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Cutthroat Pool Pocket Your Last Ball, then Opponent's Ball Drops

12/20/2018 2:17:58 PM

Cutthroat Pool Pocket Your Last Ball, then Opponent's Ball Drops

What happens in cutthroat billiards when you pocket your last ball, but then your opponent's ball also drops into the pocket after that?

For example, playing cutthroat pool, 2 balls are left on the pool table. One ball is yours, and the other your opponent's ball. Can you pocket your ball first then your opponent's ball and still win?

This question relates to the following billiard rules:

Cutthroat Pool Pocket Your Last Ball, then Opponent's Ball Drops

Replies & Comments
  1. D. Schiewerbilliardsforum on 12/23/2018 12:21:26 PM

    In cutthroat pool, you win when you are the only player with at least one ball left.

    In your scenario (if I've understood your scenario correctly), your opponent's ball comes back out, and yours stays down.

    At this point your opponent's ball is the only ball left on the table thus, your opponent has won the game.

    Relevant rule sections are:

    Cutthroat Pool - ILLEGALLY POCKETED BALLS

    [When there are illegally pocketed balls, the] opponent's group balls are spotted; no penalty. The Shooter's group of balls remain pocketed, no penalty.

    I am not an expert in this game, so please correct me if I am wrong.

    Also see similar (if not, the same) questions from others:

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Cutthroat Pool Pocket Your Last Ball, then Opponent's Ball Drops

  • Title: Cutthroat Pool Pocket Your Last Ball, then Opponent's Ball Drops
  • Author:
  • Published: 12/20/2018 2:17:58 PM
  • Last Updated: 12/22/2018 6:13:15 PM
  • Last Updated By: billiardsforum (Billiards Forum)