thickness the trip and double rings to 1/2 size to make hitting them 2x as hard. Other practice boards are similar, but can add bull (or smaller) targets in the center of each bed -- put a bull anywhere on the board -- to give you a specific target in each bed. In some places these practice boards are referred to as Champion boards.

Other times -- those things are even worse: Champion Boards Champion boards have a smaller overall diameter than standard boards, as well as having smaller double and trip areas. 13" is a common size for bristle & electronic boards. The triple and double areas and the bull are also reduced size -- usually 1/2 regular size -- in these boards. A regulation dartboard is divided into 20 numbered segments and a bullseye. Each of the segments and the bullseye are divided into different regions. These regions are called the beds of a dartboard. Each segment is divided into inner (skinny) and outer (fat) singles segments. The inner ring dividing the inner and outer singles segments is the triples area. The outer ring dividing the outer singles segment from the board is doubles. The bullseye is divided into an outer bull and an center inner bull. Different games treat the two bullseye regions differently: Bullseye 50 points. Outer bull single 25 points, Inner Bull double 50 points. In written nomenclature you will often see prefixes of S for Single, D for Double and T for Trips/Triples when describing the segment to aim for.

T20 == 60 points, etc. If you see a 25 or 50 it is a reference to a Bullseye, and D25 would be a Double Bull. Looking at the dartboard in quadrants, the 4 quadrants contain the following segments (including the overlapped segments and the top/bottom/left/right): Upper Right: 20 1 18 4 13 6 Upper Left: 20 5 12 9 14 11 Lower Right: 6 10 15 2 17 3 Lower Left: 11 8 16 7 19 3 Left: 20 5 12 9 14 11 8 16 7 19 3 Right: 20 1 18 4 13 6 10 15 2 17 3 Throwing Area The important dimensions are from the Center of the Bull to the floor, and from the Face of the board to the toe (foul) line, measured along the floor.

You can verify everything all right is by checking the distance along the diagonal from the center of the bull to the foul line. In fencing it's called the Piste, the "piece" or "strip", which is where the fencers go at it. In darts, the name isn't so grand; it's The Throwing Area. However, darts does have a cool name ... the Oche, which rhymes with Hockey. That's the name of the line. In pro tournaments it is done by placing a board on the floor at the limit of the throwing area, so you always can feel and know where the line is. In ordinary situations it is just a sticker on the floor, which the front of your foot can't be forward of it's front edge. Similar to disc golf, your feet can land in front of the throwing line -- but only after your dart has hit the board! That is a legal shot, but if your foot hits first it's a foul. Bristle Board: 5'8" to the Bull, 7' 9-1/4" to the Face. 9' 7-3/8" Diagonal.

Electronic Board: 5'8" to the Bull, 8' to the Face, 9' 9-13/16" Diagonal. The Foul Line should be 2 feet wide. The following dimensions are for a traditional steel-tip board: Inner Bull -- 1/2" diameter Double Ring -- 1/4" Wide -- 4-1/4" radius Triple Ring -- 1/4" Wide -- 6-11/16" radius In reality the BDO defines board sizes in millimeters and I converted the BDO sizes to inches for convenience. How large are the target areas?

Inner Bull ______ 0.20 in^2 Outer Bull ______ 1.05 in^2 Total Bull ______ 1.25 in^2 Trip Ring _______ 8.02 in^2 Trip Segment ____ 0.40 in^2 Double Ring ____ 12.93 in^2 Double Segment __ 0.65 in^2 Inner Segment ___ 2.32 in^2 Outer Segment ___ 3.60 in^2 Here is a ratio comparison of the target areas: Inner Bull is 1/6 the area of the Total Bull Outer Bull is 5x larger than the Inner Bull Bullseye is 6x the area of the Inner Bull Triple is 2x the area of the Inner Bull (1/3 of a Bullseye) Double is 1.5x the area of the Triple (3x Inner Bull, 1/2 Bullseye) Inner is 3.5x the area of the Double (12x Inner Bull, 6x Triple, 4x Double, 1.9x Bullseye) Outer is 1.5x the area of the Inner (16x Inner Bull, 9x Triple

