Back to basics this week with a look at simple bear-off problems. Assume for a moment that all the remaining men are on the respective one- points. Four men against four is always a double (or redouble) and a drop. Six against six is also a double and a drop - the doubled side wins 21.2 per cent of the time, not quite enough for a take.

It becomes more interesting with eight men against eight. This position is a double (or redouble) and a take. (Not a wonderful take as the taking side will lose 0.92 points per game, but that's better than losing a full point by dropping the double. With ten v ten it depends where the cube is. If it's in the centre you should double and your opponent should take. If you own the cube you should wait a roll until it is eight v eight which will happen most of the time. If your opponent throws a double you will still be in the game and get a chance to throw a double of your own as your opponent cannot double you out.

So the diagram position is eight v eight and must be a take, right? Wrong! It is not a pure eight v eight. If White rolls 1-1 he will only be able to take off three men and therefore not save himself a roll. This tiny difference makes the position a double and a drop.

What if you have six men on your one-point and your opponent has two on each of his one, two and three-points? He doubles and you drop because its six v six. Wrong again! The chances that he will roll three successive ones and the fact that 1-1 and 2-2 do not save him a roll means that you have the thinnest of takes. In such borderline cases, attention to detail can win, or save, you a lot of money.

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