In American pool...when the black ball is pot only 2-shots into the game..does it count as a loss?

Question

I had the break and on my second shot i pot the black ball do i loose??

Answer 1

yes, in any 'pool' potting the black before you have potted all of your own balls, it results in an immediate loss.

sorry!! bad luck mate

xx    

Answer 2

I assume you're talking about "stripes and solids" or sometimes called "hi-low." Where you shoot the solids and your opponent shoots the stripes (or vice-versa). Yes, in all variants of the game dropping the eight when there are still balls on the table is a loss.

In strict ABC rules (American Billiards College) sinking a ball, any ball, without calling it is a scratch.

Answer 3

Yes you lose.

Answer 4

Sorry you lose .In some or most places "Bars" if it goes on the break is the only time you can make the 8 before your suit of balls have been made

In standard rules-even that would be a re-rack or spot the 8.It would be the other players choice

Answer 5

Yep

Answer 6

I assume when you refer to American pool you're talking about 8 ball. In this case, yes, you lose. See the rules below. With the second shot, you have to intentionally pocket a ball to determine your group (solid or stripes), so it's not possible to win if you pocket the 8 ball on the second shot.

Answer 7

Yes. Potting the "black ball" (8-ball), just two shots into the game, is indeed a loss.

In a standard American Pool game of Eight Ball, it IS a loss, if you pot/pocket the "black ball" (the 8-ball) before you have cleared *your entire suit of balls (either *solids or stripes).

The Eight ball or "black ball" is to be the final/last ball pocketed/potted without scratching, to constitute a win in a standard game of American Eight-Ball.

Rack 'em up again. Better luck next time  ;-)

LAG

Answer 8

You lose!