Pile on, people... is it more weight to give up the last 2 or the 8? Last 3 or the 7? Etc...
If so, why? If not, why?
If so, why? If not, why?
Last 2 or the 8, which is more weight?
- Thread starter Michael Andros
- Start date
the 8 due to combo possibilities at any time.
8 ball is more weight. Can be made on the break, combos, etc.
Cannot be done with last 2. Forces weaker player to run the rack.
The 8 is even more weight than the last three.
Bingo... or, at least, very close... even the wild 8 as well. But I'm waiting to see what everyone else says before I toss in MHO.
You must be old, PT!
The Last Two
The Last Two is more because its there every game. If your opponent get to the 8 ball first he still has to get all the way out to win. If he makes the 8 early in the rack or on the break he still has stopped you, the pressure is there all game long.
The Last Two is more because its there every game. If your opponent get to the 8 ball first he still has to get all the way out to win. If he makes the 8 early in the rack or on the break he still has stopped you, the pressure is there all game long.
IMO the call-8 is more likely to get a somewhat weaker player out of his chair 'cause he's thinking an early carom/combo wins. Between better players the last-2 is kinda of "non spot" because both are more likely to get to those balls. I still think the call-8 is slightly bigger spot than last-2 and the wild-8 a little bigger still.
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Depends on the skill level of the two players. For cushion beaters the 8 is by far the bigger spot, because cushion beaters love to ride the money ball, and when two cushion beaters play, the balls stay on the table for many more innings.
But for shortstops and above, it's the last two, because the player giving up the weight is liable to make the 8 on the break, and because when you get up to a certain skill level, very few players try to win with combinations when they can either run out or put their opponent in jail.
Of course it also depends on whether it's the called 8 or the slop 8, especially as you go down the skill chain. For cushion beaters the slop 8 is a much bigger advantage than either the called 8 or the last 2.
But for shortstops and above, it's the last two, because the player giving up the weight is liable to make the 8 on the break, and because when you get up to a certain skill level, very few players try to win with combinations when they can either run out or put their opponent in jail.
Of course it also depends on whether it's the called 8 or the slop 8, especially as you go down the skill chain. For cushion beaters the slop 8 is a much bigger advantage than either the called 8 or the last 2.
Zactly. IMHO, the last ___ *period* is more weight than any corresponding numbered ball. The money is *always* on the table. And, if you have the last 3 or last 4. you have 3 or 4, respectively, money balls on the table. Sure, the spot-ee can make the 8 ( or 7, 6, etc ) on the break, or s**t in in, but so can the spot-er. And If I'm giving you the 8 and I make it n the break? We're playing even. And of course an argument can be ( and is ) made saying the opposite, but I grew up both giving and getting the last 2 or 3, etc, as more weight. And I don't really recall anyone back in the day saying it wasn't. It was pretty well accepted.
I agree with you Gar, on getting them outta their seats quicker offering the eight. It just *sounds* like more weight for whatever reason.
i am confused
if you are a run out 9 ball player the other guy never sees the last 2
however the 8 is on the table until it is made
and will placed in a spot (behind the one ) where it probably wont be made on the break
so please explain again
so why isnt the 8 a bigger spot.....
if you are a run out 9 ball player the other guy never sees the last 2
however the 8 is on the table until it is made
and will placed in a spot (behind the one ) where it probably wont be made on the break
so please explain again
so why isnt the 8 a bigger spot.....
i am confused
if you are a run out 9 ball player the other guy never sees the last 2
however the 8 is on the table until it is made
and will placed in a spot (behind the one ) where it probably wont be made on the break
so please explain again
so why isnt the 8 a bigger spot.....
Presuming the 8 is in the second row, you're right about it not being made on the break.
BUT, there are places/people who insist the 8 goes in the back where we were taught that it can OFTEN be made on the break.
I pretty much agree with PT, the last two and the 8, or the call 8, were about equal. You're implying a run out player will never leave the 8 or the last two, well then they probably wouldn't have the game in the first place
i am confused
if you are a run out 9 ball player the other guy never sees the last 2
however the 8 is on the table until it is made
and will placed in a spot (behind the one ) where it probably wont be made on the break
so please explain again
so why isnt the 8 a bigger spot.....
