Problem 60 (501 Essential Backgammon Problems) - Backgammon Forum - Discuss Backgammon Strategy, Clubs and Books

Two Plus Two Forums Sports and Games Chess and Other Board Games Backgammon Forum hosted by Bill Robertie.

Problem 60 (501 Essential Backgammon Problems)

Post Reply Subscribe

...

05-26-2023 , 08:23 AM

David1105

stranger

Join Date: Jun 2009 Posts: 2

Going through this book at the moment. In problem 60, 6/4 is described as a blunder because you only have two builders as opposed to 3 with 7/5. Can someone explain why this is the case? Wouldn't the third checkers on either the 5 point (with 7/5) or the 4 point (with 6/4) each be considered a builder? How does 7/5 result in a third builder?

Quote

05-26-2023 , 09:07 AM

uberkuber

veteran

Join Date: Jul 2009 Posts: 3,032

With 6/4, you have 2 builders (4-pt and 7-pt).
With 7/5, you have 3 (5, 6 and 7 points).

Quote

05-26-2023 , 09:29 AM

David1105

stranger

Join Date: Jun 2009 Posts: 2

Thanks for the explanation. I didn't realize that the 7 point would still be a builder because you would leave a blot if used to make a point on the inner board. In my head leaving the blot wasn't an option in the definition of using a builder.

Quote

05-26-2023 , 02:42 PM

stremba70

adept

Join Date: Aug 2020 Posts: 1,108

You would leave a blot on your seven point if white dances on his roll or if you make the point on which he enters. If white is on the bar he has only two rolls to return hit; it is worth the risk to make a fifth home board point and possibly be able to make white dance for a few rolls while bringing the midpoint checkers home. With any luck you can create builders with those checkers and close out white.

Obviously after 6/4 you are never going to break your six point to hit white on a lower point or make a lower point. You need to keep the spare on the six point and move 7/5 instead.

Quote

Post Reply Subscribe

...