Minimum Snooker score in ONE break.
M.Warren.
Ok..what is the theoretical MINIMUM break you can achieve in a frame of
snooker under normal match-play rules...potting ALL the balls.
Mike Warren.
p.s. for those who dont know..you get 15 reds 1 each of yellow green
brown blue pink and black. red is worth 1 point..
yellow to black worth 2-7 (yel=2 gree=3 brow=4 blue=5 pink=6 black=7)
if you pot a red you MUST pot a colour afterwards..then another red. and
so on. when all reds potted you then work up thro the colours in order
yel gree brow blue pink black.
Dafydd Price Jones
In article <4m4do4$s...@newton.cc.rl.ac.uk>, "M.Warren." <mdw95
@mailbox.cc.rl.ac.uk> writes
>Ok..what is the theoretical MINIMUM break you can achieve in a frame of
>snooker under normal match-play rules...potting ALL the balls.
>Mike Warren.
>
Minus seven.
All balls potted in one "foul" shot.
--
| Dafydd Price Jones dafy...@dafyddpj.demon.co.uk
Incipe, parve puer: qui non risere parenti,
Nec deus hunc mensa, dea nec dignata cubili est.
Virgil, Eclogues no. 4, 1.62
M.Warren.
"M.Warren." <md...@mailbox.cc.rl.ac.uk> wrote:
>Ok..what is the theoretical MINIMUM break you can achieve in a frame of
>snooker under normal match-play rules...potting ALL the balls.
>Mike Warren.
>brown blue pink and black. red is worth 1 point..
>yellow to black worth 2-7 (yel=2 gree=3 brow=4 blue=5 pink=6 black=7)
>if you pot a red you MUST pot a colour afterwards..then another red. and
>so on. when all reds potted you then work up thro the colours in order
>yel gree brow blue pink black.
>
>
you give 4 points to your opponent and you dont score if you also
potted a 'legal' ball at the same time.
Eric Morin
M.Warren. wrote:
>
> Ok..what is the theoretical MINIMUM break you can achieve in a frame of
15x1 red balls
15x2 yellow balls (one after every red)
3 green
4 brown
5 blue
6 pink
7 black
-----------
70 points
ObPuzzle:
You have 6 apples in a basket and 6 guests. How do you manage to give
each guest an apple, leaving one in the basket?
--
_/_/_/ _/ _/ Eric Morin @ SwitchView Inc.
_/ _/_/_/_/ If you only have a
hammer,
_/_/ _/ _/ _/ 884-3262 (home) you tend to see every
problem as a nail.
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_/_/_/ _/ _/ 519-746-4460 x523 (work) If you only have
a screwdriver...
Rev. Andrew Krywaniuk
: > Ok..what is the theoretical MINIMUM break you can achieve in a frame of
: > snooker under normal match-play rules...potting ALL the balls.
: 15x1 red balls
: 15x2 yellow balls (one after every red)
: 3 green
: 4 brown
: 5 blue
: 6 pink
: 7 black
: -----------
: 70 points
Aha, but what if you were to sink all the red balls on a single shot?
There is no penalty for this, and you would score 14x2 less for a total
score of only 42.
Andrew
Charles Bryant
In article <iLH9HjAY...@dafyddpj.demon.co.uk>,
Dafydd Price Jones <dafy...@dafyddpj.demon.co.uk> wrote:
>In article <4m4do4$s...@newton.cc.rl.ac.uk>, "M.Warren." <mdw95
>@mailbox.cc.rl.ac.uk> writes
>>snooker under normal match-play rules...potting ALL the balls.
>>
>Possible spoiler, but maybe not the answer you're looking for.
>Minus seven.
>All balls potted in one "foul" shot.
I would agree, except of course to say that I don't think it is possible
to pot all the balls on one shot. Assuming that the minimum number of
balls is potted by each shot, the answer is 65 + seven away. If ALL
balls are potted, the white must go down. This leaves a foul on the last
shot, which counts as seven away because the nominated ball was the
black. By deciding on how many balls it is possible to pot with one
shot, a more definite answer can be given between these two limits.
