What is required here is the simple application of the arithmetic outlined in the previous article. The only other piece of knowledge that is required is that of understanding the possible rolls of two dice.
Let's cover that first. There are 36 possible combinations of two dice. Of these 36, six are the doubles, ie 1-1, 2-2 etc. The other 30 rolls produce 15 different combinations. For example there are two ways of throwing a 2-1. It can be thrown as a 2-1 or 1-2 (if you have trouble visualizing this get out two differently coloured dice and work through all 36 possible combinations).
Of the 36 rolls, 19 of them are sufficient to bear off Black's last two men, the other 17 fail. If Black doubles and White drops, Black will win 36 points in 36 games. If Black doubles and White takes, then Black will win two points 19 times, and lose two points 17 times - a net gain of four points.
What does this tell us? Firstly, it tells us that Black is right to double. If he doesn't double he will win only 2 points (because the cube is on "1" and not "2"). Thus by doubling, Black doubles his profit. Secondly, it is correct for White to take, as opposed to losing 36 points, which he would do by dropping, he now only loses 4 points. This is a massive 89 per cent improvement on dropping. The biggest mistakes in backgammon normally involve cube handling rather than checker play.
This is one of the simplest doubling problems, because it is in the endgame and the possible results are directly calculable. From here on it gets more difficult.
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