## 13 Count

I noticed that in almost every pip count, I would be multiplying by 13 at some stage due to checkers on the mid point. This gave me an idea. The idea is that it is easier to do one multiplication by 13 rather than adding a bunch of big numbers together

1) You need to learn your 13 times table up to at least 13 × 20:

1  =  13
2  =  26
3  =  39
4  =  52
5  =  65
6  =  78
7  =  91
8  = 104
9  = 117
10 = 130
11 = 143
12 = 156
13 = 169
14 = 182
15 = 195
16 = 208
17 = 221
18 = 234
19 = 247
20 = 260

2) You need to familiarise yourself with the following four basic positions :

The idea is that checkers close to these positions will be counted as actually being in these positions and at the end, you do what is called a "mental shift" (i.e., you correct for the difference compared to these positions)

##### Position A

A checker on the 13 pt counts as one 13

##### Position B

Two checkers equidistant from your bar counts as one 13

##### Position C

A checker in your opponent’s home board or on the bar is equivalent to two 13s

No, I haven’t made a mistake in adding. You can do a simple correction at the end. The simple method to correct for counting the checkers in your opponent’s home board as two 13s is as follows

1. From the 13 count subtract the number of checkers in your opponent’s home board

2. Then subtract the pipcount of the checkers counting the pipcount in terms of your opponent’s board points. (I.e., your 24 pt is your opponent’s 1 point and will count as 1 pip; your 23 point will count as 2 pips; your 22 pt will count as 3 pips; the bar counts as 0 etc.)

So, in the above case you would have two 13s (26), minus number of checkers (1), minus pipcount from the perspective of your opponent’s home board board points (1), equals 24

##### Position D

Two opposite checkers + 1 pip equals 2 sets of 13

This gives you another way of counting a checker in your opponent’s home board if you can pair it with a checker in your home board

If you look at Example 2 bellow, the 2 checkers on white’s 3 point can be paired with the 2 checkers in his opponent’s home board. If you were to just count these 4 checkers, your 13 count would be 2 for each opposite pair which would add up to 4. In order to correct for the checkers in white’s opponent’s home board not being on his 2 point, you would have to move 1 checker back from the 22 point to the 23 point (−1) and 1 checker forward from the 24 point to the 23 point (+1). This adds up to zero and so there would be no correction. The 4 checkers add up to (13 × 4) 52 pips

Other frequent uses for Position D include:

• Pairing a 20 point anchor with an equal number of checkers on your 6 point
• Pairing an 18 point anchor with an equal number of checkers on your 8 point

### The Method

The method involves counting the 13s and then doing mental shifts. The checkers do not have to be exactly in these positions they just have to be close

##### Example 1

Step 1. Figure out your strategy as to how you will count the 13s

• The 5 checkers on the 13 pt can be counted as normal 13s as in Position A. 5 on the 13 pt = 5 sets of 13
• The 8 around the bar can be counted checkers equidistant from the bar as in Position B. 8 divide 2 = 4 sets of 13
• The 2 checkers on the 24 pt can be counted as checkers in your opponent’s home board as in Position C. 2 checkers on the 24 pt = 4 sets of 13
• That adds up to a 13 count of 13 − 13 × 13 = 169 pips

Step 2. Mental shifts

• 13 pt (Position A): No mental shifts required as all checkers are on the 13 pt
• Equidistant from your bar (Position B): To get 4 at an equal distance either side of the bar, you need to move the 3 on the 8 pt to the 7 pt (+3) and 1 from the 6 pt to the 7 pt (-1). This comes to 2 pips to add onto your total. 169 + 2 = 171
• Checkers in your opponent’s home board (Position C): The number of checkers in your opponent’s home board is 2. The pipcount (in terms of your opponent’s home board points) of the checkers in your opponent’s home board is 2. Subtract these from the running total: 171-2-2=167

##### Example 2

Let’s try a more complicated example

First you do White’s 13 Count:

Step 1. Figure out your strategy as to how you will count the 13s

• I will ignore the checkers on the 3 pt and just add them on at the end
• The 5 checkers on the 13 pt can be counted as normal 13s as in Position A. Five 13s
• The 6 around the bar can be counted as equidistant from the bar as in Position B. Six checkers on the 8 and 6 pts = three 13s
• The checkers on the 22 and 24 pts count as checkers in your opponent’s home board (Position C). Two checkers in your opponent’s home board count as four 13s
• The 13 count is 12. 12 × 13 = 156