, 6x Double, 3x Bullseye) One thing to be aware of is that more expensive bristle boards have thinner dividing rings between the beds of the dartboard....

but the dartboard stays the same size. This means there are larger beds on a more expensive board with blade dividers (such as a Nodor or a Winmau) than there are on a traditional board with wire dividers. It also means that bounce-outs are less likely to occur, because the divider or spider area is smaller and lower profile. Soft-Tip Board Target Areas There are two considerations of size when looking at an electronic dartboard. Sure, I could look at the area, like I did above for steel-tip boards. The problem is that area doesn't matter as much, in my opinion. What does matter a lot more is the number of holes in each bed. Your dart tip needs to end up in a hole to land in the board, and there are only so many of them. Another important consideration in an electronic board is the size of the spider, the area that exists between each bed of the dartboard. The spider is non-scoring and will cause bounce-outs. Unlike the spider on a steel-tip board, and the various darts that prevent bounce-outs -- there is so solution for that on an soft-tip board :-( Valley Cougar boards have a very tiny spider; basically just slim guides between the edges of neighboring segments.

The segments themsleves have a chisel edge which directs the dart hitting the guide into into the bed on each side. Certainly there is a small flat edge at each segment edge but it is darn small. Arachnid Galaxy boards have a thick rounded spider that covers the joints between segments. Bounce-outs are more likely to happen on these boards because of the large area of the board taken up by the spider. In particular, the Galaxy 3 boards seem to have even a larger spider than the Galaxy IIs -- I've had my bounce-outs increase to a frustrating number shooting on a G3 -- whereas most of my darts go home on a Galaxy II. The Dartboard Sequence The Dartboard Sequence 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 evidently dates back at least 100 years.

Some say the pattern was devised by a carpenter named Brian Gamlin in 1896, while others attribute it to someone named Thomas William Buckle in 1913, but both of these attributions are relatively recent, and neither can be traced back to a contemporary source. Also, although 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. It may be just an accident of history that this particular arrangement has been adopted as the standard dart board format. It's interesting to consider various possible criteria for choosing a circular arrangement of the first n positive integers. In order to get as "flat" a distribution as possible, we might try to minimize the sum of the squares of each k consecutive terms. For example, setting k = 3, the standard dard board sequence gives (20+1+18)^2 + (1+18+4)^2 + (18+4+13)^2 + ...

+ (5+20+1)^2 = 20478 Apparently the standard board layout described above is called the "London" dart board, and there is another, less common, version called the "Manchester" dart board, which has the sequence 20 1 16 6 17 8 12 9 14 5 19 2 15 3 18 7 11 10 13 4 for which the sum of squares of each set of three consecutive numbers is 20454, just slightly less than the London arrangement. In contrast, if we were to arrange the numbers by just inter-weaving the largest and smallest numbers like this 20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10 the resulting sum of squares of each 3 consecutive elements is 20510, so the standard dart boards are, in this sense, more flat distributions.

Needless to say, all of these arrangements are much more flat than the natural monotonic sequence 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 which has a sum of 24350. By the way, note that if the sum of the squares of every sum of three consecutive numbers for a given arrangement is S, then we can form another arrangement with the same sum simply by taking the "21-complement", i.e., subtracting each number from 21. For example, the complement of the standard London arrangement is 1 20 3 17 8 15 11 6 19 4 18 2 14 5 13 10 7 12 9 16 which has the same sum (20478) as the London arrangement. This works because if we begin with an arrangement a,b,c,d,...

having the sum S = (a+b+c)^2 + (b+c+d)^2 + (c+d+e)^2 + ... and replace each of the numbers a,b,c,... with 21-a, 21-b, 21-c,... respectively, the sum S' of this complementary arrangement is S' = [(21-a)+(21-b)+(21-c)]^2 + [(21-b)+(21-c)+(21-d)]^2 + ... = [63-(a+b+c)]^2 + [63-(b+c+d)]^2 + ... = S + 20(63)^2 - 2(63)[(a+b+c)+(b+c+d)+...] Each of the numbers from 1 to 20 appears three times in the summation inside the square brackets in the last term, so that summation equals 630, and hence S' = S.

