We were playing a game of 8-ball and this is what happened:
Player 1 had no balls besides the 8-ball left.
Player 2 had only 1 ball and the 8-ball left.
It was Player 2's shot, and in the same shot, sunk the 1 remaining ball, and after that, scratched.
I know normally if you scratch on the 8-ball, you lose, but in this scenario, who won?
2006-08-06 19:19:06 · 4 answers · asked by p42oo 1 in Games & Recreation ➔ Gambling
the game isn't over yet. if you sank the remaining ball and ALSO in that very same shot scratched it is your opponents turn to go for the eight off a scratch.
2006-08-06 19:25:15 · answer #1 · answered by The Answerer 2 · 0⤊ 0⤋
If the scratched occurred after the one remaining ball entered, but before player two had a chance to call the shot or attempt a shot on the eight ball, then it is player one's turn. If player two called the shot and said they would hit the last remaining ball and the eight ball in on the last shot and then scratched during the attempt, player two lost.
2006-08-06 19:24:53 · answer #2 · answered by Mariposa 7 · 0⤊ 0⤋
Yes, you are right about if you scratch on the 8-ball you loose, but Player 2 was not going for the 8-ball, they were shooting the ball they had left. It would now be Player 1's turn
2006-08-07 01:07:23 · answer #3 · answered by on here 3 · 0⤊ 0⤋
How is this a crazy scenario?? I play in a league and see this happen all the time.....
2006-08-07 14:21:18 · answer #4 · answered by Bensalem Strangler 2 · 0⤊ 0⤋