Step 2. Mental shifts

• 13 pt (Position A): No shift needed
• Checkers equidistant from the bar (Position B): You need to shift 3 from the 8 pt to the 7 pt (+3). So the total is now 159
• Checkers in your opponent’s home board (Position C): Running total (159), minus number of checkers (2), minus pips in terms of your opponent’s home board points (4), equals 153
• Now, add the 6 pips for the checkers on the 3 point: 153 + 6 = 159

And now Black’s 13 count:

Step 1. Figure out your strategy as to how you will count the 13s

• 13 pt (Position A): I will count the 18 pt checkers as 13 pt checkers. So, five 13s
• The 10 around the bar can be counted as in Position B: 10 checkers is five 13s. So we have a 13 count of 10. 13 × 10 = 130

Step 2. Mental shifts

• 13 pt (Position A): The 2 on the 18 pt should be on the 13 pt (10) Total is 130 10 = 140
• Checkers around the bar (Position B): You only have to make 1 shift to make all checkers equidistant from the bar. Move one checker from the 8 to the 7 pt (1). Grand total is 140 1 = 141

### Advantages of This Method

There are only four key positions to memorise

The 15 checkers on the board can often be treated as a giant cluster. Sometimes you may not get all the checkers into the 13 count. However, it is usually one or two checkers that are on low points that are easy to add on at the end

The method has advantages over standard cluster counting (although I do think cluster counting has its usefulness in certain situations). With cluster counting, you add the big numbers together and you still make mental shifts if the checkers are not precisely in the right place. The Ash Dalvi method actually includes the step of multiplying by 13 in exchange for adding big numbers together. If you learn your 13 times table by heart, you are less likely to make a mistake compared to adding big numbers together. It will also take less time

If all you need to know is who is ahead in the race, you can count the 13s in seconds. If you are two 13s ahead, you are almost sure to be ahead after the mental shifts. If you are one 13 ahead, you are still likely to be ahead but you might have to do the mental shifts

Generally, it seems like a natural method. You start with a bunch of checkers on the 13 pt and a cluster around the bar. Furthermore, general backgammon strategy involves keeping the 13 pt and trying to build points around your bar. So, you have to multiply by 13 in most positions anyway

### Other Tips

You can count an 18 point checker as a 13; it has a correction of +5

You can count an 8 point checker as a 13; it has a correction of −5

##### Example 3

Once most of the checkers are in the home board, I often combine my method with cluster counting. I can see the 5 prime is 40 pips. So, I will count all 5 checkers outside my home board as 13s and subtract 10 to correct for the 2 checkers on the 8 pt

13 × 5 = 65 − 10 = 55 + 40 (for the 5 prime with a middle point of 4) = 95

Easy peasy!

Don’t always break your 10s into 13s. If I have a 5 prime, I know the middle point multiplied by 10 is the total pip count in the prime. I will add this on to the 13 total

In the next article, I will discuss how you might incorporate checkers in your home board into the 13 count

With a little practice, you can count some seemingly complicated positions relatively effortlessly in just a few seconds. You might even find that pip-counting is almost fun.

### Home board checkers

I have discussed two ways that some of the home board checkers could be included into the count (see Position B and Position D - I repeat them again here):

##### Position D. Two opposite checkers + 1 pip equals 2 sets of 13

However, these 2 positions will not always get all the home board checkers into your count

When I started using this method, I just added home board checkers on at the end either using Jack Kissane’s cluster counting method or by adding 1 or 2 checkers on low points at the end. The idea of the 13 count system was to avoid adding big numbers and several sets of multiplication and this was achieved in Part 1. In part 1, I showed you how to get all the checkers between the 24 pt and the 7 pt into your 13 count

Note that there are only 4 very simple positions to remember to account for all of these checkers

This is arguably an improvement on traditional cluster counting methods. You can always count the checkers on high points into a 13 count. However, if they are not situated in convenient positions, they can be difficult to incorporate into a traditional cluster count

In addition, sometimes, I simply want to know who is ahead in the race. In these situations, I am happy to rely on a 13 count without corrections if there is a difference of two 13s

For this reason, I have explored a couple of methods of incorporating home board checkers into the 13 count

##### Method 1. Memorising corrections while maintaining symmetry

If you are better at memorising numbers than patterns, this method may suit you better than Method 2. I use the following positions and memorise the corrections

Position 1a. (2 checkers on the 6 point) 1 set of 13 and a correction of –1 pip

Position 2a. (2 checkers on the lowest 3 points) 1 set of 13 and a correction of –1 pip