(The same identity applies to the N+1 complement for sums of squares of every sum of k consecutive terms of a circular arrangement of the first N integers.) How would we go about finding the circular arrangement of the integers 1 to 20 that gives the smallest sum of squares of every sum of three consecutive numbers? One possible approach would be to begin with the monotonic arrangement and then check each possible transposition of two numbers to see which one gives the lowest result. Then make that change and repeat the process, at each stage always choosing the transposition that gives the steepest reduction in the sum.

This "greedy algorithm" produces arrangements with the following sums (of squares of each 3 consecutive terms around the cycle): 24350 21650 20678 20454 20230 20110 19990 19970 19950 19946 19938 19936 19930 19926 19918 Once it reaches the arrangement with the sum 19918, no further transposition of two numbers gives any reduction in the sum. Of course, this doesn't imply that 19918 is the minimum possible sum, it simply means that it is a "local" minimum. We might try to make our search algorithm more robust by considering all possible permutations of THREE numbers at each stage.

(This includes permutations of two, since some of the permutations of three numbers leave one of the numbers fixed.) Applying the greedy algorithm to permutations of any three numbers gives dartboard arrangements with the sums 24350 21542 20362 20098 19978 19954 19942 19930 Once we reach 19930, no further permutation of three numbers gives any reduction in the sum. Interestingly, this doesn't even produce as low a result as the simple transpositions, and it illustrates the fact that a local minimum need not be a global minimum. By applying permutations of three elements, the algorithm is too greedy and enters a region of the configuration space that cannot be extended by such permutations, whereas the transpositions follow a less-steep path that leads them ultimately to a lower level.

Expanding our algorithm to examine all permutations of FOUR numbers, we get a sequence of dartboard arrangements with the following sums: 24350 20678 20190 19974 19932 19918 19910 19908 19902 19900 19896 19894 Thus we arrive at the lowest sum we've seen so far, but of course this is still just a local minimum, with no guarantee that it is the lowest possible sum.

Expanding our algorithm to take the best of all the permutations of FIVE number at each stage, we get the sequence of dartboard arrangements 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 6 2 19 4 5 16 7 8 9 10 11 12 13 14 15 3 17 18 1 20 1 2 19 9 5 18 7 8 16 10 11 12 4 14 15 3 17 13 1 20 10 2 19 9 5 18 7 8 16 6 11 15 4 14 13 3 17 12 1 20 These arrangements have the sums 24350, 20406, 1999

2, and 19874 respectively.

By applying various combinations of these algorithms to various initial arrangements, we can often arrive at the ultimate sum 19874, but never to any lower sum. However, there appear to be three distinct arrangements with this sum (up to rotations and reflections). Each of them has 1 adjacent to 20, so to compare the arrangements directly we will rotate and reflect them if necessary so that they begin with 20 and 1.

With this convention, the three minimal sequences, labeled (a), (b), and (c), are (a) 20 1 11 19 2 12 16 3 14 13 5 15 10 6 17 7 8 18 4 9 (b) 20 1 11 18 2 13 15 4 14 12 5 16 9 7 17 6 8 19 3 10 (c) 20 1 12 17 3 13 14 4 15 11 6 16 8 7 18 5 9 19 2 10 The differences between the (a) and (b) sequences, and between the (c) and (b) sequences, are shown below: 0 0 0 1 0 -1 1 -1 0 1 0 -1 1 -1 0 1 0 -1 1 -1 0 0

 1 -1 1 0 -1 0 1 -1 1 0 -1 0 1 -1 1 0 -1 0 Interestingly, if we reverse the order of the lower differences and then rotate two places to the right, the result is exactly the negative of the upper differences.

