One pocket question

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Marlo

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Two players, A & B. Player A is on the hill with 7 balls, player B has -1 (owes one ball). Player B takes his turn and runs 8 balls. Player B needs 9 balls to win and there's only 8 balls on the table. Is the game over?
 
I'm new to One Pocket as well, but when looking for the rules earlier today I stumbled across this and figured it might shed some light on the subject:

Any balls which fall into a pocket other than one of the foot pockets will be spotted at the end of the player's inning. If there are no more balls on the table and a player's inning is still alive then any balls in illegal pockets will be spotted so the player can continue his inning. (An illegal pocket is any pocket other than one of the foot pockets).

For a breakdown of ALL the rules, take a peek here:
http://www.bestbilliard.com/rules/display.cfm?file=one_pocket.cfm
 
When player B has pocketed all eight of the balls remaining on the table, but still 'owes a ball', one of the pocketed balls is placed on the spot and shot from where the cueball lays (or is it lies?). That 'spotted ball' becomes the winning ball for both players (as they are now both 'on the hill).... Doug
 
Smorgass Bored said:
When player B has pocketed all eight of the balls remaining on the table, but still 'owes a ball', one of the pocketed balls is placed on the spot and shot from where the cueball lays (or is it lies?). That 'spotted ball' becomes the winning ball for both players (as they are now both 'on the hill).... Doug
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Correct.
 
Oh yeah, SB is corect. If you are in that position you need play position for the spot off your eigth ball so you have position when the ball is spotted. Been there before.
 
But isn't the ruling that balls are spotted IN BETWEEN innings and not DURING the inning?
 
Good question! What usually happens is in most games of 1h, you wouldn't run out of balls and be forced to put up a ball so a player could continue a run because when two players are playing even, there are plenty of balls on the table for a player to run out without having to spot a ball for him/her to continue a run. Also, the norm is, most players will only run a couple of balls or so and then play a safe resulting in spotting up the ball after his/her inning. This makes it appear that the answer to your question would be yes. However, in a game where there is a handicap given as described by the initial post, there aren't enough balls on the table to continue the run. Therefore, the player that is running the balls will need to play shape for the spotted ball on his/her second to last shot.

I hope all this makes sense!
 
Donald A. Purdy said:
Player B until he misses a shot.
How ya doing Hemi?
Don
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Great Don, and you? Don is right. In theory, a player could need ten or what ever balls and as long as he plays position, straight in for his pocket, he could keep shooting the spotted ball, drawing his CB to be straight in again until he has made all the balls he needs.

Obviously, this could only happen when one player is spotting the other.
 
hemicudas said:
In theory, a player could need ten or what ever balls and as long as he plays position, straight in for his pocket, he could keep shooting the spotted ball, drawing his CB to be straight in again until he has made all the balls he needs.

Obviously, this could only happen when one player is spotting the other.
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Well, I was under the impression (told by others) that if player A has made seven balls and then missed and player B owes two balls (-2) and makes the remaining eight balls. He thens spots up BOTH balls at the same time and it's still his shot. In other words, he doesn't spot up one ball, make it, draw back, spot up the 'other' ball and then make it. He is left with two frozen balls on the spot (because he was -2 when he ran the eight) and had to spot them both... What say ye ? Doug
 
Smorgass Bored said:
Well, I was under the impression (told by others) that if player A has made seven balls and then missed and player B owes two balls (-2) and makes the remaining eight balls. He thens spots up BOTH balls at the same time and it's still his shot.
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I'm afraid Bored is correct, Hemi. I was playing this guy Jimmy Vegas one time. He needed one and I needed eleven. In other words, I owed three and needed a doctor. Well, I ended up running the eight balls on the table, and proceeded to spot one of the three owed balls to continue my run. The long and short of it is that we argued for a few minutes and I won the argument. I guess it was my lucky day, because I later found out that all owed balls are paid at once and the inning continues for the shooter. It doesn't make much sense, though...seems like a penalty for a helluvan out.
 
I also agree with SB. When you owe more than one ball, they're all spotted at the same time.
 
This very question has been answered more than once on other boards. All balls owed are spotted, if possible. Obviously, if 3 balls are owed and there are only 2 available balls, the player still owes 1 ball.

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
 
And the book "Hustler Days" they mention that some players spot 15 - 5 and so on. What happens when player A has 4 and player B just ran 11 but still needs 4? Do you spot up 4 balls? rerack the whole rack? or spot up 1 at a time? or even play it like straight pool and use the last ball to break a rack of 14 to continue shooting?
 
the player that has to get 15 spots his first 4 balls after his round, like you would if they were made in another pocket.
 
Already answered !!! Spot all balls similtaneously when zero balls remain on the table.

Troy...~~~ It ain't rocket surgery folks.
efirkey said:
what if all 15 are down in the middle of a run?
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