This article originally appeared in the January 2002 issue of GammOnLine. Thank you to Kit Woolsey for his kind permission to reproduce it here.

## The Ultimate Pip Count

By Kit Woolsey
Editor's note:  The following article originally appeared in Inside Backgammon, May/June 1993 issue. It is reprinted here with permission of the publishers. The methods described in this article result in estimates of win percentage in bearoffs which are more accurate than any other formula to the best of my knowledge. Unfortunately using them is very cumbersome, and not practical at the table unless you are a human computer. I have never attempted to use them directly in actual play, although the concepts have been very helpful to me for estimating racing chances for uneven positions. If someone can find a way to simplify these formulas without losing too much accuracy so they would be practical to use at the table, we would all appreciate that.

Although the pip count is very important for evaluating a race, as we all know it isn't everything. Men off and smoothness of the bearoff can make a big difference, turning doubles which would be takeable on the pip count alone into big passes and vice versa. In the past, backgammon players have pretty much had to guess how to relate the pip count to these other factors. Even advanced methods such as Thorpe's racing formula, which counts men off and contains some evaluation of smoothness, is still a pretty rough approximation to reality.

Today, with our number-crunching computers and more sophisticated programming techniques, we can do better. The program from which I am working was written by Hal Heinrich. It is called BOINQ, and I would recommend it to anyone who is interested in the finer details of the racing game. One of the products of this program is as follows: For any given beaoff position with all the men in the inner board, the program will tell you the average number of rolls needed to bear off. For example, consider the following position:

POSITION 1

 070 ``` ``` Whitemoney game Blue

The program tells us that the average number of rolls to bear off from this position is 9.47. For the rest of this article I will be dropping the decimal point (i.e. multiplying the result by 100), and calling the resulting units pipples. Also, I will use the following notation for the above position: [012345]. Thus, we will have [012345] = 947, meaning that the position in question has 947 pipples to go.

The first question I looked at was the following: What does an optimal bearoff position look like. For this, I mean: Give a specific pip count and a specific number of men, what bearoff structure has the least number of pipples. I had the computer chug through all possible combinations of pips and men left, printing out all optimal bearoff structures. Here are a few of them:

70 pips, 15 men: [012345]
70 pips, 14 men: [001355]
70 pips, 13 men: [000247]

65 pips, 15 men: [112344]
65 pips, 14 men: [012335]
65 pips, 13 men: [001345]

60 pips, 15 men: [122433]
60 pips, 14 men: [112334]
60 pips, 13 men: [012244]

55 pips, 15 men: [222433]
55 pips, 14 men: [212333]
55 pips, 13 men: [111433]

These examples give us a good idea of what we should be aiming for when bringing our men in from the outer board in a race. Basically, the ideal structure seems to be a triangle with more men on the higher points. As the ace point starts to fill up there is more emphasis on having an extra man on the four point, probably so we won't have to put even more men on the ace point. Understanding this concept will help all players play their races more accurately. In general, the closer you can get to what looks like an optimal bearoff (the magic triangle), the better your play, with the proviso that taking men off usually has the highest priority.

The first question one might ask is, how many pipples in a pip (i.e. how many pipples of racing lead are equivalent to one pip racing lead). This will depend somewhat upon the position, of course, but a look at some of the optimal bearoffs should give us a good idea. for example:

67 pips, 15 men: [111435] = 915
66 pips, 15 men: [112335] = 903
65 pips, 15 men: [112344] = 891
64 pips, 15 men: [112434] = 881
63 pips, 15 men: [112443] = 870
62 pips, 15 men: [122334] = 860
61 pips, 15 men: [122343] = 849

As you can see, each pip seems to be worth about 10 to 12 pipples for the above examples which are representative of modeerately long races wihtout much wastage. From these examples and others, I determined that on average there were 11 pipples for each pip as long as there wasn't much wastage. This figure seems intuitively reasonable. The average roll is 8.1667 pips, so if there were no wastage at all there would be 12.2449 pipples per pip (100 / 8.1667). Since there is always some potential wastage, our figure of 11 pipples per pip looks very reasonable.

Now, how can we compare unsmooth positions with smooth positions? It looks like we are comparing apples and oranges, but we do have a common ground. It is the optimal bearoff structure. For example, consider the position [030105]. Looks pretty ugly doesn't it. Let's make the following adjustments. We will move two of the men on the six point to the five point, one of the men on the two point to the three point, and one of the men on the two point to the four point. This would leave us [011223] which certainly looks like it is an optimal structure (in fact, it is the optimal structure for 41 pips, 9 men). If we can just determine how many pipples of smoothness we have gained from these adjustments, we will be a long way toward the ultimate pip count.

