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 ofthe 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

0 70

White money game Blue

The program tells us that the average number of rolls to bear off from thisposition 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 aboveposition: [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 numberof pipples. I had the computer chug through all possible combinationsof 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 theace point starts to fill up there is more emphasis on having an extraman on the four point, probably so we won't have to put even more menon the ace point. Understanding this concept will help all players play their races more accurately. In general, the closer you can get to what lookslike an optimal bearoff (the magic triangle), the better your play, with theproviso 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 manypipples 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 theoptimal 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 theabove 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 lookslike 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 followingadjustments. 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 thetwo point to the four point. This would leave us [011223] which certainlylooks like it is an optimal structure (in fact, it is the optimal structurefor 41 pips, 9 men). If we can just determine how many pipples of smoothnesswe have gained from these adjustments, we will be a long way toward theultimate pip count.

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

[001206] = 671
[001215] = 649

I chose these two positions for comparison so as to exclude other factorsas best as possible (for example, as we shall see later, if there were menon the ace point that would affect the results). In this example, the secondposition is 22 pipples better than the first position. It is also one pipbetter, 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, andthe other 11 pipples come from the improved smoothness. Thus, we can saythat the improved distribution is worth 11 pipples.

Continuing along these lines unstacking the six point onto an empty fivepoint, 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, asthe six point gets less heavy the value of unstacking it goes down. Thisis what our intuition would tell us.

Wonderful! Now all we have to do is memorize these numbers for eachunstacking position. This is fine for computers with their virtuallyunlimited memory, but we humans are fallible -- we need some simplification.Fortunately, there is one (there always seems to be some kind ofsimplified formula if you are willling to look hard and fudge a little).In this case, the pipple difference appears to be approximately four timesthe square root of the number of men on the six point (for the rest of thisarticle I will use SQR to represent square root). This pattern (multiplying the square root of the men stacked by some constant) seems towork 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

[000250] = 500
[001240] = 473

The difference is 27 pipples. 22 of these pipples come from the two pipdifferential, so the remaining 5 pipples come from the improved distributionas the table predicts.

So far we have looked only at unstacking to an empty point. As you mightguess, the smoothing effect goes down if the point is not empty. Forexample, 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 aresix pipples worth of smoothness as opposed to 11 pipples if the five piontwere empty. In general, if you divide the result for unstacking to anempty point by two you will get the result for unstacking to a point withone man, and if you divide by three you will get the result for unstacking toa point with two men.

One more consideration: Duplication. Let's once again look at unstacking fromthe six point to the five point. Obviously this is valuable since it givesus a five to play -- if we didn't have a man on the five point we would haveto move to the ace point which is bad. Intuitively it seems as though itshould 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 comparedto 11 if the ace point were empty. These duplication situations are asfollows:

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 theduplication point, three for the second man, two for the third man, and one forthe fourth man (this addition should be done before the division due to theunstacking 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 multiplieris 4 (from the table). The Stack is of height four, so we start with4 * SQR(4) which gives us 8. This is a duplication situation and thereare two men on the ace point, so we add 7 (4+3) to get 15. Lastly thepoint 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 wedon't worry about fractions of pipples but just round conveniently). Theactual results:

[201140] = 480
[201230] = 463

Difference of 17, 11 for the pip, so 6 pipples for the smoothness. Theexample did not come out perfectly, but in general the unstacking formulaswill 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 unstackingfrom higher to lower points. Naturally we may have to unstack from lower to higher points in order to adjust the position towards anoptimal bearoff. As you might expect, the effect of unstacking heavy lowpoints (particularly the ace point) can be quite extreme. For example, let's look at the unstack from the ace point to the five point representedby the following position.

[611004] = 675
[511014] = 678

Note that we have increased the pip count by 4, which would normally beabout 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 rootof the number of men on the point to be unstacked by a constant seemsto 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

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 theoriginal 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 theposition. With a large number of pips left it will tend to be lessimportant, since there will tend to be less wastage. Similarly with fewer men onthe 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 matterof rolls than pips, which means that the number of men left becomes moreimportant. 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 offas total pips decreases and number of men left increases. So, what is themagic formula? This one didn't turn out too neat, but it is manageable.We will have to multiply two numbers together, one depending upon howmany pips to go and one depending on number of men left. For number ofpips 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 wouldestimate 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

0 49

White money game Blue

POSITION B

0 54

White money game Blue

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

This looks like a good mix of apples and oranges. The pip count is 49 forposition A, 54 for position B, but position B is obviously better both inthe categories of men off and smoother distribution. Does this make upfor 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. Sincethis is a duplication situation and we have four men on the ace point, weadd 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 theduplication = 18, divided by 2 since there is one man on the four point fora 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 take4/5 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'thave 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 improvementdue to men off is 1 * 19 = 19. Thus, the net for men off and smoothnessin this position is -59 + 19 = -40. What this means is that Position Ais 19 pipples better than a comparable optimal structure with 15 checkers offdue to the man off, and 59 pipples worse due to smoothness, for a net of40 pipples worse than the optimal structure.

Now, let's look at position B [010325]. First, let's drop a man from thesix point to the three point, leaving [011324]. From the table this shouldbe 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 thereis a man on the three point for a total of 2 (we don't deal in fractionshere -- things are hard enough as it is).

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

Putting all this together, we see that for smoothness and men off positionB is 86 pipples better than position A. Since position A is 5 pips ahead ofPosition B and there are 11 pipples in a pip, we would estimate that thereal 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 yourguess anywhere near this?

Well, let's look at the real figures to see how we did. Heinrich'sprogram 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 throwntogether 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 togive 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 youare 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 ofhow pipples, bearoff structure, and race equity are related. Here are afew 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 rollbehind 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, themore men on the board the greater the advantage, which also makes sense sinceit is more likely that the players won't miss, so it becomes more a question of rolls than pips.

POSITION 2

54 49

White money 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 themiddle. I would estimate 62.5% if these positions were playedsymmetrically, or 2.5% for each 10 pipples. Consequently, if [412052]were on roll his 34 pipple deficit amounts to a 16 pipple lead (sincebeing on roll is worth 50 pipples), which would give him about 54% chancesin the game (50 + 2.5 * 1.6). If [010235] were on roll his effective leadis 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 evenbe 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 calculationsis quite tedious, and almost certainly not worth the effort at the tableto get the exact real pip count. However, it does give us a pretty goodidea 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 prettygood estimate of the race in uneven positions and avoid the big blunderswhich can occur when you rely solely on the pip count. Also, thisdiscussion will help our checker play in races. We now know for certainthe type of position to aim for when bringing our checkers in for thebearoff. In addition, we have some information on how to choose betweenclose bearoff plays, since we will aim for what looks closest to an optimalbearoff. This knowledge will keep down the cost of any mistakes we makein 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 matchequity considerations. If this happens, you just might want to take theextra time to go through these calculations and determine your exactequity in the race. The tournament could be riding on your decision,so it is worth getting it right.