What is required here is the simple application of the arithmetic outlined last time. The one other piece of knowledge that is required is that of understanding the possible rolls of two dice.
This is the area we shall cover first. There are 36 possible combinations of two dice. Of these 36, six are the doubles, such as 1-1, 2-2 and so on. 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 a problem visualising this, get out two different coloured dice and work through all 36 possible combinations. One of the key results of this, and something that recurs again and again in backgammon, is that there are 11 possible dice combinations which will contain a specific single number. For example, 11 of the rolls contain the figure one, the other 25 do not.
Back to our problem. Let's consider 36 games. In 25 of them, Black will bear off his last two men and win the game, but in the other 11 games he will roll a one and White will win the game. Is this good enough for White to accept a double or is Black too much of a favourite?
If Black doubles and White drops, Black will win 36 points in 36 games. If Black doubles and White takes, then Black will win 2 points 25 times and lose 2 points 11 times, giving a net gain of 28 points.
This tells us that Black is right to double. If he doesn't double he will win only 14 points (because the cube is on one and not two). Thus by doubling, Black doubles his profit. Also, it is correct for White to take, as instead of losing 36 points, which he would do by dropping, he now only loses 28 points. This is a 22 per cent improvement on dropping - quite a significant difference.
This is one of the simplest doubling positions, because it is in the endgame and the possible results are directly calculable. From here on it gets more difficult.Reuse content