The approach I used is as follows: For each such adjustment, I looked at two prototype positions, before and after the adjustment. For example, suppose we have six men on the six point, zero men on the five point, and want to know the value in smoothness of dropping one man from the six point to the five point. Consider these two positions:

[001206] = 671
[001215] = 649

I chose these two positions for comparison so as to exclude other factors as best as possible (for example, as we shall see later, if there were men on the ace point that would affect the results). In this example, the second position is 22 pipples better than the first position. It is also one pip better, and as we have seen there are about 11 pipples in a pip. Consequently, 11 of the 22 pipples come from the improved pip count, and the other 11 pipples come from the improved smoothness. Thus, we can say that the improved distribution is worth 11 pipples.

Continuing along these lines unstacking the six point onto an empty five point, we have the following results:

[001206] - [001215] = 22 (11)
[001205] - [001214] = 21 (10)
[001204] - [001213] = 20 (9)
[001203] - [001212] = 19 (8)
[001202] - [001211] = 17 (6)

In each case, the first number is the actual difference in pipples. The second number is the difference due to the smoother position (the first number minus 11 since there is a one-pip difference). As we can see, as the six point gets less heavy the value of unstacking it goes down. This is what our intuition would tell us.

Wonderful! Now all we have to do is memorize these numbers for each unstacking position. This is fine for computers with their virtually unlimited memory, but we humans are fallible -- we need some simplification. Fortunately, there is one (there always seems to be some kind of simplified formula if you are willling to look hard and fudge a little). In this case, the pipple difference appears to be approximately four times the square root of the number of men on the six point (for the rest of this article I will use SQR to represent square root). This pattern (multiplying the square root of the men stacked by some constant) seems to work pretty well in all cases. The table of constants is as follows:

```          5       4       3       2       1
_____________________________________________
|
6   |   4       3       2       1       1
|
5   |           4       2       1       1
|
4   |                   3       2       1
|
3   |                           3       2
|
2   |                                   3
```
In each case, the numbers on the left hand side of the table represent the initial point we are unstacking, and the numbers across the top represent the point we are unstacking to. For example, suppose we have five men on the five point and want to see the effect of unstacking to an empty three point. The table shows the multiplier to be 2, so the result is 2 * SQR(5), which is about 4.7 -- naturally we round this off to 5. A look at one prototype confirms this example:

[000250] = 500
[001240] = 473

The difference is 27 pipples. 22 of these pipples come from the two pip differential, so the remaining 5 pipples come from the improved distribution as the table predicts.

So far we have looked only at unstacking to an empty point. As you might guess, the smoothing effect goes down if the point is not empty. For example, unstacking from the six point to the five point with one man there:

[001216] = 726
[001225] = 709

The difference is 17, 11 of which comes from the one pip, so there are six pipples worth of smoothness as opposed to 11 pipples if the five piont were empty. In general, if you divide the result for unstacking to an empty point by two you will get the result for unstacking to a point with one man, and if you divide by three you will get the result for unstacking to a point with two men.

One more consideration: Duplication. Let's once again look at unstacking from the six point to the five point. Obviously this is valuable since it gives us a five to play -- if we didn't have a man on the five point we would have to move to the ace point which is bad. Intuitively it seems as though it should be worse if we already have men on the ace point, since moving another checker there just aggravates the wastage. And so it proves. One prototype to illustrate the point:

[101206] = 699
[101215] = 672

Difference of 27. 11 for the pip leaves 16 for the smoothness, as compared to 11 if the ace point were empty. These duplication situations are as follows:

6 to 5 (ace point)
6 to 4 (two point)
5 to 4 (ace point)
5 to 3 (two point)
4 to 3 (ace point)

In these situations, the general rule is to add four for the first man on the duplication point, three for the second man, two for the third man, and one for the fourth man (this addition should be done before the division due to the unstacking to a non-empty point).

The following example illustrates all these principles. Let's compare:

[201140]
with
[201230]

We are unstacking from the five point to the four point, so the main multiplier is 4 (from the table). The stack is of height four, so we start with 4 * SQR(4) which gives us 8. This is a duplication situation and there are two men on the ace point, so we add 7 (4+3) to get 15. Lastly the point we are unstacking to already has one man there, so we divide by 2, coming up with an end result of 7 (since these are all approximations anyway we don't worry about fractions of pipples but just round conveniently). The actual results:

[201140] = 480
[201230] = 463

Difference of 17, 11 for the pip, so 6 pipples for the smoothness. The example did not come out perfectly, but in general the unstacking formulas will get you within one or two pipples for any individual unstacking, which is all we can expect for a rough calculation anyway.