Jan Kok
In article <4m6p7j$q...@nntp.ucs.ubc.ca> hob...@unixg.ubc.ca (Rev. Andrew Krywaniuk) writes:
>: 15x1 red balls
>: 15x2 yellow balls (one after every red)
>: 3 green
>: 4 brown
>: 5 blue
>: 6 pink
>: 7 black
>: -----------
>: 70 points
>
>Aha, but what if you were to sink all the red balls on a single shot?
>There is no penalty for this, and you would score 14x2 less for a total
>score of only 42.
No, make that 72 and 44 (in stead of 70 and 42), as the yellow ball
is put back one more time.
Ob Qu: Now, why did O'Sullivan not shake hands at the end of one session
last week? (the question was ask by a reader of the Volkskrant).
--
--Jan Kok E-mail: Jan...@cwi.nl | =#===-=========##=
Address: CWI (dpt. NW) | -- ,___@
P.O. Box 94079 / 1090 GB Amsterdam | -- __/\
URL: http://www.cwi.nl/~jankok | ' /_
Dafydd Price Jones
>>>Ok..what is the theoretical MINIMUM break you can achieve in a frame of
>>>snooker under normal match-play rules...potting ALL the balls.
In article <1996-04-3...@chch.demon.co.uk>, Charles Bryant
<c...@chch.demon.co.uk> writes
Yes, you're right. But I picked on the word "theoretical" in the
question, and chose to ignore the practicalities.
Mind you, there could well be an arrangement of the balls on the table
from which it's possible to pot the lot. If we assume a game in which an
indefinite number of shots can be played before the one in question,
then that arrangement of the balls could be reached.
Another argument suggests that a player who possesses enough strength
could theoretically give the ball a whack which is strong and accurate
enough to get every ball down.
Cheers,
Dafydd.
Matthew Daly
In article <31869C...@switchview.com> Eric Morin <eric....@switchview.com> writes:
>ObPuzzle:
>You have 6 apples in a basket and 6 guests. How do you manage to give
>each guest an apple, leaving one in the basket?
SPOILERS
I suppose you can give five apples to five of your guests, and then
give the basket with the remaining apple to the sixth guest....
-Matthew
MATTHEW PRIESTLEY
Matthew Daly (da...@PPD.Kodak.COM) wrote:
: >You have 6 apples in a basket and 6 guests. How do you manage to give
Put the sixth guest in the basket and make sure he stays there :)
--
___
/ /\ MATTHEW PRIESTLEY
/ /::\
/ /:/\:\ prie...@students.uiuc.edu
/__/:/-/:/\
\_ \:\/:/\ \ http://www.cen.uiuc.edu/~priestle
\ \::/-/ /
\_ \:\/ / You can write "HOT" in snow.
\ \:\/
\__\/ Rage is optional.
Pete Mitchell
In article <4m4do4$s...@newton.cc.rl.ac.uk>,
M.Warren. <md...@mailbox.cc.rl.ac.uk> wrote:
>Ok..what is the theoretical MINIMUM break you can achieve in a frame of
>snooker under normal match-play rules...potting ALL the balls.
I'm not positive what "match-play" rules are, but the minimum points if
you ran the break would be 15*3 + 27 = 72 points (sinking the yellow
for each cherry, then running the colors). Was that the question?
- Pete
--
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
: "There but for the grace of God, go I."
Pete Mitchell :
p...@mv.mv.com : No, wait...
: Shit, that IS me! :-)
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
Dave Ewart
M.Warren. (md...@mailbox.cc.rl.ac.uk) wrote:
: >Ok..what is the theoretical MINIMUM break you can achieve in a frame of
: >snooker under normal match-play rules...potting ALL the balls.
: >Mike Warren.
: >yellow to black worth 2-7 (yel=2 gree=3 brow=4 blue=5 pink=6 black=7)
: >if you pot a red you MUST pot a colour afterwards..then another red. and
: >so on. when all reds potted you then work up thro the colours in order
: >yel gree brow blue pink black.
