The 321 count is a colourless pip counting method that produces a *relative* pip count. It’s a sort of hybrid descendent of Zare’s half-crossover count and Urquhart’s colorless count. It is so easy and works so well that it feels like magic. It’s the first one I learned, and the one I still use most often over the board.

The method has 2 stages — a simple one that only has addition and multiplication (so numbers only go up) to determine the relative pip count within 5 or so pips, and a slightly more tricky one that corrects the first stage to get an exact relative count. The second stage is known as a unit adjustment, which we go over in a later article.

Consider the position below:

We recognize the position as a standard holding game position. We’re on roll and are considering doubling. From our lesson on doubling in holding games, we know that we have a good double if we have 4 or fewer checkers on our midpoint, and we are leading the race by at least 14 pips. Let’s find out the pip count and see if we should double.

**1. **The first step is where the method’s name comes from. Starting from your home board (the 6- to 1-pts), you count the number of **total** checkers in your quadrant and multiply that number by 3. You then count the number of checkers in the next quadrant (your outfield, from the 12- to 7-pts), multiply that by 2, and add it to your total. Finally, you count the number of checkers in the next quadrant out (Gary’s outfield, from the 18- to the 13-pt) and add that to your total:

In our example board, we have 3×10 + 2×5 + 8 = 48. Note that at each step we are counting both our checkers and Gary’s checkers. That’s what makes this a colourless method, and what makes it feel like magic.

**2. **The next step is to double this total. 48×2 = 96.

**3.** Now we add the number of checkers in **the diagonals**. These are also called the octets or triads in different locations. They are the clockwise halves of the quadrants:

Starting in our home board and counting counter-clockwise, we add 2 + 3 + 4 + 7 checkers to our previous tally of 96, which adds up to 112.

**4.** To obtain the relative count, we take 105 and subtract our tally from it, and multiply the remainder by 3x. In our case, we end up with 105 – 112 = −7; −7 x 3 = −21. A negative number indicates that we are leading the race, and it’s by about 21 pips. (The real count is −18 pips. Pretty close!)

As this is substantially larger than 14 pips, we probably don’t need to do a Stage 2 unit adjustment to perfect the count, so we can comfortably offer the cube!

#### What if there are checkers on the bar?

Something we haven’t explicitly described is how the relative pip count is changed when there are checkers on the bar. The checkers need to be counted with their relevant triads. If *your checker* is on the bar, you need to pretend it’s in the 24-pt to 22-pt triad. That checker turns out not to be counted. On the other hand, if *your opponent’s checker* is on the bar, it contributes to the 3-pt to ace-pt triad. It ends up being added to the first step (multiplied by 3!) and also added to the homeboard diagonal. **This is very easy to forget**; however, you will find that you aren’t often doing counts with checkers on the bar. If you are considering the cube, you are usually doing so based on blitz value, and not necessarily race equity.

To recap, the 4 steps are:

- Add the quadrants multiplied by 3x , 2x, and 1x
- Double the total
- Add the diagonals
- Subtract our tally from 105 and multiply the remainder by 3x. A negative number means we are leading, a positive number means we are trailing.

This is one of those things that seems like a lot of steps until you try it out. Pretty quickly you will be doing the whole thing in 15s or fewer.

##### Further reading:

- Colourless ‘321’ Counting video by David Gallagher, the lecture that first introduced me to this method, on the Backgammon is Beautiful YouTube channel.

Next lesson: Criss-cross pip count

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