This is because the (a) and (c) arrangements are the 21-complements of each other (as defined above). The (b) arrangement is "self-dual", i.e., it is its own complement. We also note that (a) and (b) differ by the transpositions (3,4) (6,7) (9,10) (12,13) (15,16) (18,19) whereas (b) and (c) differ by the transpositions (3,2) (6,5) (9,8) (12,11) (15,14) (18,17) Thus the three minimal sequences differ from each other by permutations of six numbers, and no permutations of just five or fewer numbers can transform one of these to the others using the greedy algorithm, if we require the sum to drop or remain constant on each permutation. But if we allow permutations of six numbers it becomes possible to oscillate between these three arrangements in steps with constant sums. This is an interesting example of "symmetry breaking".

At lower "energies" (permutations of fewer terms) every sequence of arrangements progresses to one of several different possible stable limiting arrangements, but at higher "energies" (permutations of more terms) these asymptotic arrangements can transform into each other, so the sequence can oscillate between them. (Of course, if we allow permutations of all 20 terms at once, then any arrangement can be transformed to any other in a single step.) Despite the extensive numerical evidence, and the apparently unique symmetry of the (a), (b), and (c) arrangements, one could still question whether our search algorithm based on permutations of five elements is guaranteed to find the global minimum. To prove that the three arrangements (a),(b),(c) presented above are indeed the absolute minimal solutions, note that the sum of the sums of three consecutive elements must be 630, which is three times the sum of the integers from 1 to 20. If we didn't require integer values, the minimal solution would be given by uniformly distributing this, so each sum of three consecutive terms would be 31.5, but since we require integer values, this is ruled out.

We could consider arrangements such that every sum of three consecutive terms is either 31 or 32, but it's easy to see that this cannot lead to an acceptable solution. Notice that the two consecutive 3-sums for the four elements n1,n2,n3,n4 are n1+n2+n3 and n2+n3+n4, so if the two 3-sums are equal, it follows that n4=n1, and hence this is not an acceptable solution (the 20 elements are distinct). Similarly we can show that two 3-sums can't alternate more than twice. Hence the flattest possible arrangements that are not ruled out by these simple considerations must have more than two distinct values for the 3-sums Indeed the solutions with 19874 consist of the 3-sum values 30, 31, 32, and 33 with valences 6,4,4,6 respectively.

These 3-sums for the (a), (b) and (c) arrangements are as shown below (a) 32 31 32 33 30 31 33 30 32 33 30 31 33 30 32 33 30 31 33 30 (b) 32 30 31 33 30 32 33 30 31 33 30 32 33 30 31 33 30 32 33 31 (c) 33 30 32 33 30 31 33 30 32 33 30 31 33 30 32 33 30 31 32 31 By examining each sequence of the values 30, 31, 32, and 33, checking to see which ones correspond to 3-sums of the integers 1 to 20,

 we find that indeed the only viable sequences are those corresponding to the arrangements (a), (b), and (c).

Thus these are the circular arrangements of the integers 1 through 20 such that the sum of squares of every 3 consecutive terms has the smallest possible value, namely 19874. (If we evaluate the sum of squares of every three consecutive elements of these 3-sum sequences we find that they yield 178614, 178618, and 178614 respectively.) The (a), (b), and (c) sequences each consist of three interleaved arithmetic progressions. If we designate the position of each number by the integers modulo 20, then the positions of the values are as shown in the table below.

positions modulo 20 values (a) (b) (c) 3k+1 -3k-3 6k 6k k = 0 to 6 3k+2 6k 6k+3 -3k-3 k = 0 to 6 3k+3 6k+3 -3k-3 6k+3 k = 0 to 5 By the way, to find the arrangement that maximizes (rather than minimizes) the sum, it's fairly intuitive that we would cluster the largest numbers together as tightly as possible. This leads to the arrangement 20 19 17 15 13 11 9 7 5 3 1 2 4 6 8 10 12 14 16 18 which has the sum 25406.