: >
I would guess that it is 44. You pot ALL reds at the same time, 15, then
play the yellow, 17. Then all the colours, 44.
Yes?
Dave.
--
Dave Ewart (Imperial Cancer Research Fund, Oxford)
ew...@europa.lif.icnet.uk
Gary Duke
In article <4m6p7j$q...@nntp.ucs.ubc.ca>, hob...@unixg.ubc.ca (Rev. Andrew Krywaniuk) writes:
> : > Ok..what is the theoretical MINIMUM break you can achieve in a frame of
> : > snooker under normal match-play rules...potting ALL the balls.
>
> : 15x2 yellow balls (one after every red)
2 yellow
> : 3 green
> : 4 brown
> : 5 blue
> : 6 pink
> : 7 black
> : -----------
> : 70 points
>
> Aha, but what if you were to sink all the red balls on a single shot?
> There is no penalty for this, and you would score 14x2 less for a total
> score of only 42.
>
> Andrew
--
Gary Duke, Maths Dept, King's College London, Strand, WC2R 2LS
Email: G.A....@kcl.ac.uk, UDA...@kcl.ac.uk "Gods play games with the
fates of men. But first they have to get all the pieces on the board,
and look all over the place for the dice."(Soul Music, by T Pratchett)
Eric Morin
At 19:04 30/04/96 -0400, you wrote:
>remember you can pot all 15 at once from break off(granted VERY
>unlikely..probably nearly impossible)..but its allowed in the rules..(i did say
>theoretical;)...also you could go in-off from the last black..so minimum is
>37 .;)
>Mike.
Good one!
I take it you are required to spot the yellow after potting it the first time then?
PS: I wonder if it really is "theoretically" possible to sink all the reds in one shot, given
that...
1. there is a maximum force the balls will tolerate and remain on the table
2. you have to shoot on an angle to avoid the colored balls
3. no colored balls can go in
M.Warren.
ew...@europa.lif.icnet.uk (Dave Ewart) wrote:
>M.Warren. (md...@mailbox.cc.rl.ac.uk) wrote:
>: >Ok..what is the theoretical MINIMUM break you can achieve in a frame of
>: >snooker under normal match-play rules...potting ALL the balls.
>: >p.s. for those who dont know..you get 15 reds 1 each of yellow green
>: >brown blue pink and black. red is worth 1 point..
>: >yellow to black worth 2-7 (yel=2 gree=3 brow=4 blue=5 pink=6 black=7)
>: >if you pot a red you MUST pot a colour afterwards..then another red. and
>: >so on. when all reds potted you then work up thro the colours in order
>: >yel gree brow blue pink black.
>: >
>: >
>
>I would guess that it is 44. You pot ALL reds at the same time, 15, then
>play the yellow, 17. Then all the colours, 44.
>
>Yes?
>
>Dave.
>--
>Dave Ewart (Imperial Cancer Research Fund, Oxford)
>ew...@europa.lif.icnet.uk
>
my initial answer was 37 ..cos you COULD go in-off on the last black..
therefore the 7 for the black wouldnt count...then it hit me.
Of course...the minimum break potting ALL the balls is 0 if we enter
the twilight zone and ALL the balls go in off the break!!;))
Mike Warren.
p.s. I once saw a computer simulation on tomorrows world..where they
had simulated NO friction at all...so off the break eventually ALL
the balls got potted;)
M.Warren.
Ok...the question was potting ALL the balls...therefore pot the
last black and make sure the white is potted as well...the 7 therefore
for the black doesnt count..making minimum 37..that was my first
answer...then it occured to me...if we are assuming the 15 reds get
potted at once..then we COULD have the scenario where ALL the balls get
potted off the break.(and so incurring a foul)..making minimum break 0.
(do-do do-do do-do do-do):))
MIke Warren.
