THE PIP COUNT OF MONTE CRISTO By Michael Crane © 2007 
I had planned to call this article, The Pip Count of Michael Crane, but it didn't have the same effect as the title it now bears! I'm not suggesting that the Count of Monte Cristo actually played backgammon; I'm just being clever with words ;)
Mind you, this is a serious article and one which will improve your backgammon tremendously. Of course; if you can already do your own pip count quickly in your head and not forget it after you've done your opponents' pip count, then this article is not for you. But, if pip counting is alien to you or you find the mathematics of counting all those checkers on all those points just too much work; then read on for some handy shortcuts. During any game of backgammon, knowing the pip count can be a great advantage as many decisions throughout the game are based upon knowing it. If you don't know the pip count when needed you're going to be at a disadvantage  you might well lose the game or match through your ignorance; it's that important! Just to get you in the mood, here are four little pip counting problems for you. Just scroll down until you can see just below the bottom two boards and no further . . . 
See if you can work them out. When you have done so click here to continue. 
Wherever two checkers are lined up opposite each other they will always have a pip count of 25; it doesn't matter what points they occupy, it will always be 25 pips. If there is more than one checker on a point then each one counts as 25. Now, look back at Diagram 1 and you'll see that the quickest way to count is that we have three blocks of opposites, two of which total 50 (1&24 and 13&12) and one block that totals 50 (19&6); therefore, quick as a flash we see that the pip count is 100. In fact, when you've finished this article and come to understand pip counting short cuts you'll have worked it out as 4 x 25 = 100 in about 2 seconds at the most. This is much faster than doing it the long and labourious way: 24+19+19+13+12+6+6+1 = 100. Pips can also be counted in BLOCKS 

A block of 10 checkers will always total 30 pips.


A block of 8 checkers will always total 20 pips.


A block of 6 checkers will always total 12 pips.


As can be seen from the point numbers these totals only work for checker blocks connected to the 1point. When the blocks are higher up the board, adjustments have to be made. Whenever there is a vacant point below the block, the pip count is increased by as many checkers in the block as there are vacant points. Look at Diagram 9. As you already know a block of ten checkers equals a pip count of 30 pips. 

With just the 1point vacant the pip count of 30 is increased by 1 x 10 (vacant point x checkers in block) = 10 + 30 = 40. So, very quickly you can count any block anywhere. If the block had been eight and there were two vacant points then the count would have been 2 x 8 (vacant points x checkers in block) = 16 + 20 = 36 pips. So, for each vacant point add the number of checkers in the block to the base number. This is true of blocks straddling the bar or the outer tables  anywhere you can find a block of checkers.


Take another look at Diagram 2. You should now see this as two separate blocks. One of opposites with two checkers on each (19&6) and one of a block of 10 (5 to 1 point). The opposites total 50 (2 x 25) and the block totals 30; therefore very quickly we can calculate 80 pips. NB: A block of 6, as shown in Diagram 2 always add up to 42. 

Now look again at Diagram 3. Quite clearly we have a block of ten (8 to 4 point) that will have a base of 30, plus three vacant points @ 10 pips each = 60 total for the block, plus 6 for the extra checker on the 6point making 66 so far; but how do we calculate the remaining four checkers? To make things a little clearer let's have a look at a section of Diagram 3 on its own. This is shown in Diagram 3a 

Here we can now see an OFFSET OPPOSITE. The single checkers on the 11 and 9points form a triangle with the two checkers on the 15point; this is akin to moving the two single checkers onto the 10point and adding up the opposites to a total of 50 pips making a grand pip count of 66 (block) + 50 (opposites) = 116 pips.
When doing a pip count look for blocks and opposites as these are the easiest of checkers to count. Offset opposites are sometimes a little 'out of sync' inasmuch as you might need to make a mental adjustment now and again to contrive one but generally this isn't too hard after a little practice.

Diagram 3a 
Look once again at Diagram 4 and see if you can work out the pip count the easy way using blocks and opposites. This is very tricky using blocks, and there are no useful opposites. The easiest way to count this position is by using the CENTRE POINT block counting method; but before we get to that, let's look at counting blocks with holes in them in Diagram 10! 

