an interesting question

an interesting question

Posers and Puzzles

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r

Joined
08 Feb 04
Moves
1881
24 Feb 04

this is my first attempt at starting a thread, but here goes 🙂🙂😀😀
how many different combinations of 6 darts will it take to get to 301 (for example 5 bulls eyes [50/bull for 250] and a triple 17 [51 points])

r

Joined
20 Mar 04
Moves
1920
20 Mar 04

i would say 8 - 10 different ways to get to 301

d

Joined
04 Mar 03
Moves
4274
20 Mar 04

Interesting puzzle and I would like to try it...but I know nothing of the scoring in darts (other than what you let me know in your example). Is it this simple:

each dart can be worth 50 (bullseye), 0 (miss the target entirely),
or 1x, 2x or 3x where x is an integer between y and z

I would just need to know what are the values of y and z.

Thanks!

r

Joined
08 Feb 04
Moves
1881
20 Mar 04

Originally posted by ddebened
Interesting puzzle and I would like to try it...but I know nothing of the scoring in darts (other than what you let me know in your example). Is it this simple:

each dart can be worth 50 (bullseye), 0 (miss the target entirely),
or 1x, 2x or 3x where x is an integer between y and z

I would just need to know what are the values of y and z.

Thanks!
what i would like to know is how many different combinations there are with only _6_ darts to get 301 points.

the circle in darts is divided into 20 sections. and then each part of the circle (except for the middle) is divided in to 4 sections in order from outside to inside (double, single, triple, single). the middle bulls-eye for most are 50 points but some games its 25 and 50 (outer and inner) but for this the whole area is 50 points.

HAPPY HUNTING!! 😀

Now With Added BA

Loughborough

Joined
04 Jul 02
Moves
3790
21 Mar 04

Originally posted by rattlerz28
what i would like to know is how many different combinations there are with only _6_ darts to get 301 points.

the circle in darts is divided into 20 sections. and then each part of the circle (except for the middle) is divided in to 4 sections in order from outside to inside (double, single, triple, single). the middle bulls-eye for most are 50 points ...[text shortened]... its 25 and 50 (outer and inner) but for this the whole area is 50 points.

HAPPY HUNTING!! 😀
In other words, there are 81 targets which count as 'different' for the purposes of this puzzle.

r

Joined
08 Feb 04
Moves
1881
21 Mar 04

yes. but for the purpose of this i want to know how many times 301
points can be achieved with only 6 darts? iknow the obvious one (5 bulls and a triple 17) and if your real cute ( 5 triple 20's and a single 1!) how many more after that.
HAPPY HUNTING!!!!!! 😀😀😀😀

Barefoot Chessplayer

central usa

Joined
22 Jul 03
Moves
61132
22 Mar 04
1 edit

Originally posted by rattlerz28
yes. but for the purpose of this i want to know how many times 301
points can be achieved with only 6 darts? i know the obvious one (5 bulls and a triple 17) and if you're real cute (5 triple 20s and a single 1!) how many more after that?
HAPPY HUNTING!!!!!! 😀😀😀😀
five 3*(triple-)19 (285) and a 16 or 2*(double-)8 are two.
four 3*20 (240) plus any of: 2*20 and 3*7, 3*19 plus 4, 3*18 plus 7, 3*17 plus 10, 3*16 plus 13, 3*15 plus 16, 3*14 plus 19, or bull plus 11 (each adding up to 61) are eight more.
four 3*19 (228) plus any of: 3*20 plus 13, 3*19 plus 16, or 3*18 plus 19 (73) (three more).
four 3*18 (216), 3*15, and 2*20 make one more.
four 3*17 (204), 3*19, and 2*20 are yet another.
three 3*20 (180) plus any of: two 3*19 plus 7, two 3*18 plus 13, two 3*17 plus 19, or two bulls plus 3*7 (121) also do it (four more combos).
three 3*19 (171) plus any of: two 3*20 and 10, or bull and: 3*20 and 20 (or 2*10), 3*18 and 2*13, 3*16 and 2*16, 3*14 and 2*19 (130) also comprise 301 (six more).
three 3*18 (162) plus any of: two 3*20 and 19 or two bulls and 3*13 (139) are two.
that brings us to three 3*17 (153) plus: two 3*20 and 2*14, two 3*19 and 2*17, two 3*18 and 2*20, two bulls and 3*16, or bull, 3*20, and 2*19 (148) (another five).
lastly, three 3*16 (144), two bulls, and 3*19 (157)--probably as far as we can go.
i presume we are talking unique combinations (shot-order-independent), so five 3*20 plus 1 is the same as two 3*20, 1, and three more 3*20.
can anyone come up with more? that above plus the two original come to 35, i believe.