Jan Kok
In article <3187E8...@switchview.com> Eric Morin <eric....@switchview.com> writes:
...
>PS: I wonder if it really is "theoretically" possible to sink all the reds
>in one shot, given that...
>
>1. there is a maximum force the balls will tolerate and remain on the table
>2. you have to shoot on an angle to avoid the colored balls
>3. no colored balls can go in
>--
it would certainly require the help of the opponent to position
all reds in front of the pots, such that two or three tours would
knock in all the reds, but then still: expecting that at least near
every corner pot the queue ball cannot just skim the red(s), as
it will probably have to make a right angle next (well, not necessarily),
it will lose a lot of impetus, so the starting velocity needed could
be too high. We don't want the queue ball to go through the side cushion
:-). Or : one collision will already give the queue ball a substantial
vertical speed, and it will leave the table.
Now, what about trick shots. Has it been done?
Speaking about the help of the opponent, a favorite remark of mine
is that the maximum highest break is not 147 (see FAQ, re: teasers)
but 155.
Thomas D Bending
> : > Ok..what is the theoretical MINIMUM break you can achieve in a frame of
> : > snooker under normal match-play rules...potting ALL the balls.
>
> : 15x1 red balls
> : 15x2 yellow balls (one after every red)
> : 3 green
> : 4 brown
> : 5 blue
> : 6 pink
> : 7 black
> : -----------
> : 70 points
>
> Aha, but what if you were to sink all the red balls on a single shot?
> There is no penalty for this, and you would score 14x2 less for a total
> score of only 42.
Not quite - the calculation you quote counts a yellow potted after
each red, but omits the yellow potted as the first of the clearance
colours. The minimum break should be 44 points.
Thomas Bending JANET: t...@uk.ac.qmw.maths
<a href="http://www.maths.qmw.ac.uk/~tdb/homepage.html">WWW homepage</a>
Richard Goodman
"M.Warren." <md...@mailbox.cc.rl.ac.uk> wrote:
>"M.Warren." <md...@mailbox.cc.rl.ac.uk> wrote:
>>Ok..what is the theoretical MINIMUM break you can achieve in a frame of
>>snooker under normal match-play rules...potting ALL the balls.
>>p.s. for those who dont know..you get 15 reds 1 each of yellow green
>>brown blue pink and black. red is worth 1 point..
>>yellow to black worth 2-7 (yel=2 gree=3 brow=4 blue=5 pink=6 black=7)
>>if you pot a red you MUST pot a colour afterwards..then another red. and
>>so on. when all reds potted you then work up thro the colours in order
>>yel gree brow blue pink black.
>>
>>
>you give 4 points to your opponent and you dont score if you also
>potted a 'legal' ball at the same time.
>
>
all fifteen reds go in
pot the yellow
then all the colours
total 44
Another snooker question
In one break of four shots you pot the yellow on the first three shots and miss the fourth shot scoring six points.
How can this be done?
David Burn
Richard Goodman wrote:
[snip]
> Another snooker question
> In one break of four shots you pot the yellow on the first three shots and miss the fourth shot scoring six points.
>
> How can this be done?
SPOILER:
The last red and all the colours are on the table, but your opponent has
just played a foul shot which leaves you snookered on the red. You
nominate the yellow as a free ball and pot it, simultaneously potting
the red (2 points). You now pot the yellow as a colour following the
last red (2 points) and pot it again as the first colour in sequence (2
points, making six in all).
How you miss your fourth shot after doing all this is a mystery to me!
David Burn
db...@lnhdent.agw.bt.co.uk
M.Warren.
>>Ok..what is the theoretical MINIMUM break you can achieve in a frame of
>>snooker under normal match-play rules...potting ALL the balls.
>>Mike Warren.
>
>you ran the break would be 15*3 + 27 = 72 points (sinking the yellow
>for each cherry, then running the colors). Was that the question?
>
>- Pete
>
so the answer is 0 .(cos theorectically you could pot all the balls
off the break.(very unlikly tho;))..