We are still using a block of ten with a base of 30 so that's where we start. Adjustments are made for each vacant point or gap where each one is multiplied by the number of checkers above it in each case; thus: 

Ten block 
30


Four gaps of 10 
40 

One gap of 8 
8 

One gap of 4 
4 

One gap of 2 
2 

Total pip count 
84 

Ah, if only backgammon positions were so easy all of the time! Unfortunately they aren't so you'll have to work at this method of counting until it is second nature to you. It will occur, I promise. If you aren't happy with this method of block counting, try the next one. . 
CENTREPOINT BLOCK COUNTING METHOD As the name implies, blocks are counted using the central point of the block (blocks spanning an odd number of points are easiest but even ones can be calculated). Look at Diagram 11. Firstly we need to identify the central point, in this easy example it is quite obviously the 4point. Now, imagine levelling off all the checkers onto the one, central point. Now, any idiot can multiply ten by four can't they? So, just by identifying the central point and using your brain a little you can count blocks in a twinkling of the eye. . 
Diagram 11 
As I said, this method works easily for odd blocks but even ones require a little more brain power  but don't panic; it's not that hard! Look at Diagram 12. The central point for this block is 8.5; so simply multiply the checkers (8) by 8.5 to get 68; simple isn't it? Mind you, this example is as quick using the centrepoint method (20+48=68) but I'm sure you get the idea. With a little practice and lateral thinking odd blocks, even blocks and blocks with gaps can all be counted quickly using the central point method. 
Diagram 12 
Blocks and Opposites are not the only short cut counting methods. Here's two further examples which I reproduce from Paul Magriel's excellent book, Backgammon with acknowledgements to Paul and his publishers (X22 Publishing for the new, reprinted softback version. Availabe from The Backgammon Shop).
COMPARISON METHOD This method of finding the pip count gives us the same result as the direct method. It provides a running total of the pip count. The comparison method is preferable to the direct method [and the block and opposites methods] because it eliminates the need to figure out and then compare two separately totalled sums. Using the comparison method, you compare the number of checkers that you and your opponent have on corresponding points. You subtract the number of men your opponent has from the number of men you have. Then you multiply this difference by the number of the point you are comparing. You now have a pip count for a particular point. If you have more checkers on the point than your opponent, you will end up with a plus pip count; if you have less, you will end up with a minus pip count for that particular point. As you compare each pair of points in turn, you keep a running total of the pip counts for each point, subtracting a pip count when you have less pips than your opponent, and adding when you have more. When you have finished comparing the sets of points, the final running total will indicate the complete pip count for the entire position. If it is plus, you are behind; if it is minus, you are ahead. Let’s use the comparison method to determine the pip count in Diagram 13. . 

The final total indicates that Black is +6, or 6 pips behind in the race against White. See Diagram 13a. Experience has shown that the comparison method is usually quicker and more reliable than the direct method  especially when the positions of the two players are similar. 
Diagram 13 
Diagram 13a
MENTAL SHIFT METHOD This method is a modification of the comparison method. In this case you mentally move men of one player to make a position identical with the other player. Then you count the number of pips you had to move each checker to make the positions identical. If you have to move your men toward your home board, you add the number of pips; if you have to move your men away from your home board, you subtract the number of pips moved. . 
In Diagram 14, we have indicated how a few of Black’s men may be rearranged to yield identical positions with White. In order to make the positions identical, we must move Black’s men a net total of 7 pips forward, so therefore Black is 7 pips behind in the race. Calculation: 2+21+1+3=7 
Diagram 14 
The advantage of this method is obvious  with practically no work or calculation, you get the total pip count in a matter of seconds. For those who hate having to remember and total long strings of numbers, as the author does, this is a real blessing.
The disadvantage is that there is no set way of determining which checkers to move to make the positions identical. Furthermore, there may be many different ways to shift the men to arrive at identical positions. To determine when to use this method, study various positions and find the one where the fewest checkers need be moved to make Black’s and White’s positions identical. When the two positions are entirely dissimilar, the number of men to be shifted may well make this method more trouble than it is worth. With practice, however, it is often surprising how few checkers have to be moved to equate the positions. Paul Magriel wrote the above segment in 1976 and it is still good advice today. Many backgammon players and authors have invented their own system. Here's one from a Biba member, Richard Howes: THE RICHARD HOWES COUNT At first glance this method might look very complicated, but, if you can multiply by six then you shouldn't have any difficulties. The count is done in two steps: . 
1. Start with a pip count of 15
2. Deduct 1 for each checker in the home board on the points indicated by a minus sign (). Thus in Diagram 15, 15  4 (4 x 1) = 11. 3. Continue adding to the pip count as indicated by the plus signs (+); 11 + 10 (3 x 1) + (2 x 2) + (1 x 3) = 21. 4. Multiply this total by six; thus 21 x 6 = 126 pip count so far. 
Diagram 15 
5. Starting in your own board, add the amount indicated for each checker occupying that point; thus in Diagram 16, 126 + 26 (2 @ 4) + (2 @ 5) + (4 @ 1) + (2 @ 2) = 152 total pip count.

