One pocket question

  • Thread starter Thread starter Marlo
  • Start date Start date
Another situation comes up when not playing even. Example, playing 8-7.....
Each player needs 1 ball (7-6). Two balls on the table. If a player makes his ball AND the opponemts ball on the same shot, the game goes to the shooter.

Troy [/B][/QUOTE]

That has never come up for me, Troy, but where did you get that rule? I tend to agree with it. It makes more sense than the one that I have heard, which is game goes to the first ball that dropped. Now that could make for an ugly argument. It's amazing as many games as I've played that it's never happenned.
 
That has never come up for me, Troy, but where did you get that rule? I tend to agree with it. It makes more sense than the one that I have heard, which is game goes to the first ball that dropped. Now that could make for an ugly argument. It's amazing as many games as I've played that it's never happenned.

I get that from none other than players such as Mike Massey, Grady Matthews, Willie Jopling, Ronnie Allen, Bucktooth, and others who have been around the game a long time.

You're correct, it hardly ever comes up though.

Troy
 
Player A has 7 balls, Player B needs 9 balls (owes one ball)... Player B runs the table and then takes an intentional foul because he's not out and cannot spot up a ball in the middle of his inning. No balls are spotted in the middle of one's inning, that's why Player B has to take an intentional foul if he runs the table.
 
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