Position 3a. (2 checkers on the 5 and 6 points) 2 sets of 13 and a correction of –4 pips

Position 4a. (3 prime on the highest points) 2 sets of 13 and a correction of +4 pips

Position 5a. (4 prime on the highest points) 3 sets of 13 and a correction of –3

Position 6a. (5 prime on the highest points) 3 sets of 13 and a correction of +1 pip

Position 7a. (Closed board) 3 sets of 13 and a correction of +3 pips

##### Method 2. Memorising unsymmetrical positions without having to make corrections

If you are more comfortable memorising patterns than numbers, Method 2 may be more useful to you

Position 1b. (2 checkers on the 6 point and 1 on the 1 point) 1 set of 13

Position 2b. (2 checkers on the lowest 3 points plus 1 one extra on the 1 point) 1 set of 13

Position 3b. (2 checkers on the 5 and 6 point and 1 on the 4 point) 2 sets of 13

Position 4b. (2 checkers on the 6, 4 and 3 points) 2 sets of 13

Usually when your opponent makes your 5 point, it is considered bad. However, on the bright side, it can make pip counting easier. That thought helps me to remember this pattern

Position 5b. (4 prime on the highest points with an extra checker on the 3 point) 3 sets of 13

Position 6b. (4 on the 6 point and 3 on the 5 point) 3 sets of 13

Position 7b. (3 on the 5 and 6 points and 2 on the 3 point) 3 sets of 13

##### Example 4

White’s 13 Count

Step 1. Figure out your strategy as to how you will count the 13s

The 3 checkers on the 13 pt can be counted as normal 13s as in Position A (Part 1)

• 13 pt (Position A) — Three 13s

The 8 in the home board are 3 sets of 13 (Position 6b)

• Position 6b — Three 13s

The 4 checkers in white’s outer board will be counted as 4 sets of 13 (Position A Part 1)

• Four 13s. The reason I am counting them as 13s is because I can know that the correction for an 8 pt checker is –5. Because there are 4 checkers, the correction will be simple –20 (for the 4 checkers) + 2 (for the checker on the 10 pt). I could have counted them as equidistant from the bar (Position B Part 1). However, the mental shifts look more time-consuming using that method
• The 13 count is 10. 10 x 13 = 130

Step 2. Mental shifts

• 13 pt (Position A) — no shift needed
• The four checkers in my outer board 130 - 20 + 2 (Position A) = 112
• Checkers in my homeboard (HB 6b) - no mental shifts required

This gives a total of 112 pips

Black’s 13 Count

Step 1. Figure out your strategy as to how you will count the 13s

The 2 checkers on the 13 pt can be counted as normal 13s as in Position A (Part 1)

• 13 pt (Position A) — Two 13s

The 3 on the 6 pt, 2 on the 7 pt and 1 on the 8 pt can be counted as equidistant from the bar (Position B, Part 1)

• Equidistant from the bar (Position B, Part 1) — Three 13s

The checker on the 24 pt can be counted as two 13s (Position C, Part 1)

• Two 13s

The remaining 6 checkers on the 3, 4 and 5 pts can be counted as two 13s (Position 4b)

• Two 13s

The 13 count is 9 — 13 x 9 = 117

Step 2. Mental shifts

• 13 pt (Position A) — no shift needed
• The checker on the 24 pt (Position C, Part 1) 117 — 1 (for 1 checker) — 1 (for it being on the
opponent’s 1 pt) = 115
• The 6 checkers that are equidistant from the bar (Position B, Part 1) + 1 for the checker on the 8 pt. 115 + 1 = 116
• The 6 checkers in the home board (Position HB 4b); the 2 checkers on the 5 pt need to be moved to the 6 pt to match Position HB 4b which is 2 pips. 116 - 2 = 114 which is the total pip count

##### Example 5

Let’s start with Black’s pip-count this time

Black’s 13 Count

Step 1. Figure out your strategy as to how you will count the 13s

The 3 checkers on the 13 pt can be counted as a normal 13 as in Position A (Part 1)

• 13 pt (Position A) — One 13

The 2 checkers on the 21 pt can be counted as two 13s each (Position C, Part 1)

• 2 checkers (Position C) in opponent’s home board — Four 13s

Six checkers in black’s homeboard (2 on the 6, 5 and 4 pts) can be counted as 2 sets of 13 (Position 4b)

• Position 4b — Two 13s

The remaining 6 checkers (3 on the 8 pt, 2 on the 7 pt and 1 on the 6 pt) can be counted as equidistant from the bar (Position B, Part 1) — 3 sets of 13