Diagram 16 
.
Here is another example for you to work out on your own. Cover up the text after the board and see if you can get the correct pip count for both sides. When you have worked it out click HERE to see if you were correct.

Diagram 17 
Count for Black: Count for White: So, did you understand Richard's method? It looks cumbersome but, once you understand what to add and subtract it really is quite straight forward. So far all the methods I've shown are only useful for part of the game. The pip count worked out using these methods are not much use when both sides are bearing off due to a lot of wastage on the lower points. Often a game or match can be decided on a correct doubling decision during this crucial period, therefore, we move to a higher level. THE THORP COUNT This method was invented by the famous Edward O. Thorp for making doubling decisions in pure race money games. It is a modification of the basic pip count which takes into consideration some elements of checker distribution.The Thorp count provides a cube decision for races in money games. The formula works as follows :
Repeat steps 1 to 4 for your opponent, but do not carry out step 5. This will be his corrected pip count (B). Now you determine the Thorp metric by calculating A  B.
In this next example, Diagram 18, the pip count is even at 94 each; so, in a pure count it is level ... but what is the Thorp count? White is on roll: 

Diagram 18 

So, it isn't a double, but it is a take. Snowie confirms, No Double/Take. As far as I am aware the Thorp Count isn't 100% accurate but it is useful to have in your counting repertoire. A more accurate method has been devised by Tom Keith. For each player, start with the basic pip count and: add 2 pips for each checker more than 1 on the one point; Increase the count of the player on roll by oneseventh (rounding down). A player should double if his count exceeds the opponent’s count by no more than 4. 

A more indepth look at this type of pip count adjustment can be seen at Backgammon Galore.  
Our next method is one devised by Bob Hoey, he's called it ...
THE HALF ROLL METHOD The half roll count is to determine the equity in a race and is not an accurate pip count; nonetheless it is effective for determining who is leading in the race. Bob uses the average roll to work out the count. 

Since the average roll is about 8.3 pips or thereabouts, one could say that any checkers on the 7 to 10points are half a roll from coming in. Checkers on the 11 to 14points are a roll each from coming in. Checkers on the 15 to 18points are a roll and a half from coming in. Checkers on the 19 to 22points are two rolls from coming in and a checker on the 23, 24 or the barpoint is two and a half rolls from reaching home. It is a much quicker calculation as to who has more half rolls to get home. Having determined the race to home by this method, then one has only to consider the distribution in the home board. Also take a half roll from my your opponent's count if he is on roll (if he were doubling, that is). 
Diagram 19 

Another method is one (probably) devised by Mark Denihan with some assistance from Mark Driver. This one uses crossovers and sixes to determine the count. THE CASTING OUT CROSSOVERS METHOD First of all you need to know how many crossovers are required to get all your checkers into your home board. A crossover is a move of 6 pips; therefore to move a checker from your opponent's home board will take 3 crossovers, from his outer board, 2 crossovers, and from your outer board, 1 crossover. A checker on the bar is 4 crossovers. When you have worked out how many crossovers you need, you must then work out how many checkers will occupy your 6point after all the crossovers have been made. A checker on the bar is assumed to reenter on the 19point and will finish up on the 1point. You then add them to your crossovers and multiply the total by 6 for an initial pip count. The next step is to mentally move the remaining outer checkers into your home board and count the number of checkers on each home board point working down from the 5point to the 1point, totalling the pip count as you do. Add this pip count to your initial pip count and this is the total pip count. Let's see how it's done using Diagram 20. Working out the pip count for both sides. . 

