I've stumbled across what I believe to be an interesting result - feel free to argue that it isn't!
Say we simplify the game of darts such that we only play with two darts. Then, according to my calculations, there are $1323$ possible ways to 'checkout' (finish on a double), the lowest being $2$ (Double $1$), the highest being $110$ (Treble $20$ + bullseye).
Interestingly (or not!), there are exactly $882$ possible ways to checkout from the even numbers: $2,4,6,...,110$; and $441$ possible ways to checkout from the odd numbers: $3,5,7,...,107$ ($107$ is the maximum odd number checkout). Therefore there are exactly twice as many ways to checkout from the even numbers when compared to the odd numbers.
A coincidence?
Probably worth mentioning that this trend vanishes when you extend to the full three dart version of the game: $82047$ ($45444$ even + $36603$ odd).