I was recently posed a question by a friend –
How many possible finishes exist within the darts game 501 which include 3 (or more doubles) and using no more than 9 darts?
For those unfamiliar with 501 the premise is as follows.
- Each player has unlimited darts which are thrown in a set of three.
- Each player begins with the score 501
- The aim is to reduce the score to zero in as few darts as possible
The ultimate aim of each player is to achieve a 9 dart finish. A nine-dart finish is a perfect leg, using only nine darts, the fewest possible, to checkout from 501. It is notoriously difficult to achieve, even by the game's top professionals. It is considered to be the highest single-game achievement in the sport, similar to a maximum 147 break in snooker or a 300-point game in bowling.
Example Answer
My answer was the following
- Darts 1,2,3 each hit Treble 20 (180 score leaving 321 remaining)
- Darts 4,5,6 each hit Bullseye (double) 150 score leaving 171))
- Darts 7,8,9 each hit Treble 19 leaving zero
How many further methods exist of scoring 501 with 9 darts and using three doubles in order to do so?
Edit to add Fundamentals of Scoring in Darts
A dartboard is divided evenly into 20 segments comprising the full 360 degrees of the board. The segments are numbered 1 to 20 representing the score for hitting each segment with a dart. Within the segments designated bands allow the score to be doubled or trebled.
Therefore the maximum score with a single dart is 60, achieved by hitting the treble 20. The lowest score is 0 (a miss) or 1 for hitting the 1 segment. In addition two centre circles exist worth 25 and 50 points respectively. The 50 point ring is classed as a double and is known generally as the bullseye.
With any single dart, barring a miss, a player has the chance to hit one of 62 point scoring zones, however the maximum score per dart cannot exceed 60.
Best Answer
Assuming that a treble does not count as one of the three doubles, let's look at the maximum score: which would be $510 = 6 \cdot 60 + 3 \cdot 50.$
Thus we must distribute "negative $9$" points among the 9 shots.
If any of the doubles is not a $50$ point double, it can at most be a $40$ point double; therefore all three double shots must be bull's eyes.
So the 'losses' must come from the trebles. $9$ is divisible by $3$ (fortunately!), so we have the 3 following options (in whatever shot order):
$3$ of $20$ trebles, $3$ of $19$ trebles, $3$ of bull's eye doubles.
$4$ of $20$ trebles, $1$ of $19$ treble, $1$ of $18$ treble, $3$ of bull's eye doubles.
$5$ of $20$ trebles, $1$ of $17$ trebles, $3$ of bull's eye doubles.