## Introduction to Bearing Off

In this first article on bearing-offs we are going to look at the final stages (or endgame) of a backgammon game. As a reminder you cannot bear off any checkers until you have all fifteen of them in your home board. If, during the course of bearing off, you have a checker hit and sent to the bar, it must make the return journey from the bar all the way back into your home board before you can bear off any more checkers.

We will look at the type of bear-offs that can occur and then at some of the very basic checker strategy. In subsequent articles we will look in depth at an area of the game that might seem simple but which is actually quite complex.

### Bear Off Types

There are really four types of bear-off:

- Without Opposition (a simple race)
- Against an anchor (your opponent holds one point in your home board)
- Against a checker (or checkers) on the bar
- Against a back game (your opponent holds two – or more – points in your home board)

As with all backgammon games effective use of the doubling cube is critical to success and we will make sure we cover the use of the doubling cube in future articles

### The Basics of Bearing Off

For the time being we won’t worry about the opposition and we’ll assume we’re in a simple race. You can’t bear your checkers off before you’ve borne them in and therefore our first priority is to get them into our home board.

During the normal course of a game you will have made some points in your home board and the only decision that you then have to make is where to try to put the other checkers as your bear them in.

It might seem intuitively obvious to spread them evenly across the your home board as shown below:

Black wins with all but the following rolls: 11, 21, 31, 41, 32. 9 of his 36 rolls lose, the other 27 win. White will never be able to use the doubling cube to his advantage.

In a cross-section of 36 games, if black doubles and white drops, white will of course lose 36 points. If white takes he will:

Win 9 games x 2 points = +18

Lose 27 games x 2 points = -54

Net loss: -36

We see from this that when the taking side cannot use the cube, and there are no gammons, they need 25% winning chances to take. This 25% figure is by far the most important thing to know about the doubling cube. In this case, white is indifferent between taking and passing.

Let’s also introduce an important concept here:

**An efficient double is as good as a win!**

By “an efficient double” I mean a position where the taking side is indifferent between taking and passing. The long-run effect is the same. Being 75% to win and having access to the cube is the same as being 100% to win when you can’t use the cube. For that matter, being 76% or 77% or 80% with the cube is the same as being 100% to win without the cube.

II. The value of the cube- Backgammon Game

Let’s change the above position slightly.

Black looks better off, doesn’t he? He now wins on 32 and 41 as well – 31 winners rather than 27, 86.1% win chances rather than 75%

Actually he is worse off! Notice that the cube is now on white’s side. Black was effectively 100% to win the earlier position, because he could turn the cube.

In this case, white will lose 86.1% and win 13.9%, for a net average loss of 0.722 points. Owning the cube is worth 0.278 points to him.

It is not possible when the cube is turned to know what its future value will be. We will use an estimate that it increases wins for the side owning the cube by 10%.

This changes the taking equation slightly. We now need to solve the equation:

2 * [ (win% * 1.1) – (1 – win% *1.1) = -1

solving this give us 22.73% wins needed to take.

So take the following position:

Computer analysis tells us that black wins 76.5% of games here. Yet white should take. Using our 10% rule of thumb, white will actually win 23.5% * 1.1 or 25.9%. So by taking, he loses on average 2 * ( .259 – .741) or .964 points, 0.036 better than passing. (Actually his advantage is a bit larger than that, because the 10% rule may understate white’s chances in this position a bit.)

III. Doubling in match situations-

In a match, every point is not equal. The percentage odds needed to take a double will vary with the score.

We’ll start with a simple example. We’ll give more complicated examples in future articles.

Let’s say you are trailing 2-0 in a 3-point match. The next game will of course be the Crawford game. The trailer will have to win two consecutive games. (If he wins a gammon in the first game, he’ll have to win the next game from a score of 2-2. If he wins a single game, he will trail 2-1 and double in the next game at the first opportunity.) The odds of winning two games in a row are 50% * 50%, or 25%. So the trailer is 25% to win the match.

Now, let’s assume that you are trailing 1-0 in a match to 3, and your opponent doubles. What odds do you need to take? If you pass you will trail 2-0 Crawford, and have 25% odds to win the match. If you take, you will redouble immediately so the game will settle the match. Therefore, you should take if you are 25% or better to win the game (and therefore the match).

The next section is a bit complicated, especially for those who don’t speak algebra. I recommend taking the time to understand it. The concepts are very important.

What if you are the side leading 1-0, and are doubled to 2? For simplicity assume that:

a) There are no gammons

b) The leader at 2-1 Crawford is 70% to win the match. (We’ll explain how we arrive at this in a future article)

To know what odds you need to take, you need to solve the equation:

Match-winning chances if I pass = Match-winning chances if I take

Since passing would make you tied and therefore 50% to win, the equation becomes:

.50 = Game-winning chances * 100% + Game-losing chances * 30%

Let’s use G for the game-winning chances. Of course, losing chances = 1 – G

So G + [0.3 * (1-G)] = 0.5

Or G + 0.3 – 0.3G = 0.5

Subtract 0.3 from each side and get

G – 0.3 G = 0.2

0.7G = 0.2

divide each size by 0.7 and get

G = 0.2 / 0.7 = 28.7%

This is not as complicated as it seems. A simpler way to express it (and we’ll omit all the algebra) is:

Win chances = loss when you take and lose / total amount at risk

Here the loss is 20% (from 50% by passing, to 30% by taking and losing) and the total amount at risk is 70% (from 100% by taking and winning, to 30% by taking and losing). So the win chances needed to take are 20% / 70%, same as in the first example.

Does this matter? Is this information important? Let’s do some more math.

First, how many times do your matches reach a score where one player needs 2 and the other needs 3? If you play 3pt matches, this will happen anytime the first game ends in one point being scored – so pretty often. In longer matches, it will still happen a fair amount of the time.

Now, let’s look at two positions:

This is a double and a borderline take or pass in a money game. However, if white leads 1-0 in a match to 3, it would be a huge mistake to take. Black wins the game 77.8% of the time. So if white takes, white will win the match 22.2% of the time when he wins the game, plus 30% of the remaining 77.8%, for a total of 45.5%. A loss of 4.5% compared to the 50% he can get by passing.

This position is the same for black, but we made white a bit better. Now it is a borderline double for money. However, it would be a significant mistake not to double. Black wins the match about 49.6% of the time if he doubles, but 47.6% if he does not.

So does the math of when to double in matches matter? Only if it matters to you whether you can increase your match-winning chances by 2-3% every time these kinds of situations come up, just by understanding this one aspect of how to use the doubling cube.