# Compute minimum amount of 60s to obtain a 9-darts finish in 501 double darts out

In 501 double out darts, it is possible to reach the final score of 501 with only 9 darts. The dartboard has fields 1 - 20, and also double and triple fields and the semi-bull (25) and bullseye (50). The last dart needs to hit a double field, i.e. either of:

[2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 50]

I already know all possible 9-dart-finish combinations (you can look them up on the internet). I have noticed that you need a minimum of three 60s to obtain a 9-darts finish. Now I wonder: is there a way to prove this? Or is there at least a way to see that you need to hit a 60 at least once in order to obtain a 9-darts finish? Because in 301 darts you can obtain a 6-darts finish without ever hitting a 60.

Would anyone care to show an example calculation?

• Can't you replace a 60 and N with a 60-N and 2N ? May 9, 2021 at 11:20

Proof you need at least one 60:

The minimum scoring at each throw should be at least $$501-8\times57=45$$. Thus, the only double that suffices this condition is bullseye $$50$$. Now we have a problem getting $$451$$ points in 8 throws.

The score $$451$$ isn't a multiple of $$3$$, so you cannot cover it with triples. The largest points not divisible by 3 is again bullseye $$50$$. However, $$451-50 = 401 > 57\times7$$.

Edit: proof you need at least three:

By analogy, the minimum scoring at each throw is $$501−6×57-2×60=39$$. So you can finish the game either with $$40$$ or $$50$$.

With 40: You need to cover $$501-2×60-40=341$$ points with 6 throws. It's not divisible by 3, so one throw have to be at least 50, $$(341-50)/5=58.2 > 57$$.

With 50: You need to cover $$501-2×60-50=331$$ points with 6 throws. $$331\equiv 1 \mod 3$$, so we need at least one throw with a remainder 1 (the largest one is 40; $$(331-40)/5 = 58.2>57$$) or two throws with a remainder 2 (the largest one is 50; $$(331-100)/4 = 57.75>57$$),

• You haven't allowed for triple 18: $451-54$ is greater than $57\times 7$. May 9, 2021 at 11:38
• You cannot cover 451 with only triples. You need point that not divisible by 3. 54 is divisible by 3 by design. May 9, 2021 at 11:46
• @VasilyMitch: why is the minimum scoring at each throw not at least 501 - 8 * 60 ?
– Luk
May 9, 2021 at 11:52
• Because you are not allowed to throw 60, right? The next points after 60 is 57. I am going to add proof for at least 3 60s. May 9, 2021 at 11:56

The highest score with one dart is a triple $$20$$ $$(60)$$ and the second highest is triple $$19$$ $$(57)$$. The highest score with a final dart is a bullseye $$(50)$$ and the second highest is double $$20$$ $$(40)$$.

Suppose you want to finish with a non-bullseye double, and no more than two triple $$20$$s. Then your maximum score is $$2 \times 60 + 6 \times 57 + 1 \times 40 = 502$$ which is $$1$$ too much, and you cannot reduce this nine-dart total by exactly $$1$$ since any change reduces any of the triples by $$3$$ or the double by $$2$$, and a change from a triple to a double or single or bullseye is at least $$7$$.

Suppose you want to finish with a bullseye $$(50)$$, and no more than two triple $$20$$s. Then your maximum score is $$2 \times 60 + 6 \times 57 + 1 \times 50 = 512$$ which is $$11$$ too much, and you cannot reduce this nine-dart total by exactly $$11$$ since any change reduces any of the triples by a multiple of $$3$$, and a change from a triple to a double or single is at least $$17$$, while a change from a triple $$20$$ to a second bullseye leaves an impossible-to-close $$1$$ and a change from a triple $$19$$ to second bullseye leaves an impossible-to-close $$4$$.

So there is no nine-dart finish without at least three triple $$20$$s.