Pocket Opponent's Last Ball and Your Last Ball Also Falls In
6/13/2008 3:42:14 PM
Pocket Opponent's Last Ball and Your Last Ball Also Falls In
I have a question about a unique situation in Cutthroat pool:
What happens when you legally pocket your opponent's last ball and in the same shot, your last ball also drops into a pocket afterward?
For example, let's say that there are two balls left on the table, your ball and an opponents ball.
On your shot you knock your opponent's ball in the called pocket first, but then the cue ball continues moving and knocks your last ball in as well.
The result is a table with no balls on it at all except for the cue ball.
Does anyone win?
Or does everyone lose?
Chase TaylorThis question relates to the following billiard rules:
Pocket Opponent's Last Ball and Your Last Ball Also Falls In
Replies & Comments
billiardsforum on 1/13/2019 2:39:02 AMNeed to know how many total players there are.
If there is a third player who was "out" earlier, that player would also get to bring up one of their balls, and thus, be back "in" the game as well. You would be "out".
See these posts for info:
user1617240329 on 3/31/2021 6:39:07 PM
Why?
It doesn't sound like a scratch or foul was committed.
- billiardsforum on 4/4/2021 9:46:19 AM
@user1617240329 - Good point, not sure how I missed that the first time around.
So I guess the shooting player wins the game in this case because no foul was committed and the shooting player's ball was technically the last one remaining on the table.
Pocket Opponent's Last Ball and Your Last Ball Also Falls In
- Title: Pocket Opponent's Last Ball and Your Last Ball Also Falls In
- Author: Chase Taylor
- Published: 6/13/2008 3:42:14 PM
- Last Updated: 1/13/2019 2:30:11 AM
- Last Updated By: billiardsforum (Billiards Forum)
