View Full Version : Giving weight in One Pocket
05-03-2003, 05:09 PM
I've seen people say they were playing 11 - 9 in One Pocket. Is that actual games won? So the one guy is basically giving up 2 games?
I only ask cause 11 games of One Pocket would take all night to play, wouldn't it? I mean some of those games just become defensive shot after defensive shot.
05-03-2003, 06:33 PM
11-9 means one player has to make 11 balls in his or her pocket before the other makes 9.
11 games would take a long time, most one-pocket I have seen is played by the game $5 to $1,000 a game. Most one pocket games are very defensive, but strong players do run out when given the chance. /ccboard/images/graemlins/tongue.gif <font color="blue"> </font color>
Thanks Carlton, I have had the same question for a long time.
11-9 in one pocket sounds unlikely if the players know what they are doing; Let's see, there are 15 balls in a rack, so perhaps 11-4 is a posibility, or 9-6, and any other combo that equals 15. If the players were doing a race 11-9 in one rack, as soon as the player going for 11 got 7, the player going for 9 would only have 8 to shoot at. /ccboard/images/graemlins/confused.gif
LOL, that was a funny, right?
05-04-2003, 01:55 AM
if both players were to play even,they would both need eight balls in order to win ... wouldn't you agree?
8 and 8 is ?
each player would 'owe' balls from the start of the game.in this case maybe the better player would owe 3 balls and the weaker player 1 then the game proceeds normally after all balls are paid ?
Balls are spotted to create the 11-9 weight. I've seen alot of players matched up 9-7, 10-7, ect. It is extremely common, I believe especially in $$$$$ games.
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