Wags said:
The shooter and the opponent each need one ball and there are two balls on the table. The shooter plays a ball to his pocket and sinks it but also then makes the other ball in his opponents pocket in the same stroke of the cue.
Both players have 8 balls. The opponents ball actually went into his pocket first.
Who wins? What happens?
Well, technically, this could only happen if the two winning scores add up to 15, such as 8-7, 10-5, etc.. It can't happen that both players have 8 balls.
That aside, this is a very interesting question. I'd almost have to say that what counts is who's turn it is. If it is player A's turn, and he sinks his ball on his turn, he wins. Other than that, considering this situation can only come up during a match where one player recieves a spot.. It would have to depend on what the players agreed to before hand.
This will come up once every 10 years or so, so the most likely situation is that player A realizes he might make player B's ball also, stops, and ask what the rule is before he shoots the shot. Player B, of course, almost certainly having a monetary interest in the shot, will say that either: Player B wins, or the game is a draw.
I would say that the rule SHOULD be, if player A is shooting:
A. Player A wins..
or
B. Game is a draw and replayed.
IMHO, player B should NEVER be awarded a win on an inning where his opponent sunk his own winning ball.
Russ