So far, we have just examined smoothing the position out by unstacking from higher to lower points. Naturally we may have to unstack from lower to higher points in order to adjust the position towards an optimal bearoff. As you might expect, the effect of unstacking heavy low points (particularly the ace point) can be quite extreme. For example, let's look at the unstack from the ace point to the five point represented by the following position.

[611004] = 675
[511014] = 678

Note that we have increased the pip count by 4, which would normally be about 44 pipples, yet the pipple count increases by only 3. Therefore, by making the adjustment we are gaining 41 pipples from the improved smoothness. Once again, it turns out that multiplying the square root of the number of men on the point to be unstacked by a constant seems to do the trick. This time, the table looks as follows:

```          2       3       4       5       6
_____________________________________________
|
1   |   8      11      14      17      20
|
2   |           6       8      10      12
|
3   |                   5       7       9
|
4   |                           5       6
|
5   |                                   4
```
The numbers on the left hand column represent the point moved from, the numbers across the top represent the point moved to. For the above example, (unstacking from 6 men on the ace point to the empty five point), we would have 17 * SQR(6). Since the SQR(6) is about 2.4, you can see that the result is pretty close to the 41 pipples we found in our prototype.

Once again, the above table assumes unstacking to an empty point. This time, if the point is occupied the change is not as extreme. Unstacking to a point with one man there results in about 4/5 of the original pipples, and unstacking to a point with two men there results in about 3/5 of the original pipples.

We have yet to take into account one more very important factor -- number of men left. The importance of this will depend considerably upon the position. With a large number of pips left it will tend to be less important, since there will tend to be less wastage. Similarly with fewer men on the board it will tend to be less important for the same reason. However, with fewer pips left and many men on the board it becomes more a matter of rolls than pips, which means that the number of men left becomes more important. To illustrate this, let's look at a few optimal bearoff positions:

70 pips, 15 men: [012345] = 947
70 pips, 14 men: [001355] = 944
70 pips, 13 men: [000247] = 943

60 pips, 15 men: [122433] = 840
60 pips, 14 men: [112334] = 830
60 pips, 13 men: [012244] = 824

50 pips, 15 men: [233331] = 760
50 pips, 14 men: [222422] = 737
50 pips, 13 men: [122332] = 720

These examples illustrate what we would expect -- more emphasis on men off as total pips decreases and number of men left increases. So, what is the magic formula? This one didn't turn out too neat, but it is manageable. We will have to multiply two numbers together, one depending upon how many pips to go and one depending on number of men left. For number of pips to go, the base numbers are as follows:

75: 1
70: 2
65: 5
60: 10
55: 16
50: 23
45: 33

For number of men left, the multiplier is:

15: 0
14: 1
13: 1.6
12: 2
11: 2.3
10: 2.5

If the number of pips is in-between, we just interpolate intelligently. For example, suppose we have 62 pips left with 12 men remaining. I would estimate 8 for the main multiplier, so we would be 16 pipples better than a similar position with 15 men left.

Now, let's run through a full example which will illustrate how these concepts can be applied. Consider the following positions:

POSITION A

 049 ``` ``` Whitemoney game Blue

POSITION B

 054 ``` ``` Whitemoney game Blue

Position A: [412052]
Position B: [010325]

This looks like a good mix of apples and oranges. The pip count is 49 for position A, 54 for position B, but position B is obviously better both in the categories of men off and smoother distribution. Does this make up for the five pips or not? Make your guess as to what the real relationship between these two positions is and then we'll go through the arithmetic.

First, let's look at position A, which is [412052]. The first adjustment we will make will be to drop one man from the five point to the four point, resulting in [412142]. From the table we will have 4 * SQR(5) = 9. Since this is a duplication situation and we have four men on the ace point, we add 10, coming to 19.

Next adjustment, we drop another man from the five point to the four point, resulting in [412232]. This give us 4 * SQR(4) = 8, plus 10 for the duplication = 18, divided by 2 since there is one man on the four point for a total of 9.

Next adjustment, we drop one man from the ace point to the two point, leaving us [322232]. From the table this is 8 * SQR(4) = 16, and we take 4/5 of this since there is a man on the two point for a total of 13.

Finally, we move a man from the ace point to the four point, giving us [222332]. From the table this is 17 * sqr(3) = 29, and we take 3/5 of this since there are two men on the four point for a total of 18.

