DCC related question - gambling handicaps

B

Buckster_uk

AzB Silver Member
Silver Member
Hi guys,

Have been around pool for a long time but have never really learnt about the whole "giving balls" and handicaps gambling aspect that are used at the DCC.

Can anyone give me a brief rundown of what getting the 7-9 etc mean?
 
If the players are racing to a certain number of games, the stronger player would have to win 9 games before the weaker player won 7 games.

If this were one pocket, the stronger player would need to make 9 balls before the weaker play makes 7 balls.
 
Buckster_uk said:
Hi guys,

Have been around pool for a long time but have never really learnt about the whole "giving balls" and handicaps gambling aspect that are used at the DCC.

Can anyone give me a brief rundown of what getting the 7-9 etc mean?
Click to expand...

The way I read this, it's the 7-out, which means you win if you make the 7, 8, or 9. If you're getting the 7, you win if you make the 7 or the 9. If you get the 8, it's the 8 or the 9. Games on the wire, you have to make a certain number of games before your opponent. Getting the break, you break all the time, your opponent doesn't break at all.
In heads up action, there are all kinds of handicaps. Balls, games on the wire, the breaks, and any combination thereof.
 
So if you get the 7, you have to reach the 7 or 9 to win. What happens if you come back to the table and pot the 8, do you still need to pot the 9?
 
Pushout said:
The way I read this, it's the 7-out, which means you win if you make the 7, 8, or 9. If you're getting the 7, you win if you make the 7 or the 9. If you get the 8, it's the 8 or the 9. Games on the wire, you have to make a certain number of games before your opponent. Getting the break, you break all the time, your opponent doesn't break at all.
In heads up action, there are all kinds of handicaps. Balls, games on the wire, the breaks, and any combination thereof.
Click to expand...

Yes, and kazoo is correct when referring to one pocket.

But in 9 Ball, it would never be referred to as 7-9. Rather it would be either.

1. The 7 ball...in which case you win when making either the 7 or 9 or
2. The "7 out" which includes the 7, 8 and 9...BUT you have to agree whether the 7 AND the 8 are "wild" i.e. you win if you pocket either on the break. Sometimes only the 7 and 9 win on the break unless you agree otherwise up front. You win if you pocket the 8 too...but not on the break.

That distinction needs to be agreed upon up front.

Regards,
Jim
 
av84fun said:
Yes, and kazoo is correct when referring to one pocket.

But in 9 Ball, it would never be referred to as 7-9. Rather it would be either.

1. The 7 ball...in which case you win when making either the 7 or 9 or
2. The "7 out" which includes the 7, 8 and 9...BUT you have to agree whether the 7 AND the 8 are "wild" i.e. you win if you pocket either on the break. Sometimes only the 7 and 9 win on the break unless you agree otherwise up front. You win if you pocket the 8 too...but not on the break.

That distinction needs to be agreed upon up front.

Regards,
Jim
Click to expand...

I've always played as "wild" meaning wherever it goes the ball wins, otherwise you have to call the pocket for those balls, break or afterwards (no slop). But if you make it in that pocket even on the break you win.
 
I understand most of the handicapping methods, but the one I still don't get is this:
Last night on TAR's free feed Tater was playing Bobby Pickle some 9 ball. Tater was getting the 6 ball and the last three in an 8 ahead set.

I understand the part about Tater getting the 6, but what does "getting the last three" refer to?

Thanks,
Jeremy
 
thats the same as getting the 6 out. potting the 6, 7, 8 or 9 wins the game for that player.
 
he can win if he makes the 6 ball, the last 3 part means he wins if he makes any of the last 3 remaining balls on the table.

example lets say the games been going on........the 6 is off the table, there has been combos or balls made on the break........and like the 2, 4, 5 and 9 are all thats left. The shooter getting the spot is at the table. Once he makes the 2, there will only be 3 balls left (the last 3). He can win by making any of them b4 his opponent. Its a bigger spot because in the exact same situation, if the shooter had been getting the 7 and 8, he could only win by making the 9, because the 7 and 8 had already been played somehow.

thats my understanding anyway.
 
last x: if there are x or fewer balls on the table, any make is a win for the shooter with the spot. i really prefer giving this to giving wild balls.

-s
 
scottycoyote said:
Its a bigger spot because in the exact same situation, if the shooter had been getting the 7 and 8, he could only win by making the 9, because the 7 and 8 had already been played somehow.

thats my understanding anyway.
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i see this as a smaller spot, because he doesn't get any BS wild wins early in the game on on the break. Also, most people who give this type of spot are getting out consistently from the last 5 (x+2).

-s
 
steev said:
i see this as a smaller spot, because he doesn't get any BS wild wins early in the game on on the break. Also, most people who give this type of spot are getting out consistently from the last 5 (x+2).

-s
Click to expand...

good point, ive always looked at it that if the balls are indentified you can go to extra trouble to play them in combos and safes, etc.....but you might be right hadnt thought about it like that

i was going along the line of thought that the 7 and 8 werent wild though......called 7 8 versus last 3. Wild 7 and 8 is definitely a bigger spot
 
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