There are many algorithms to help you get an absolute count for a position. Of the lot, I think the kangaroo count is the easiest. It was invented by Nack Ballard, and follows this poem:

Each line of the poem represents a step in the procedure. It’s spiritually related to the counts we have already learned about, but here we are only counting the checkers that belong to a single player. Let’s go through it step by step and see how it works. Let’s present a position and do the pip count for our checkers using this method:

**1. First you double the far side men.** We only have 5 checkers on the opponent’s side of the board (on the 16-pt and 13-pt). We take that number and double it. 5 × 2 = 10.

**2. Add big diagonal, double again.** We add our checkers on the diagonal quadrants that include our opponent’s ace-pt.

^{1}In this position, there are 0 checkers to add. We add 0 to 10 and double it. (10 + 0) × 2 = 20.

**3. Add small diags, times 3, plus 30.** We add our checkers on the diagonal triads that include our opponent’s ace-pt. Here, we are counting the 5 checkers between the 4-pt and 6-pt, and the 2 checkers on the 16-pt. We add that to our count (20), multiply by 3 and add 30. (20 + 7) × 3 + 30 = 111.

^{2}

If we pause here for a moment, the number we have arrived at, 111, is an approximate absolute count of our checkers of 111. The exact count is 110, which is very close. So what is the need for step 4? As in the relative counts, we can do a unit adjustment to obtain the exact count:

**4. And shifting to the middles is perty.** The checkers on the 3-pt cancel out the checkers on the 13-pt, and the checkers on the 6-pt cancel out the checkers on the 16-pt. The lone checker remaining that isn’t centered in its triad is on the 4-pt. This checker is closer to the bear-off tray, so it removes one pip from our count. 111-1 = 110, the exact count.

##### Further reading:

- Nack Ballard’s Kangaroo Count, an article by Tom Keith on Backgammon Galore.

- A mnemonic to help you remember which quadrants to choose is that we want the ones that would contribute the most pips. Those always include the farthest point on the board. ↩︎
- Personally, I prefer to add 10 and multiply by 3. You get the same answer but it feels easier to my brain. So (20 + 7 + 10) × 3 = 111. ↩︎

Next lesson: Cluster counting 1: 10 checkers

## Leave a Reply

You must be logged in to post a comment.