Indeed this is the highest sum I've found using the greedy algorithm with permutations of 2, 3, 4, and 5 elements (selecting the highest rather than the lowest at each stage), although it's interesting that there are many initial arrangements from which this algorithm does not lead to this global maximum. In general if H(n,k) and L(n,k) are the highest and lowest sums of squares of every k consecutive elements in a circular arrangement of the first n positive integers, are the values of H(n,k) and L(n,k) well known and/or easily computed? Another possible way of "optimizing" the arrangement of the numbers 1 through 20 on a dart board would be to minimize the sum of the squares of every sum of TWO (rather than three) consecutive numbers. In general, I think the minimum sum of squares of every sum of two consecutive numbers in a cyclical arrangement of the integers 1 through N is S_min(N) = N^3 + 2N^2 + 2N - j where j is 1 if N is odd, and j is 2 if N is even. For the particular case N=20 this formula gives a minimum sum of 8838.

For even N the minimum arrangement has the odd and even numbers restricted to separate halfs of the cycle, as illustrated below for N=20 1 3 5 7 9 10 8 6 4 2 19 17 15 13 11 12 14 16 18 20 For odd N the minimum arrangement is very simple, as shown below for N=19. 1 3 5 7 9 11 13 15 17 19 18 16 14 12 10 8 6 4 2 This raises some interesting questions. Given any circular arrangement of the integers 0 through n-1, let S denote the sum of squares of every sum of two contiguous numbers, and let v(n) denote the number of distinct values of S for all n! possible arrangements.

Following is a table of the number of distinct values of v(n) n v(n) ----- --------- 1 1 2 1 3 1 4 3 5 8 6 21 7 43 8 69 9 102 10 145 Hugh Montgomery, A. M. Odlyzko, and Bjorn Poonen developed a very nice approach to this problem, showing that the general term with n>6 is given by / (n^3 - 16n + 27)/6 if n is odd v(n) = ( \ (n^3 - 16n + 30)/6 if n is even A whole family of interesting sequences can be produced by generalizing the definition as follows: Given any circular arrangement of the integers 0 through n-1, let S denote the sum of the qth powers of every sum of k contiguous numbers. Then let v(q,k,n) denote the number of distinct values of S for all possible arrangements. With this nomenclature, the previous sequence is denoted as v(2,2,n).

Of course, we have v(1,k,n) = 1 for all k and n, because the sum of the 1st powers is independent of the arrangement. We also have v(q,1,n) = 1 because the sum of any fixed power of the individual numbers is also independent of the arrangement. Also, for fixed values of q and n, the function v is PERIODIC in k. Another generalization is to add some constant integer j to each of the numbers 0 to n-1. Thus, the general function has four indices, v(q,k,j,n). Notice that v is independent of j for q<3, but for larger values of q, j becomes significant. Can v(q,k,j,n) be expressed in closed form as a function of the indices? Which other integer sequences are contained in this family? Which continuous functions (e.g., sin(x), cos(x), exp(x), etc) can be approximated by sequences of this form? Darts 501 - Darts and Maths Darts and Maths! Your browser does not support HTML5 video.

Video Synopsis When Philip New, the head of Year 6 at Woodland Middle School, set up a darts club, he noticed a significant change in their attitude to maths and their ability to do it. In this programme Philip explores darts as a vehicle for teaching different areas of KS2 maths. First, he gives his pupils a breakdown of the dart board followed by quick-fire questions to develop their mental calculation skills. From this, Philip moves to the interactive white board and asks the children to calculate his scores and share how they do their calculations. He also develops a range of other activities based on collecting and recording data and adapts the dart board to tailor lessons according to objectives, abilities and the time available. Celebrity darts player, Bobby George, shares a few tips with the pupils and demonstrates his own mental arithmetic skills. Phil feels that using darts has brought a new dimension to his maths teaching and is pleased to see the children smiling and enjoying what they're doing. This programme resources and support materials were available via Teachers TV website but as from 29th April 2011 the website was disbanded.