BBB - Regardless of the break ( and half the time, the spot-ee doesn't rack the 8 behind the 1 and, even if they do, as you know, it sometimes goes on the snap anyway ), the 8 can be s**t in or, go on a combination. If the spot-er does any of those, they're then playing even. With the last 2 ( or 3, 4, etc ) that can never happen. The money is always on the table. And as Jay pointed out in another thread, if I have the last 4 and the 5, 6, 8 and 9 are on the table, all FOUR are my money and if I s**t or combo any of them, I win. That is WAY more weight than any one ball ( IMHO of course :grin: ).
You're right on the money, Gar... I wouldn't even play on a bar box, let alone give up weight on one. They're against my religion. :grin:
I hear ya. Thing is most of the big action in my area(Ok,Tex,Ks,Mo) over the years has been on the cracker-box. I've seen guys like Matlock, Buddy and James Walden out-run stupid spots on Valley's in the past.
It's pretty well accepted that the wild 8 is more weight than the last 2, and that the call 8 is closer to the last 2. Logic will tell you the same. That is, with the wild 8, you can potentially win with 2 balls (8 or 9) on any shot. With the last 2, you can only win when there are 2 balls left. AND, if one of them is the 8, you would win anyway. So, the last 2 could ONLY better a better spot when the 8 is made early by your opponent. Think about how many times your opponent has made the 8 by combo or carom early in a game. Very few. Simple math will tell you that the wild 8 will be better over the long haul.
A simple thought experiment will illustrate that the last X is never as good as the corresponding wild ball(s). If the last 2 was better than getting the 8, then the last 3 would be stronger than getting the 7 and the 8. And the last 4 would be stronger than getting the 6, 7, 8. In this last example, you have 4 chances to win every shot with the wild balls. But you could only win late in the game with the last 4. Consider if your opponent missed the 4 ball and left you a combo on the 4-6, 4-7, or 4-8. You would win with the wild balls, but not the last 4. But, what if your opponent played a mediocre safety on the 4, but the 6, 7 or 8 was hanging somewhere? You still couldn't win with the last 4 without running balls, but you COULD win with the wild 6, 7, 8. There are many other scenarios to contemplate, but virtually all of them result in the wild balls being a better spot.
Lastly, if you ask anyone that's gambled a lot, they'll confirm that they would rather give up the last X, or a call ball, than one or more wild balls. If someone tells you different, that probably means that their skill set is mediocre, they gamble with ATMs, or they haven't really gambled enough for statistics to prove itself.
-td
PS - here's a thread from a decade ago on this. The last 2 is almost the weakest spot you can get. Times haven't changed...
http://forums.azbilliards.com/showthread.php?t=103835&highlight=spot [Note, "the breaks" can improve the spot marginally.]
A simple thought experiment will illustrate that the last X is never as good as the corresponding wild ball(s). If the last 2 was better than getting the 8, then the last 3 would be stronger than getting the 7 and the 8. And the last 4 would be stronger than getting the 6, 7, 8. In this last example, you have 4 chances to win every shot with the wild balls. But you could only win late in the game with the last 4. Consider if your opponent missed the 4 ball and left you a combo on the 4-6, 4-7, or 4-8. You would win with the wild balls, but not the last 4. But, what if your opponent played a mediocre safety on the 4, but the 6, 7 or 8 was hanging somewhere? You still couldn't win with the last 4 without running balls, but you COULD win with the wild 6, 7, 8. There are many other scenarios to contemplate, but virtually all of them result in the wild balls being a better spot.
Lastly, if you ask anyone that's gambled a lot, they'll confirm that they would rather give up the last X, or a call ball, than one or more wild balls. If someone tells you different, that probably means that their skill set is mediocre, they gamble with ATMs, or they haven't really gambled enough for statistics to prove itself.
-td
PS - here's a thread from a decade ago on this. The last 2 is almost the weakest spot you can get. Times haven't changed...
http://forums.azbilliards.com/showthread.php?t=103835&highlight=spot [Note, "the breaks" can improve the spot marginally.]
I just can't imagine not having 9' GCs right down the street. Spoiled I tell ya. Spoiled rotten!