What happens when all the same color balls (solids or stripes) go in on the break?

Question

I played on a pool league (TAP league, 8-ball) a few years ago and some of us pondered what would happen according to the rules if a pool player made either all solids or stripes on the break (but not the 8 ball).

We figured (with our rules anyway), since any balls made on the break does not automatically determine which color you had to make for that game, then you would be forced to go for whatever the other color was still on the table, essentially putting your opponent on the 8-ball as soon as one was made. I would figure that either player wouldn't want that situation so they would both most likely choose to stalemate (unless a player was confident they could run the table), but I was wondering if there was any precedent or rule for this situation.

Oh yea, and spare me any responses like "It's probably not possible so don't worry about it!". Just asking out of fun curiosity.

Thoughts? Stories? Please share!

Answer 1

Looking at the rules page of the World Pool-Billiard Association, your understanding is right. For the break, Groups are not yet determined.

3.3 (b) No ball is called, and the cue ball is not required to hit any particular object ball first.

After the break, you can determine your group by calling a ball and pocketing it. But the rules explicitly say that you can't strike the eight-ball when the table is open.

3.4 Open Table / Choosing Groups Before groups are determined, the table is said to be “open,” and before each shot, the shooter must call his intended ball. If the shooter legally pockets his called ball, the corresponding group becomes his, and his opponent is assigned the other group. If he fails to legally pocket his called ball, the table remains open and play passes to the other player. When the table is “open”, any object ball may be struck first except the eight ball.

So your only option would be playing a safety (possibly resulting in a stalemate) or trying to run the table.