White - Pips 140
Black - Pips 130
Code:
bar/24 12/10 11/10 Eq.: -0,038
0,488 0,166 0,005 - 0,512 0,161 0,008 CL -0,022 CF -0,038
bar/24 12/11 2/1*(2) Eq.: -0,123 ( -0,084)
0,475 0,166 0,003 - 0,525 0,193 0,013 CL -0,087 CF -0,123
bar/24 11/9 2/1* Eq.: -0,401 ( -0,363)
0,412 0,132 0,002 - 0,588 0,264 0,033 CL -0,338 CF -0,401
Not grapping the opportunity to seize a containment point is extremely bad.
The pips are of no issue here:
White - Pips 156
Black - Pips 130
Code:
bar/23 12/11 2/1* Eq.: +0,422
0,590 0,211 0,004 - 0,410 0,114 0,005 CL +0,276 CF +0,422
bar/23 24/23 12/11 Eq.: +0,206 ( -0,216)
0,541 0,191 0,004 - 0,459 0,128 0,005 CL +0,144 CF +0,206
A lot of checkers have to pass each other, whereas both sides have little or no connectivity. This will very likely not take place without hit(s).
Black has bad coverage after hitting loose, 6 to 4:
Code:
bar/24 12/11 4/3 2/1* Eq.: -0,075
0,484 0,174 0,003 - 0,516 0,191 0,014 CL -0,061 CF -0,075
bar/24 12/10 11/10 Eq.: -0,105 ( -0,029)
0,475 0,165 0,005 - 0,525 0,171 0,009 CL -0,060 CF -0,105
This illustration is somewhat flawed because the checker has been turned into semi-dead. That means there are two semi-killed checkers, which offers a very good reason to use one of them for practical purposes. So the value increase of hitting will be less dramatic if we take this into account. But enough value will be left to consider the facts that:
– black wants to use his builder on the 6-point to make the 5-point, instead of the 1-point
– in the original position black's 4s are duplicated.
In the next position black needs a 3 instead of a 4 to move off the 24-point:
Code:
bar/24 12/10 11/10 Eq.: -0,140
0,462 0,174 0,006 - 0,538 0,182 0,010 CL -0,088 CF -0,140
bar/24 12/11 2/1*(2) Eq.: -0,151 ( -0,012)
0,470 0,160 0,003 - 0,530 0,199 0,015 CL -0,111 CF -0,151
So duplication is of greater relevance than the 5-point thing.
White's 5w and 7-point are stripped. If black doesn't hit in the original position, white hopes to throw a 4 or 6 sothat she can move her rear checker. Otherwise, she likely has to leave two blots while breaking a point, or has to semi-kill in her homeboard. However, this doesn't seem much of a concern, probably because this happens in only 30% of the cases. In the following position white is given flexibility by moving from 7 to 5:
Code:
bar/24 12/10 11/10 Eq.: +0,155
0,532 0,195 0,005 - 0,468 0,148 0,007 CL +0,108 CF +0,155
bar/24 12/11 2/1*(2) Eq.: +0,090 ( -0,065)
0,526 0,183 0,003 - 0,474 0,173 0,011 CL +0,054 CF +0,090
Black seems to have something similar. In case he hits on his 1-point and is hit back, he will hope he will enter on the 20-point. However, there is a fifty-fifty chance that he will enter on the 24-point. That means there will be an extra checker stuck behind. In the following position white occupies the 18w-point, sothat black's 6s are blocked too:
Code:
bar/24 12/10 11/10 Eq.: -0,534
0,354 0,140 0,003 - 0,646 0,212 0,013 CL -0,374 CF -0,534
bar/24 12/11 2/1*(2) Eq.: -0,746 ( -0,212)
0,330 0,116 0,002 - 0,670 0,250 0,019 CL -0,490 CF -0,746
However, this drop is mainly caused because of the fact that the duplication of 4s weights a lot heavier. This is shown when I move 6 to 4 again in the previous position:
Code:
bar/24 12/10 11/10 Eq.: -0,606
0,337 0,134 0,003 - 0,663 0,222 0,013 CL -0,423 CF -0,606
bar/24 12/11 4/3 2/1* Eq.: -0,654 ( -0,048)
0,343 0,129 0,003 - 0,657 0,247 0,018 CL -0,446 CF -0,654
Still, in the other position hitting was best, so this argument has some relevance.
Conclusion:
Generally it seems that the merit and demerit of the hit&blot move are pretty equivalent: it's good when the blot is covered, but it could also happen that a third checker gets stuck on the 24-point. In this case however, the duplication of 4's swings the decision to the safe play.