nine-ball spots
- Thread starter tradr48
- Start date
If you're shooting and you have multiple balls on table, a win for you will be shooting down to the 9 (clear all lowet balls) or making the 9 early.
If you come to the table and only the 9 is up then you must make it to win.
This spot is generally considered the least amount of weight that one can receive in 9 ball.
If you come to the table and only the 9 is up then you must make it to win.
This spot is generally considered the least amount of weight that one can receive in 9 ball.
When two balls are left they are both your money balls. When there are more then two balls you can only win on the nine.
Unlike getting the 8, you don't have an extra ball you can make early.
With all due respect, gentlemen, allow me to help with the explanation.
Your explanations in bold seem very confusing.
"The last two" is a spot—as Lonestar states— that is considered very weak.
How it works is:
the player receiving the spot only has to make the highest-numbered ball before the nine ball,
regardless of which ball it is.
So at any given time, if the player getting the spot is shooting, and, let's say, the 4, 7, and 9 balls remain,
the player merely has to pocket the 7 (the highest-numbered ball before the 9).
As macguy alludes to, one doesn't know what the highest-numbered ball before the 9 will be,
which essentially "protects" the spotter from the other player simply trying to "ride the cheese" on that extra money ball.
Again, it's not much of a spot, as spots go.
If you give this spot to someone and they engage in argument because they get confused after your explaination,
then DON'T PLAY THIS PERSON. If they are too stupid to understand this dynamic, they aren't worth your time.
Last edited:
Now I want my lost money back on early combos ! I didn't know the early pocketed highest non 9 was also a win for someone getting the last 2. But I do confess openly to being stupid and shouldn't gamble at pool.
I still say that in my mind once the 4-7 combo drops that the shooter still has "2" balls up, hence my confusion.
I still say that in my mind once the 4-7 combo drops that the shooter still has "2" balls up, hence my confusion.
Yes, Lonestar. You seem to understand the issue now.
Early combos don't mean a thing [except with the 9-ball of course] with this spot because..........
..........The highest-numbered ball before the 9 can change during the course of the game
if those combinations, slops, and other game-playing dynamics occur.
It's only when you get down to the last few balls of a game that this "last two" spot makes itself evident.
Last edited:
Yes, anything legally played on the 9-ball remains germane for either player.
It's only when you get down to the LAST TWO balls left on the table where the spot is DECIDED.
When the person receiving the spot legally sinks the SECOND-TO-LAST-BALL, that player wins.
Of course, the person giving the spot still would have to run out to the 9-ball in order to win.
ONE PARTICULAR GAME SCENARIO WITH THIS SPOT:
Player A gives Player B "last two."
A breaks and sinks two balls (4-ball & 8-ball).
[At this point the highest-numbered ball before the 9 is the 7-ball].
A keeps shooting and makes the 1, 2, then makes a 3-7 combo.
[NOW, at this point, the highest-numbered ball before the 9 is the 6-ball]
A continues his run....... A makes the 3, 4, and then misses the 5-ball.
B steps up to the table with the 5, 6, and 9-ball remaining. B makes a 5-6 combo.
GUESS WHAT? The "last two" balls on the table are the 5 and 9 balls.
B makes the 5-ball and WINS.
Why? Because the 5 was the highest-numbered ball before the 9.
Again, you're not going to know how the spot plays out until the last few balls of any particular game.
Last edited:
Thank you boogeyman, very good, clear explanation.
The above example is a clearer example of winning/spotting 'last two'.
Your previous explanation/example was still not clear:
So at any given time, if the player getting the spot is shooting, and, let's say, the 4, 7, and 9 balls remain,
the player merely has to pocket the 7 (the highest-numbered ball before the 9).
When the 4,7, and 9 remain...only the 9 wins. Once the 4 is gone, either the 7 or 9 wins. At least how we play 'last two' around here.
Since the next to last ball is not identified until it is next to last, it only comes into play when only two balls remain on the table. So shooting or playing safe on the 'next to last' ball has more consequence.
It's NOT a win to pocket another ball if you are getting the last 2.
The guy will only win if there are two balls left and he makes one of them.
If only the 7 and 9 are left, he makes the 7, he wins. If he makes a 7-8 combo when the 7 8 9 on the table, he still needs to make the 7 or the 9 legally to win since after the 8 drops, there are now "last 2 left". If he played with the last 3, with the 7 8 9 left, a 7-8 combo wins the game.
I never play with spots or give them so maybe I'm wrong here, but when I hear "last 2" that does not mean you make a combo on the 8 and you win. That will be giving someone the 8.
Playing last two where you can combo in the 2nd-last ball during the rack to win would be an even bigger spot than the Wild 8 because you'll always have a wild ball even if the 8 is off the table. Like others have said, last 2 only applies when there are two balls left on the table. Make one of them and you win.
Well it only took 12 hours to wrap this up, lol. Not that I was confused, hehe...some folks are playing "last few" and calling it the last two.
I was just about to say that when I saw you had written it. The last two in terms of a spot is very little. In fact it is not a spot anyone will ever ask for. It is a spot you offer because you know it is worthless. I almost feel the same about giving the 8. If a guy wins significantly with the 8, he could probably have won even. It is not enough to change the outcome unless the players play almost even to start with and even then it will not change it by much.
In fact, when you are giving the last two you could play for hours and count on one hand the number of games that the guy won as a result of the last two where he would not have won anyway making the nine.
I used to give spots all the time I liked it. I guess I was what is refereed to as a "Lamb Killer". It is much easier giving spots to weaker players then it is playing an equal player. The weak player can almost be depended on to pretty much dog it off and play even below their speed against a good player.
It was funny, they would boldly throw around those spots expecting you to say no. You always offered less then what they asked for but they would stick to their guns asking for the 7 & 8 or 6 and actually trap themselves. In reality, they didn't want to play at all and now they are stuck.
Last edited:
lol. This was a long drawn out description for understanding a pretty simple spot. "Last 2" means you win if you make one of the last 2 balls on the table - regardless of what those 2 balls are (obviously one will be the 9). "Last 3" means you win if you make one of the last 3 balls on the table. It's not rocket science. If the 4-7-9 are on the table and you make a 4-7 combo - there are still 2 balls on the table, so you didn't make one of the "last 2" - you made one of the last 3. The "last 2" on the table would be the 4 and the 9. So if you make one of those, you win.
Yeah, I think I just restated what everyone else was saying. But now I'll say something else -
Don't gamble if you don't know the rules.
-td
Yeah, I think I just restated what everyone else was saying. But now I'll say something else -
Don't gamble if you don't know the rules.
-td
Last edited:
Yes, sir. The only thing I would add/clarify is, no combos unless its on the 9 ball.