View Full Version : 8 ball rule question from Archer vs Strickland
12-21-2007, 09:00 PM
In the final game of the Archer vs Strickland match Earl broke and made a ball but didn't have a shot. He called a strip that was on the rail close to the bottom right pocket. To make the shot it looked like he had to curl the cue ball around a solid. He clipped the solid on the way down the table, but definitely got the cue ball to the rail. Archer took the cue ball in hand.
I thought that in an open table you could play any ball in a combination and hit a solid first even if you were calling a stripe. I don't understand how this was a foul.
12-21-2007, 10:21 PM
I believe that prior to the beginning of the game, they stated that a house rule or special rule, in effect for this game, would be that what ever ball you made on the break was going to be your group. The table wasn't open. This was to increase the difficulty of the game for the players.
12-21-2007, 11:59 PM
It had to be something like that IMO...sid
12-22-2007, 06:17 AM
I take it you guys never played in the APA. They follow that format. Personally, I prefer the open after the break rule.
Why punish the breaker when he finally gets a good break, makes a ball or two or three and then he has no shot to make one of his balls or, worse, he can't even hit his ball.
In the APA giving up BIH is no big thing but at the pro level it is devestating. The player not only usually runs out but then he may win the next several games playing winner breaks.
Look at all the IPT players who made six packs. I watched Robb Saez run six and he made it look so easy.
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