[ Post Response ] [ Return to Index ] [ Read Prev Msg ] [ Read Next Msg ]

BGonline.org Forums

Naccel 2 -- post #6x (clarification of block)

Posted By: Nack Ballard <nack2000@sbcglobal.net>
Date: Friday, 22 January 2010, at 10:21 p.m.

In this extra post, I'm combining two clarifications. The first is a recap of the Blue and White counts in the submitted position discussed at the end of post#6. The second is a clarification of the location of "blocks."

Midpoof, and six-stack plus 1 pip2(1)

1X1X2X '2O3X '4X2X ' '2O

 ' '2O2O '2O1O3O1O ' '2X

Poof, and Midgold7

White's count is 2(1) and Blue's count is 7, so White leads by nearly 5 supes (29 pips, to be exact). It therefore looks like a strong money double or redouble, though with the existing midpoint contact Blue can still take.

The remainder of this post is a question-and-answer discussion of the "block." The block can be reviewed here.

Jim said:

Nack – can I just check that I am looking at this correctly – it looks to me as if the block only applies to certain sets of points – namely those which divide exactly by 3 ? (4 sets of points on one side and 4 on the other). Had in your first example the four blue checkers in white’s home board been on the n16 and n15 points you would not have been able to use the block but would have done something else?

Astute of you to notice.

Strictly speaking, you can count any two points on the board (in the same quadrant or in different quadrants) by adding the point numbers together and dividing by 3. However, if you choose those points at random, there is only a 33% chance you will obtain an integer count. (In the case of two consecutive points in a quadrant, your chances improve to 40%; and if you additionally exclude the Super, you're 50%.)

Your example of n16 + n15 gives you a count of 31/3 = 10(2). If the other eleven checkers had already been counted (or canceled), this would actually be a good way to obtain the leftover count. Otherwise, it leaves you with the burden of a baby-pip count to carry forward.

A proper "block" is one that yields an integer count.

A six-point quadrant, from high point to low, is composed of the Super, the high block points, the pair point, and the low block points. Each of these will visually stand out, and even jump off the board at you, the more you practice.

In short, the block points flank the pair point. In your example, n15 is a pair point so it can't be part of a block. Likewise, with n3 in the diagram below: Blue's blocks exist on either side of the vacant pair point (the only point in that quadrant where Blue can have a pair).


2O ' ' ' '5X '3X ' ' ' '

2X ' ' ' '5O2O2O '2O2O5X

Blue's two Blocks flank the Pair point10


Post Response

Your Name:
Your E-Mail Address:tchow8@hotmail.com

If necessary, enter your password below:

Save Password: Yes No



[ Post Response ] [ Return to Index ] [ Read Prev Msg ] [ Read Next Msg ]

BGonline.org Forums is maintained by Stick with WebBBS 5.12.