For anyone wanting further information regarding any of the 3,500 15min video that Teacher TV broadcast please visit the UK Department of Education website. click here . My thanks for Teachers TV for their permission to continuing broadcast this video. If you wish to download the interactive checkout the teacher uses plus supporting material you can from here. When visiting the site scroll to the bottom to primary games 1. You then have the option to evaluate the program or buy it. There are options for schools and personal use. 30th April 2011 / page updated June 2015 The Dartboard If you are new to darts you may have wondered why the numbers are arranged in this fashion. Well the numbering system used in the modern day dart board is accredited to Brian Gamblin a carpenter from Bury in Lancashire.However this cannot be categorically be proved and a more plausible person is also suggested (Read the History of Darts) The numbering system on a standard dartboard is designed in such a way as to reduce ‘lucky shots’ and reduce the element of chance. For further details about the board click here.

Know the Dartboard So now you know the board is arranged in a way that reduces lucky shots and punishes bad shots. Just look at what is either side of the 20; 5 and 1 so if you miss the 20 you are likely to hit a 5 or a 1. Many social players therefore throw most of their scoring darts towards the left hand side of the board from the 12 down to the 19 because they believe they have a higher percentage of hitting a higher score than throwing for the twenty / treble twenty and in many cases they would be correct. Regular players and the professionals you may have seen on TV never adopt this strategy / method of play. They will throw for the higher scoring trebles and regularly hit them this is purely down to practise. If you want to be good at the game not only do you need to practise and set yourself goals and targets to achieve. You also need to learn dart maths! The Dartboard Scoring Segments Darts Maths! ‘I hate maths’ I hear you say. Well we use maths everyday in a number of forms and you probably have never given it a second thought.

Every time you look at a clock you are using maths: every heard someone ask for the time and a reply coming back quarter past one? Similarly well you may know that is the same as 15 minutes past one and dart maths can be as easy as this. Depending on your age you may have heard of the following: 1. Addition + 5. Percentages % 6. Averages Ave Well the first four of these are used in every game of darts and the last two are used to judge how good a dart thrower is at scoring and hitting checkouts. A basic game of darts starts with each thrower starting from a number ending in ’01 such as 501, 301 etc. The reason why the start numbers end in ’01 is so that a dart thrower must hit at least one odd number to enable him or her to finish the game on a double. Each player throws three darts in turn and the sum of the score they hit is subtracted from the starting or the remaining total. The game must be finished by hitting a double equalling the amount they have left i.e.

If a player is on 20 and they hit a double 10 (2x 10 =20) then they win the game. If their score more than the 20 say they hit a double 15 (2x 15 =30) they bust and go back to the score they were on before they took their throwing turn. For the purpose of this game we are going to play 301. How to score Each player throws three darts into the board and the total number is added together then subtracted from the starting number or remaining total Example First throw: Dart thrower A hits 16 , 3 , 17 giving a total of ? 16 + 3 + 17 = 36 Now can you take the total scored (36) from the starting number 301? 301 - 36 = 265 This now leaves you now 265 to score, but before I go any further I will explain how to add and subtract, you may already know but if you don’t here is how to do it.

Lets take the three dart scores again 16, 3, & 17 by aligning them up under each other it makes adding them together easy. We align them up in HUNDREDS (H), TENS (T) & UNITS (U) and in this case we have do not have any HUNDREDS e.g. First of all we arrange the numbers under as follows: Fig 1 We then add together then UNITS column 6 + 3 +7 = 16. We place the 6 under the === sign and carry the 10 or 1 TEN over to the TEN’s Column: Fig 2 We now add the TEN’s Column including the carried 10 TEN 1+1+1 = 3 Fig 3 Now how to subtract this 36 score from the starting score 301. Remember how we added numbers together, well we subtract these in a similar fashion. First we arrange the two numbers in the same HUNDREDS (H), TENS (T) & UNITS (U) fashion but we put the number we are want to subtract from first followed by the number we wish to subtract.

