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Checker Problem |
The Best Spot for This Blot
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| From Backgammon Times, Volume 2, Number 4, Fall 1982. |
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5–5 in match to 7: Black to play 4-1. |
If Black could have stayed out of range of White's man on the 2 point with one or both of his outfield checkers, he might have chanced the throw of a 6 that White needed to escape. Two men would have the opportunity of returning White's blot to the bar. But in order to keep two men back, Black would have to leave them both vulnerable to indirect hits from the 2 point.
For a moment Black even contemplated slotting on the 8 point, but quickly rejected this in view of White's formidable board.
Because the race was so close, Black decided against a conservative play from the 11 point. Instead, he hit on the 2 point with a single man from his 6 point. Then, to provide three different direct numbers that would cover (if his blot escaped being hit as White reentered), he played the 1 from the bar point to the 6 point.
It is not easy to fault Black's play. If White fails to throw a 2, Black can close the point with an 5, 4, or 2, as well as 6-3, 3-3, 3-1, or 1-1. Of these thirty-three rolls, twenty-eight also form a full 6-point prime. 6-1 lifts the blot making the 1 point. Only 6-6 continues to leave the direct shot on the 2 point.
Nevertheless, Black's play was wrong! Leaving the blot on the 2 point or continuing with it to the 1 point makes a great difference in the outcome if White hits from the bar.
Of course White hits with 12 rolls on the 2 point and only 11 on the 1 point, but this is not the principal reason for making the correct play.
A hit on the 2 point places Black on the bar while White is in position to escape and even to hit additional blots if Black should fail to enter. Furthermore, whatever number comes up with the 2 can be played safely by White.
Not so when the hit is on the 1 point. If the other number is 6, White will have to place an unprotected man on his own 1 point, vulnerable to a hit from Black's man on the bar. Except for double 1, with any number that hits on Black's 1 point White will be stuck there, needing another 1 followed by a 6 to escape. And finally, if Black goes to the 1 point and White throws the magic 2-6, Black, unencumbered with the need to use one of his numbers to reenter, will have two direct numbers bearing on the 8 point.
If White comes in with any other 2, Black will again be able to use both his numbers. This could easily save the game by pointing, hitting and bouncing, simply hitting loose a second time, or even playing conservatively now to win the race.
Of course, it is attractive to stay on the 2 point because of the likelihood of constructing a full six-point prime. However, with the opponent's man on the 1 point, a five-point prime from the bar point to the 3 point is almost as effective.
| More articles by Jerry Nathan |
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