8 ball break

Philthepockets

Philthepockets

AzB Silver Member
Silver Member
Is this ruling correct?

If you break in 8‑ball and all seven balls of one group (all solids or all stripes) drop on the break, the ruling is actually very simple — and it’s the same across all major rule sets.


🎱​

Pocketing balls on the break does not give you ball‑in‑hand or any special continuation. The break is always considered its own turn. Once the balls stop moving, your inning ends unless you pocketed the 8‑ball (which is a separate rule).

So even if you made all seven solids (or all seven stripes):

  • You do not continue shooting
  • Your opponent comes to the table
  • Your opponent is automatically assigned the other group
  • Your opponent must shoot at the remaining seven balls of their group
 
If a ball is made, you continue shooting!
I don’t know what is the ruling if all balls of a group are made on the break, whether you shoot the 8 or the other group, my guess is that we will never know as this probably can’t happen with normal legal rack

My guess it that since it is an open table, you can shoot the 8 for the win
 
What rule set is that?

I've never heard of an 8-ball rule set where the player who breaks does not continue to shoot if he made at least one ball on the break.

The only difference lies in whether the table remains open after balls of a single suit are made on the break or if he is automatically assigned the suit he made...
 
Philthepockets said:
Is this ruling correct?

If you break in 8‑ball and all seven balls of one group (all solids or all stripes) drop on the break, the ruling is actually very simple — and it’s the same across all major rule sets.


🎱​

Pocketing balls on the break does not give you ball‑in‑hand or any special continuation. The break is always considered its own turn. Once the balls stop moving, your inning ends unless you pocketed the 8‑ball (which is a separate rule).

So even if you made all seven solids (or all seven stripes):

  • You do not continue shooting
  • Your opponent comes to the table
  • Your opponent is automatically assigned the other group
  • Your opponent must shoot at the remaining seven balls of their group
Click to expand...
They must play by different rules in Canada, everywhere else that I have seen when you make at least 1 ball on the break and do not foul you get to continue shooting and your opponent stays in their chair.
 
Heres a different AI take on the same question.

📘​

Under WPA, BCA, and APA rules:

  • The table is open after the break, no matter how many balls drop.
  • Groups are not assigned until a player legally pockets a called shot after the break.
  • But if one group is completely gone, the remaining group is automatically assigned to the incoming player.
Since the breaker cannot shoot again (the break turn is over), the opponent gets the table with their group already determined.
 
If this (all 7 of one suit on the break) ever actually happened, surely a reasonable person would agree that the breaker has choice between the 8 ball or taking the other balls. Unless playing take what you make, then they're on the 8 ball.

Ah, this is addressed in the WPA rules (2025-09-15) - that AI is also wrong, there's no automatic assignment:

4.4 OPEN TABLE / CHOOSING GROUPS
...When the table is “open,” any object-ball may be struck first except the 8-ball. Hitting the 8-ball first when the table is open is a foul unless a group has been completely pocketed. In this situation, the shooter may temporarily claim that group, and thereby be shooting the 8-ball, possibly for a win.

The "temporarily" implies (I think) that if they shoot at but miss the 8 ball, it's still open; opponent can then attempt the 8.
And note that this could happen after the break, if all of one suit gets pocketed before any called shot is made. Also unlikely...but possible.

Interesting. But yes I know WHO CARES it's (probably) never going to happen.
 
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