Fig 4 So here we subtract the units first but 6 from 1 leaves a negative number and therefore we need to borrow a TEN to complete the calculation but as you can see we have a 0 in the TEN’s column therefore we need we need to borrow a ten from the HUNDRED column. Fig 5 Now you can see the HUNDRED’s column has reduced from 3 to 2 because we have taken Ten from it and now the Ten’s Column reads 10 not 0. We can now borrow a Ten from the TEN’s Column to place into the UNIT’s so we can subtract 6 Fig 6 6.1 Now the UNIT’s Column reads 11, the TEN’s Column reads 9 and the HUNDREDS column reads 2. We now proceed to subtract the 6 from 11 = 5 Fig 6 This is basic mathematics and I know arithmetic is now taught differently in school however this was how it was taught back in the 1970’s! It works! if you are an adult and find this hard to follow please. Click here. Now back to the game!

After throwing 36 in his first throw player A is left to score 265. In the next three throws his dart throwing accuracy improves and he starts to hits a few doubles and trebles which helps reduce his score quiet fast. Here is what he scored with his next three darts: Double 20, Single 18 and a treble 13 so what has he scored? Double 20 = (2 x 20 = 40) or (20 + 20 = 40) Single 18 = (1 x 18 = 18) or ( = 18) Treble 13 = (3 x 13 = 39) or (13+13+13 = 39) We add the totals of each dart together to as before: In his next throw, player A scores the following : Double 19 Treble 15 and Double 17 Can you work this one out? Here is his total score 117 but I will leave it to you to workout how I got there. Now subtract this from player A’s remaining total of 168 = 51.

I will again leave you to work this one out. Now the game of darts must be won by hitting a double that equals your remaining score. A bull’s eye (50 points) counts as a double outer bull (25). Therefore player A may wish to throw for a number 1 to leave 50 and if he hits the bull he wins the game, but there are many ways he can finish. Can you work them out? Here are a couple of ways to get you started. The player could hit a single 10 and a single 1 to leave 40 which is double 20 i.e. (2 x 20) He could hit a treble 15 (45) to leave 6 which is (2 x 3) There are many more ways and for a bit of fun see if you can record them all using only 3 darts! Times Table We have only touched on two other maths problems here so far. When we came to higher scoring Multiplication and Division come into play.

You may have heard the words ‘Times Table’ before in a maths lesson at school or may have built your own ‘Times Table’ sheets which may have looked something like this: I found this a handy device to help me learn and memorise basic Multiplication. So if I wanted to know what 5 x 9 totalled I would move down to the 5 and across to the 9’s column and before long I didn’t need the sheets at all. In darts you only need to know up to your 3 Times table up 20 and of course know that 2 outer bulls (2 x 25 = 50) Free to use for schools, colleges, students and teachers I have now added a free downloadable Times Table Sheet 25 x 25. Here are all the scoring numbers on a dartboard. Bullesye = 50 Outer Bullseye = 25 Ok so, you know how to add up, subtract, multiply and divide. Any decent player would not use a calculator or use H T U to add up and subtract fast. A decent dart player can work out basic maths and know how to throw for high checkouts faster than many maths teachers!

It is a question of learning. If you have ever watched a professional darts player on TV, when he or she is looking to win the game they know what to hit to checkout high scores Example: 143 This can be achieved a number of ways but because most dart players practise throwing for the treble 20 most of the time, if a combination shot can be achieved using a treble 20 more often than not they will use it. Therefore 143 = Treble 20 (60) Treble 17 (51) Double 16 (32) to win. But are there other ways to go? The answer yes, but I will leave you to work them out use the above table to help you if you need to. In the sport of darts there are a number of three dart finishes that can be hit, the highest being 170. This can be achieved by hitting Treble 20 (60) Treble 20 (60) and the bullseye 50 (which is counted as a double outer bullseye).

