"Going Downstairs" in Professional Darts

A week ago I flicked on the TV and caught the tail end of a quarter final in the World Darts Championship. One of the pundits (with lots of gold chains round his neck) was saying "if you're not hitting twenties, go for 19s, and if that doesn't work try 18s, even 17s if you have to!". The message was that if you start missing the treble 20 then you should switch it up and go for lower numbers until you found a treble you could hit more reliably. This got me thinking, do darts players actually do better if they switch to aiming at treble 19 after missing a couple of treble 20s? To do so they would surely have to get more accurate after switching to 19s than they were when aiming at 20s, to compensate for the lower score for each dart. I mentioned this to Dave Millican at work and he suggested that 19 might be a safer target for a player with an accuracy problem because the segments either side of it aren't such low scores as those surrounding the 20. This is a good point, the 20 is flanked by 5 and 1, whereas the 19 has the slightly friendlier 7 and 3 for company. Although the same logic doesn't support what the pundit was saying about going for 18 if 19 wasn't working for you, as the 18 has the 1 and the 4 either side which is worse than the 20!

A quick search on Google Scholar revealed a very interesting paper written by researchers at Stanford who have looked into this question and found some interesting results. They calculated the optimal place to aim on the dartboard for players of varying levels of skill. Skillful players consistently land their darts close to the point that they're aiming at, so when aiming at a fixed point their darts land in tight groups; in other words, the standard deviation of the distance between the landing point and the aiming point is small. Rubbish players' darts land all over the place; they have a high standard deviation. Tibshirani, Price, and Taylor found the best place for any player to aim at by maximising the integral of the score over the area in which their darts are likely to land, weighted by the likelihood that their darts land there, for any given aiming point. As you might expect, they found that a very good player, with a standard deviation of only a few millimetres, should aim at the treble 20 for maximum points. A rubbish player, with a very high standard deviation, should aim very near the centre of the board, to minimise the chance of missing the board altogether! The interesting bit is what happens in between very good and rubbish. As standard deviation increases from zero, the optimum aiming point moves slightly up and to the left, to favour hitting the 5 instead of the 1 on the occasional wild dart. Then, when standard deviation increases beyond 16.4mm, the optimum aiming point jumps to treble 19! As the standard deviation further increases, perhaps after the third pint, the optimal aiming point curves upwards and then around to the right until it settles just to the left of the bullseye.

Figure 1: movement of the optimum aiming point as a player's standard deviation increases (reproduced with permission from A Statistician Plays Darts , Tibshirani, Price, Taylor - JRSS Series A, Vol. 174, No. 1, 213-226, 2011)

So it seems that our pundit was right in one case, it is worth "going downstairs" for the treble 19 if your aim on treble 20s degrades past a certain point. But in terms of maximising expected score, it's not worth switching to 18 if your aim on 19 isn't good, nor to 17 if your aim on 18 isn't good. Of course there may be psychological factors, and a certain bias due to a player's dart distribution being skewed at an angle and therefore making her better at hitting trebles at one angle than another (this is also investigated in the paper).

In terms of whether darts players do the right thing during matches (statistically speaking) there is still a question mark over switching to treble 19. Players obviously only switch to 19 if they start missing treble 20s, in other words their internal estimate of their standard deviation rises and they react by switching to 19. The question is how does this internal estimate compare to the true value, and how close to the ideal threshold of 16.4mm do they switch between 19 and 20? This is much more difficult to answer than our first question. To do so we would need dart by dart position data for a player over many many games, including many occasions where the player switched to 19, which most players don't do that often. Perhaps I should write a video analysis algorithm to watch and datafy TV footage of darts and then crunch the numbers to see which players' mental estimates are closest to the truth. My guess would be that some players tend to switch too soon because of one random dart going awry when in fact this is not sufficient evidence that the underlying standard deviation has actually changed. On the other hand, maybe the player has a good idea of not only what his aim is like at the moment, but whether he's feeling more tense and therefore about to get worse, enabling him to preemptively switch to 19 before hitting the 1!

The Quickest Way to Rack Pool Balls

Whilst playing pool in the college bar years ago my friend Chris posed an interesting question: after removing the balls from the end of the table and randomly dropping them into the triangle, what is the minimum number of 'moves' necessary to rearrange the balls into the correct pattern? Where a 'move' is swapping the positions of two balls.

It turns out that if the black ball starts off in one of the centre positions (marked with an 'X' in Fig 1) then you canĀ always get to the correct pattern in 3 swaps or fewer. If the black starts off in one of the outside positions then you can still get to the correct pattern in 3 swaps 75% of the time, the remaining 25% require one additional swap. An explanation of this result is given below...

The correct pattern, as defined by the World Eightball Pool Federation, is shown on the left in Fig 1:

The official pattern (left) and symmetrical variations of it (right).

Fig 1: The official pattern (left) and symmetrical variations of it (right).

To make our lives easier, we will also allow the three patterns on the right of the figure, as these are all symmetrical, in colour or side, versions of the official pattern. The table is symmetrical, so these 4 are all equivalent. We also note that turning the triangle 1/3 of a turn either way is very easy, so we will not count this as a 'move'. This means that all 3 rotated versions of each of the above 4 patterns are counted as correct. In total then we have 12 patterns which we consider to be 'solved'.

Let's number the positions, starting from 1 at the bottom left corner, and counting to the right along the bottom row, up to 5, then left to right along the second row, from 6 to 9 etc... If we represent each red ball as a '1', each yellow ball as a '0', and the black as a '2', then we can write down the pattern of balls in the official arrangement (left of Fig 1) as a list of numbers, or vector, namely:

[1 0 0 1 0 0 1 0 1 1 2 0 0 1 1]

If we swap the bottom left red with the yellow on its right (positions 1 and 2), we get a slightly different pattern:

[0 1 0 1 0 0 1 0 1 1 2 0 0 1 1]

If we have these two vectors on a computer we can easily find which elements don't match by using the 'not equal to' operation, in Matlab this is denoted by '~=', so for example:

[1 0 0 1 0 0 1 0 1 1 2 0 0 1 1]


[0 1 0 1 0 0 1 0 1 1 2 0 0 1 1]


[1 1 0 0 0 0 0 0 0 0 0 0 0 0 0]

In this way we can identify which, and how many, balls differ in position from one pattern to the other. One swap changes the positions of two balls, so if we divide the number of differences between any two patterns by two we get the number of swaps necessary to get from one pattern to the other. We have to round this number up, if it's not an integer, because if two patterns differ by the positions of three balls (one of which must be black) then two swaps will be required to change one into the other. We can't do one and a half swaps!

It is a fairly simple operation to compute all patterns that are possible with seven '1's, seven '0's and one '2'; there are 51,480 of them. Using the above method we can find the number of swaps necessary to change each of the 51,480 possible patterns into each of the 12 solutions. Taking the smallest of these 12 numbers for each pattern we find the answer to our question:

If the black is in one of the central positions (which happens in 3 of every 15 patterns) then the pattern can always be solved in 3 swaps; if the black is not in the centre then the pattern can be solved in three swaps 75% of the time, and always in 4 swaps!