Our final position look approximately like an optimal bearoff, so we don't have to do any more shifting. Our total improvement due to smoothing is: 19 + 9 + 13 + 18 = 59.

Now, let's look at men off. Our optimal position has a pip count of 53, and we have 14 men on the board, so by interpolation, our improvement due to men off is 1 * 19 = 19. Thus, the net for men off and smoothness in this position is -59 + 19 = -40. What this means is that Position A is 19 pipples better than a comparable optimal structure with 15 checkers off due to the man off, and 59 pipples worse due to smoothness, for a net of 40 pipples worse than the optimal structure.

Now, let's look at position B [010325]. First, let's drop a man from the six point to the three point, leaving [011324]. From the table this should be 2 * SQR(5) = 5.

Next, let's drop a man from the four point to the three point, leaving [012224]. From the table this is 3 * SQR(3) = 5, divided by 2 since there is a man on the three point for a total of 2 (we don't deal in fractions here -- things are hard enough as it is).

This looks like an optimal bearoff, so our improvement due to smoothness is about 7. Now, how about men off. Our resulting position has 50 pips and there are 11 men left, so we have 2.3 * 23 = 53. Thus, our net for men off and smoothness in this position is -7 + 53 = +46.

Putting all this together, we see that for smoothness and men off position B is 86 pipples better than position A. Since position A is 5 pips ahead of Position B and there are 11 pipples in a pip, we would estimate that the real difference is 86 - 55 = 31 pipples, or, to put it another way, Position B really has almost a three pip lead over Position A. Was your guess anywhere near this?

Well, let's look at the real figures to see how we did. Heinrich's program spits out the following:

Position A: [412052] = 7.83
Position B: [010325] = 7.49

Difference = 34 pipples. Not bad for a rough estimation with a few thrown together formulas.

One of the other outputs from Heinrich's program gives the probability of winning with any bearoff against any other bearoff. We can use this to give us an idea of how many pipples equal each percent of equity. Since 100 pipples represents an average of one roll, it is logical that if you are on roll with a 50 pipple deficit the race is virtually even. By examining the race with symmetrical positions we should have a good idea of how pipples, bearoff structure, and race equity are related. Here are a few samples:

[012345] = 974; equity = 60.1%
[122343] = 849; equity = 61.5%
[000345] = 844; equity = 60.5%
[234321] = 748; equity = 65.7%
[000235] = 732; equity = 61.5%

What this means is that, for example, if both players had the position [012345], then the player on roll will win the race 60.1% of the time, whici amounts to about 2% for each 10 pipples, since if you are on roll behind 50 pipples it is an even race. As you can see from these examples the relation of equity to pipples is not constant, but depends upon the position. The fewer pipples left, the greater the advantage, which makes sense since we are getting closer to the end of the game. Also, the more men on the board the greater the advantage, which also makes sense since it is more likely that the players won't miss, so it becomes more a question of rolls than pips.

POSITION 2

 5449 ``` ``` Whitemoney game Blue

As a practical example, let's look at our sample position: [412052] vs. [020132]. Comparing it with the above examples it looks about in the middle. I would estimate 62.5% if these positions were played symmetrically, or 2.5% for each 10 pipples. Consequently, if [412052] were on roll his 34 pipple deficit amounts to a 16 pipple lead (since being on roll is worth 50 pipples), which would give him about 54% chances in the game (50 + 2.5 * 1.6). If [010235] were on roll his effective lead is 34 + 50 = 84 pipples, which looks like about 71% equity (50 + 2.5 * 8.4). This would certainly be worth a double from the center, and might even be worth a recube. Would you have guessed that without this analysis? The actual figures are:

[412052] vs. [010325] = 53.6%
[010325] vs. [412052] = 70.3%

How can we put this information to use? Going through these calculations is quite tedious, and almost certainly not worth the effort at the table to get the exact real pip count. However, it does give us a pretty good idea of how men off and smoothness of the position relate to racing chances, and a player who understands these concepts will be able to make a pretty good estimate of the race in uneven positions and avoid the big blunders which can occur when you rely solely on the pip count. Also, this discussion will help our checker play in races. We now know for certain the type of position to aim for when bringing our checkers in for the bearoff. In addition, we have some information on how to choose between close bearoff plays, since we will aim for what looks closest to an optimal bearoff. This knowledge will keep down the cost of any mistakes we make in the bearoff. Also, the day may come when you are in the finals of a big tournament contemplating what to do with a 4-cube with complex match equity considerations. If this happens, you just might want to take the extra time to go through these calculations and determine your exact equity in the race. The tournament could be riding on your decision, so it is worth getting it right.