If I am able to shoot in the 8-ball and my last object ball in a single shot and I call the 8 ball in a pocket, do I win? Does the order in which the 2 balls leave the table matter?
One official source: WPA rulebook.
Rule 3.8c says that the shooter loses if he pockets the eight ball in an uncalled pocket.
Furthermore, Rule 3.6 says that every shot must be called (except for the break) AND says that "The eight ball may be called only after the shot on which the shooter's group has been cleared from the table."
Therefore, you cannot possibly call the eight ball on a shot as you described. And since you did not call it, you lose.
(Side note about a common misconception in pool - If you look at rule 3.6a, if you're shooting at the eight ball you only lose on a scratch if you also pocket the eight ball. In other words, if you scratch but the eight ball doesn't go in, the game continues.)
According to Wikipedia, you actually lose
Any of the following results in a game win:
- The opposing player illegally pockets the 8 ball (e.g. before clearing all of that player's object balls, does so on the same shot as the last such object ball, or the 8 falls into a pocket other than the one that was designated)
Technically in my experience in tournament play, if you pocket a ball of yours in before you pocket the 8-ball but on the same shot, it is the same as scratching on the 8-ball shot because technically you aren't supposed to hit any other ball before you hit the 8-ball with the cue ball. In this case, you do not lose because scratching on an 8-ball shot is not an automatic loss. The second player receives a chance to position and pocket any remaining balls. If you pocket the 8-ball and then your last ball, this is also a lost game because you pocketed the 8-ball before all of your balls were pocketed. That defines your loss.