MIke.
match play rules say you dont score if you foul off the break...a foul
is if you hit any other ball except red..OR if ANY other ball including
the cue ball (except red) gets potted.
Ross Parker
Eric Morin <eric....@switchview.com> wrote:
Ignoring the snooker stuff... (BTW the argument for a score of zero
sounds the most convincing to me - for what its worth)
>ObPuzzle:
>You have 6 apples in a basket and 6 guests. How do you manage to give
>each guest an apple, leaving one in the basket?
Possible spoiler below...
Give five guests an apple each and give the last guest the basket with
the last apple in it?
Cheers,
Ross.
Lovett Shahar
minimum score is 1 point.
you pot a red and in the same show the white ball in it's track pots all
the rest of the balls.
let see u go under that.
shalov.
Amiga lover / CHASE OUT
Eric Morin (eric....@switchview.com) wrote:
: ObPuzzle:
: You have 6 apples in a basket and 6 guests. How do you manage to give
: each guest an apple, leaving one in the basket?
: --
Give all guests an apple each - and put one guest in the basket!
--
================================
Daniel Roach
Cm2b...@bs47c.staffS.Ac.uk
Amiga owner, Rem and Indie music fan!
"Thunder only happens when it's raining"
9dani...@gmail.com
Pot all 15 reds from the break. Then pot a yellow = 17. Pot all the colours 27 + 17 = 44. Job done.
denkbaar
>
> Pot all 15 reds from the break. Then pot a yellow = 17. Pot all the colours 27 + 17 = 44. Job done.
>
p p p 1C
p p X p
p XX p p
1D p p X
...
5 suits (including NT) * 7 bidding levels = 35.
Every suit/NT bid is followed by 8 others, makes 9 per suit/NT bid.
So that would be 3 + 35*9 = 318, of which the last 2 passes (after 7NT
rdbl) are meaningless, so one could argue for 316.
Is that correct?
d.
Mark Brader
>
> p p p 1C
> p p X p
> p XX p p
> 1D p p X
> ...
>
> 5 suits (including NT) * 7 bidding levels = 35.
> Every suit/NT bid is followed by 8 others, makes 9 per suit/NT bid.
>
There are two errors. First, after the 35th and last redouble there
have to be *three* passes to close the auction, so that's 319 calls.
Second, this is only the longest possible *legal* bidding sequence.
Since a player is allowed to accept an insufficient bid by the
preceding player, the auction could actually go in indefinitely.
West: 4 spades.
North: Pass.
East: 4 hearts. (Oops.)
South: Pass. (She knows that 4H is insufficient but holds a heart stack
and is happy to defend against that contract.)
West: 4 spades. He realizes why South passed and won't leave 4H in.
Now repeat ad infinitum.
(Yes, of course this would never happen in real life -- but then, neither
would the auction Denkbaar describes.)
--
Mark Brader, Toronto | "If we gave people a choice, there would be chaos."
m...@vex.net | -- Dick McDonald
My text in this article is in the public domain.
denkbaar
> There are two errors. First, after the 35th and last redouble there
> have to be *three* passes to close the auction, so that's 319 calls.
> Second, this is only the longest possible *legal* bidding sequence.
> Since a player is allowed to accept an insufficient bid by the
> preceding player, the auction could actually go in indefinitely.
>
> West: 4 spades.
> North: Pass.
> East: 4 hearts. (Oops.)
> South: Pass. (She knows that 4H is insufficient but holds a heart stack
> and is happy to defend against that contract.)
> West: 4 spades. He realizes why South passed and won't leave 4H in.
>
> Now repeat ad infinitum.
>
> (Yes, of course this would never happen in real life -- but then, neither
> would the auction Denkbaar describes.)
Thanks, Mark.
duncan smith
>
> Pot all 15 reds from the break. Then pot a yellow = 17. Pot all the colours 27 + 17 = 44. Job done.
>
clearance. There's also an argument for a solution less than 44.
Duncan
Anders D. Nygaard
/Anders