• Three 13s

The 13 count is 10. 10 x 13 = 130

Step 2. Mental shifts

• 13 pt (Position A) — no shift needed
• The two checkers on the 21 pt need correcting 130 — 2 (for the number of checkers) — 8 (for each being on the opponent’s 4 pt) (Position C, Part 1) = 120
• Six checkers in black’s homeboard (2 on the 6, 5 and 4 pts) (Position 4b) — no correction
• The remaining 6 checkers ( 3 on the 8 pt, 2 on the 7 pt and 1 on the 6 pt) counted as equidistant from the bar (Position B, Part 1)
• The 5 checkers on the 7 and 8 pt need to be shifted 1 pip in for the checkers to be symmetrically equidistant from the bar — 5 pips

120 + 5 = 125 which is the total pip count

Now white’s pip-count

White’s 13 Count

Step 1. Figure out your strategy as to how you will count the 13s

The checkers on the 15 pt can be counted as normal 13s as in Position A (Part 1)

• 13 pt (Position A) — One 13

The 2 checkers on the 8 pt can be paired with the 1 checker on each the 5 and 6 pt to be counted as equidistant from the bar (Position B, Part 1)

• Equidistant from the bar (Position B, Part 1) — two 13s
• The other 9 checkers in white’s home board can be counted as a 4 prime with an extra checker as in Position HB 5b — three 13s

The checker on the bar can be counted as two 13s (Position C, Part 1)

• Two 13s

The 13 count is 8 — 13 x 8 = 104

Step 2. Mental shifts

• 13 pt (Position A) — the checker on the 15 pt needs to be moved 2 pips be on the 13 pt. 104 + 2 = 106
• The checker on the bar (Position C, Part 1) 106 — 1 (for 1 checker) and 0 (for it being on the bar) = 105
• To get the checkers on the 8, 6 and 5 pts equidistant from the bar (Position B, Part 1), you can move the 6 pt checker forward by 1 pip. 105 + 1= 106
• In order to get the remaining 9 checkers in white’s home board into Position HB 5b, you need to move 2 checkers on the 2 pt and 2 checkers on the 3 pt back by 1 pip (ie by a total of 4 pips). 106 - 4 = 102, which is the total pip count

That was quite easy to do and it didn’t take much time. However, I included this position deliberately to demonstrate a position that I would definitely change counting method

Anyone who is familiar with cluster counting will immediately see that if you remove a checker from black’s 6 pt, the remainder of the checkers on black’s home board will add up to 40 pips. It is a 5 prime position with 1 checker either side of the 4 pt instead of actually being on the 4 pt (ie the middle pt which is 4 x 10 = 40). In addition, if you are used to counting in 10s, you will know that the checker on the bar and the checker on the 15 pt will add up to 40. So all you have to do is add the 2 checkers on the 8 pt and the checker on the 6 pt to 80 to get 102. Similarly, in Example 1, for black’s pip count, I would count the 10 checker that make the 5 prime as 50 pips and just include the remaining checkers into my 13 count

It is important to be flexible when pip-counting. If the layout of the checkers is more amenable to another pip-counting method, you should switch or combine methods

##### Which Method is Best?

The method you choose depends on how your mind works and the layout of the checkers in a given position. After practicing including home board checkers into my 13 count, I’ve decided that an eclectic approach to pip-counting is best for me. I use both of the above methods as well as traditional cluster counting methods and sometimes I just add the pips on low points onto my total at the end of my 13 count

The method is flexible because you adapt it in accordance with how the checkers are situated. Whatever strategy of counting you see first is probably the best strategy

It can be advantageous including home board checkers into the 13 count if all you want to know who is ahead in the race. As mentioned earlier, if you are two 13s ahead, you can be fairly comfortable about being ahead in the race. An exact pip count may not be necessary

For the home board checkers, I would simply advise you to use the strategy to pip-count that is easiest for you. I wouldn’t say that the 13 count is particularly an improvement on other methods for the purposes of counting pips on low points. As mentioned earlier, the purpose of this technique was to reduce the addition and multiplication of large numbers and count the checkers on high points easily. For most people, the checkers in the home board aren’t difficult to count anyway

There are many pip-counting methods. They all have some advantages and disadvantages. Some of the disadvantages of a system may only be disadvantageous to some people but not to others

I find this method suits me because I have a tendency to make mistakes when I add large numbers together and to forget one number while calculating a second number to add to it

I suggest practicing with the techniques in Part 1 first and then include the techniques in Part 2