Now that we’ve described the entire method, let’s practice it in some real situations. We will go through a few examples together, and we provide other exercises in the further reading links at the end of the lesson.

### Early game

In the position above, we can count our checkers on the far end of the board ① using our midpoint cluster. If there were a 4th checker on the 13-pt, all 6 of those checkers would be 100 pips; without that 4th checker, we have only 87.

On our side of the board, we have 10 checkers almost lined up in a 5-prime ②. The center of this 5-prime is the 7 pt; we start with 70 pips, but we need to mentally adjust the 2 checkers on the 6-pt to the 7- and 9-pts. Removing 4 pips for those, we have 66 pips.

66 + 87 = 153 pips. Boom, in just 2 steps!

How about our opponent? Their two checkers within our home board ③ are 43 pips (20 each for being in our board, and 3 more from mentally shifting away from the 20-pt).

The heavy stack of checkers on their 13-pt ④ (our 12-pt) can be counted by mentally shifting 5 checkers to their 7-pt and counting them in pairs worth 20 pips each. Together, they add up to 99 pips.

There are 3 checkers remaining ⑤: 2 on the 4-pt and 1 on the 6-pt. We can mentally shift them all to the 5-pt, for 14 pips.

43 + 99 + 14 = 156 pips.

### Bear off

On our side of the board, we have a triangle hiding in plain sight ①. Its “3-checkers tall” location is on the 4-pt. So we start with 40 pips.

Our remaining checkers ② are on the 4-pt and 6-pt. We can mentally shift all 5 of them to the 5-pt, for 25 pips. We need to account for one more checker being on the 6-pt than the 4-pt, so 26 pips.

40 + 26 = 66 pips.

Gary doesn’t have a triangle, but he has a 5-prime centered around his 4-pt ③. Here we also start with 40 pips.

4 of his remaining checkers (on the 2-, 3-, 7- and 8- points) are symmetric about the 5-pt as well ④, so we can add 4 × 5 = 20 to our count. The last checker is on the 1-pt ⑤, adding just 1 pip, so 21 pips.

40 + 21 = 61 pips.

### Backgame

Without cluster counting, our checker distribution, with 5 checkers deep in Gary’s homeboard ①, would be very daunting to count. 3 × 22 + 2 × 24 = 66 + 48 = … it takes a lot of mental math with awkward numbers.

Instead, we start with 100 pips (5 checkers × 20 pips each). Next, we adjust for the depth of the different checkers. 2 are 4 pips past the 20-pt, 3 are 2 pips past the 20pt, so we add 8 and 6 pips, respectively, for 114 pips total.

Next we count the 4 checkers on the 13-pt and 7-pt ②, worth 40 pips.

What’s left are 6 checkers that surround the 6-pt ③. They would be symmetric around that point if one of the 5-pt checkers was on the 7-pt, so we can take 6 × 6 – 2 = 34 pips.

114 + 40 + 34 = 188 pips.

This concludes our lessons on cluster counting. There are many cluster to memorize to further speed up your counting, but from these lessons you should have the essentials in your toolbox. Hopefully you’ve seen how fun cluster counting can actually be!

##### Further reading:

- A pdf (I believe by Art Williams) with exercises and solutions:

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