But you can not checkout 169, 168, 166,165, 163,162 & 159 with three darts! These are commonly known as bogey numbers because you can checkout higher numbers but there is no way of checking out these in three darts. All good dart players will know this and they will try to avoid leaving these numbers if possible. I mentioned earlier that a basic game of darts uses addition, subtraction, multiplication and division in all games but to see how good a dart thrower is we refer to statistics! ‘What is that!?’ I hear say, another boring maths lesson! Well, yes, but many adults find the statistics hard but for this purpose I will keep it very basic. The two most commonly used statistics used in darts are dart averages and percentage doubles hit. We will start with the latter first. What is a percent? Well a percentage is a term that can be described as a rate, number or amount in each 100.

If your teacher says you have got 98 questions correct out of a 100 then this can be described as you achieved 98% of the questions correct and to work a percentage out we just need to divide one number into another and multiply the answer by 100. Here are some examples Again you are set 10 questions you answer 8 questions correctly so what is this as a percentage: Well if we divide 8 / 10 = 0.8 if we now multiply this by 100 it now equals 80 therefore we can say you have achieved 80% Here is another you score 4 out of 8 attempts at goal when you play football so what is your percentage success rate? Well let’s see: 4 / 8 = 0.5 and if we multiply this by 100 we get 50 therefore your success rate is 50% But now back to darts, sometimes we see some statistics showing the frequency a player hits a target he is aiming for and the most common in darts is the doubles percentage hit.

To calculate this we basically look at how many darts are thrown for a double and how many are hit therefore if a player throws 1 dart and hits the double first time he or she is said to have a success rate of hitting a double 100% of the time (remember 1/ 1 x 100 = 100 or in this case 100%) But if he takes 2 darts to hit the double then he or she has hit 1 out of 2 and expressed as a percentage = 1 / 2 = 0.5 x 100 = 50 or 50% So during a course of a match the player has to win several games or in dart terms 'legs' to win a match. So he or she may have thrown 20 darts at double and hit a total of 7 this works out as 7 / 20 = 0.35 x 100 = 35%. So it can be said the success rate of the dart player hitting his or her doubles is 35%. Darts Checkouts and Average Calculators Click here...

Dart Averages It is one of the most referred to statistics in the game of darts. It gives a good indication of good a dart thrower when compared to others even if they don’t play against each other. Here is how you work out dart averages. Basically you just divide the number of darts thrown into the score achieved. Therefore if a player wins a game of 301 and he or she has taken 12 darts to do this then their dart average is 301 /12 = 25.08 this sometimes can be expressed as a throw and therefore this number is then multiplied by 3 (3 darts in a throw) = 75.24. This is very simple but what if you didn’t win the game what is your average? Well let’s look at the loosing player. He / she also threw 12 darts but only achieved 255 points so again we divide 255 / 12 = 21.25 or 63.75 per throw.

Averages are usually a sum of a number of games not just one but the same principle applies: you divide the total amount scored by the number of darts thrown. Generally speaking a dart thrower with a higher average is likely to win more games but this is not always the case because who ever throws first has an advantage and also a player may play a very bad game followed but several good ones. There are a number ways you can use darts to help your maths and in turn this helps your darts. Being confident about your maths will give you a edge over players that struggle with arithmetic. Here are a few questions you can test yourself on but please don’t email your answers or ask for the answers as I am not your maths teacher! How many ways can you score 26 with 3 scoring darts How many ways can you checkout 100 finishing on a double but not using more than 3 darts. What is the highest score you can hit using three scoring darts? what is the highest checkout you can hit with 3 darts? What is the lowest checkout you can do using all three 'scoring darts?

Hitting three different green segments of a darts board what is the highest number you can achieve in three darts? Hitting three different only red segments and not counting the bull what is the lowest number you can hit? If a player plays a game of 501 and he averages 60 points per throw with his 3 darts how many darts will it take him to finish the game ? (whole number only) Give two examples of checking out 82 with two darts. You must finish on a double! What is the sum or total of all the scoring segments on a dart board? Useful links: Another great resource site for helping you understand or teach maths can be found at the 'Interactive Resources Site.' click here

Question: What number is between 9 and 11 on a regulation dartboard?
Answer: 14