Mickey makes a good point. If people would just embrace the World Standardized Rules, represented in America by the BCA, there would be much fewer arguments. Now, I do understand why an 8-ball on the break is considered a win on coin-op tables. It's not worth another dollar to spot the 8-ball, so the convention became that an 8-ball on the break would be a win. Like Mickey says, just make sure you agree ahead of time what rules you want to play under. M.D.-BCA Instructor/Referee.
2006-12-29 06:41:48
·
answer #1
·
answered by straight_shooter526 6
·
0⤊
0⤋
Yes. My dad is on a team and he wins by sinking in the 8 ball first
2006-12-29 06:36:44
·
answer #2
·
answered by Pinay 2
·
0⤊
0⤋
Your first question should be ....."Before we start what are the (bar) rules?"......every place or city has their own set of rules....here if you call the ball in the corner pocket and it even "rides" the rail just before it goes in it is a "miss".....so before you start...ask about the rules we will be playing under....and for what its worth if everyone would just use BCA rules it would stop all of the "hassles" that come with "bar rules"....they are not set in stone (official books).....so watch it.......especially in a "bar" with someone you don't know and he has been there awhile and a bottle in his hand.....good luck on the tour!!!
2006-12-29 03:31:43
·
answer #3
·
answered by Mickey Mantle 5
·
1⤊
0⤋
Ike D is right. It is a win as long as the cue ball does not go in as well.
2006-12-28 22:46:28
·
answer #4
·
answered by T_Jania 3
·
0⤊
0⤋
yes
2006-12-29 09:00:52
·
answer #5
·
answered by Anonymous
·
0⤊
0⤋
Yes, as long as you don't scatch also.
2006-12-28 22:41:19
·
answer #6
·
answered by Al Dave Ismail 7
·
1⤊
0⤋
It sure does,providing you do not scratch
2006-12-28 22:46:30
·
answer #7
·
answered by AVENGER 2
·
0⤊
0⤋
Yep, it sure does!
2006-12-28 22:40:02
·
answer #8
·
answered by capnemo 5
·
0⤊
0⤋
no u lose
2006-12-28 23:41:24
·
answer #9
·
answered by champ 1
·
0⤊
0⤋
ABSOLUTELY!
2006-12-29 08:38:25
·
answer #10
·
answered by allamerican 2
·
0⤊
0⤋