Diagram 20 



THE RUNNING TOTAL DIFFERENCE METHOD Here's one method that seems so simple you're probably wondering why I've left it 'til last. Well, it is simple until you're hit and placed on the bar. Look at Diagram 21, the starting position. . 
Don't bother doing a pip count, trust me; the pip count for each player is 167 pips. All you have to do in this method is keep a running total of the difference between you and your opponent. For example, you roll 31, therefore you are 4 pips ahead; your opponent rolls 65 and you drop down to 7 pips behind, you roll 55 and go into the lead with 13 pips; and so on and so on. Now, you might just think that this is by far the best method you've come across; it's so easy, all you need to do is add and subtract; and the biggest number you'll have to deal with is thirtysix. 
Diagram 21 
Don't fall for it. It is nigh impossible to keep up to date; being hit back onto the bar, dancing with half a roll, both of these will complicate matters  but if you don't believe, have a go! Another method using the starting pip count is ... 

THE RUNNING TOTAL METHOD Just like the method above except this time you keep your own pip count in your head rather than the difference. Not much use and far more difficult than the difference method. Don't even try this one! There are other short cuts to pip counting, but for the moment there should be enough here for you to develop a system that suits you. No system is better than the others, the best system is perhaps a hybrid of one or more methods; find the one that suits you the best and then practice, practice, practice. KEEPING COUNT So, you've mastered counting, but have you figured out how to keep the score in your head whilst counting your opponent's pips? This is perhaps the hardest thing to do; you get your own count, you get your opponent's count, and then you find you've forgotten your own count! Because it is illegal to write down the pip count devious methods have been invented by some clever people using an aid that no one can take away from you  your hands! Here are three examples: 

PALM UP  PALM DOWN Place your right hand face down on your leg and count 1 to 5 with 1 being your thumb; turn your hand over and count 6 to 9 starting with the little finger and ignoring your thumb. Do the same for the tens with your left hand. To record your pip count all you need do now is identify your tens finger and your singles finger and there you are. Perfectly legal. THE CURLY FINGER Count one side, and then store the answer on your fingers in your lap to save having to remember it while counting the other side. It’s really easy to do this. Record the tens on your left hand, one finger down for 10, two for 20 etc.; one down curled up for 60, two for 70, etc. Same for the units on your right hand. Once again to record your pip count all you need do now is identify your tens finger and your singles finger and there you are. Another easy one. THE BOARD & FINGER This sounds like one of those trendy pubs that have mushroomed up in the last few years (awful places in my opinion). But this is a really easy method and you don't have to contort your fingers in the process. Quite simply make use of the points on the board. Use a thumb to indicate tens and a finger to indicate units; so for a pip count of 86 place your left thumb near the 8point (80) and your right finger near the 6point (6); and there you are, eightysix. When the count is greater than 129 (thumb near 12point, finger near 9point) then use your knuckle to indicate tens and keep the finger for the units. With this method there isn't any need to cross your hands over for awkward counts just keep fingers for units and thumbs or knuckles for tens and you can go up as high 252 (knuckle near 24point, finger near 12point). So, we are now at the end of this article on pip counting. If you've followed and understood what I've been on about, counting over the board will become second nature to you. It really isn't difficult and is very worthwhile  especially if you know how to do it and your opponent doesn't. Before we finish, here are a few tips and extra short cuts. 

TIPS


I would like to acknowledge the following for their assistance: Robin Clay, Paul Magriel & X22 Publishing, Richard Howes, Edward Thorp, Patti Beadles, Bob Hoey, Mark Denihan, Mark Driver, Stephen Turner and Tom Keith. Also, your are recommended to visit Backgammon Galore for loads of articles related to pip counting and backgammon in general. 
