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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?

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  • $\begingroup$ Can't you replace a 60 and N with a 60-N and 2N ? $\endgroup$ – David P May 9 at 11:20
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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; $(331-100)/4 = 57.75>57$),

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  • $\begingroup$ You haven't allowed for triple 18: $451-54$ is greater than $57\times 7$. $\endgroup$ – TonyK May 9 at 11:38
  • $\begingroup$ You cannot cover 451 with only triples. You need point that not divisible by 3. 54 is divisible by 3 by design. $\endgroup$ – Vasily Mitch May 9 at 11:46
  • $\begingroup$ @VasilyMitch: why is the minimum scoring at each throw not at least 501 - 8 * 60 ? $\endgroup$ – Luk May 9 at 11:52
  • $\begingroup$ 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. $\endgroup$ – Vasily Mitch May 9 at 11:56
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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.

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