While that sounds good in theory, here is the catch.
There are 15 balls in the rack. IF you can knock TWO OF THEM uptable EVERY SHOT(and that won't happen), then there is a GUARANTEED SEVEN 1/2 INNINGS FOR SCOTT. And you certainly cant be that effective(2 every time). And Scott isn't an idiot- when you give up a major spot you ALREADY know how to not leave a guy down low to shoot them away.
Do you think scott can pull off something spectacular 1 in 3 tries? Yep. You give him 7 tries by wasting innings where something else is more productive in the first place. How many in those 7 tries can he pull something off?
Yes, it is a good idea to let the balls go uptable, but it isn't JUST so easy to get them there. It moves UP for sure on the priority list, but is still beneath safety, aggressive play, GETTING TO THE HILL, and good moving.
And Scott, in case no one has said it, is pretty much the Cream of the crop when it comes to playing an uptable game. IF (with emphasis) you get them there before he pulls something off.
Dip has to fire at his hole as usual, and lay down whitey. Emphasis on layin it down. Every shot-lay it down, lay it down...so interjecting "simply put them up table" does not work - I have played against some VERY strong players, and some professionals myself. When given the chance, you do move them up, but it isn't a matter of "hey, just poke them up table!" Scott has a higher perccentage to make a long straight back that Dippy does to make easier shots. And two-railers, and kicks, And once he sees 3-4 balls go uptable it is Scott's obligation to look for opportunities to prevent/ counteract it. And he CAN!!
And IF you manage to get them uptable in the first 15 innings, I think Scott is WAY favored to lay something down over Dippy. Let's say 5-1?
5 to 1
18 to 4
If Scott is 5 to 1 better than Dip, at an 18 to 4 ratio, then you come up with him scoring 20 to Dippy's 4. Minus one scoring inning, when Dip is at 3, Scott is at 15. So then you have a guy that is 5-1 favorite even up, needing 3-1 odds. in 6 games that equals 5 wins, 1 loss. Multiplied by the 3 to 1 that equals 5 to 3 a winner for Scott. Every cycle of the odds equate to a two game jump for Scott.
petie said:
I still say he should knock all the ball up table as fast as he can and this will turn the odds in his favor. You take the long run away from Scott. A friend of mine suggested that all Dave has to do to win is to come out of his stall
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