THE PIP-COUNT OF MONTE CRISTO
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:
 Diagram 1
 Diagram 2
 Diagram 3
 Diagram 4
If you took more than a couple of seconds working out Diagrams 1 & 2 then you really need to read this article; and if you took more than five seconds each for Diagrams 3 & 4 then keep reading. The answers:
Diagram 1: 100
Diagram 2: 80
Diagram 3: 110
Diagram 4: 119
Did you spend all your time counting every checker on every point and adding them all together? This is called the Direct Method and is very tedious. Well you didn't have to do it that way, there are several shortcuts to assist you in this thankless task. One of the easiest counting shortcuts is:
OPPOSITES
This is when two checkers are positioned exactly opposite each other in the same table (home or outer) as in:
 Diagram 5
Wherever two checkers are lined up as above then 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 24+19+19+13+12+6+6+1 = 100.
Another short cut method for quick counting is in:
BLOCKS
This is when a block of checkers are used to determine a base number and then adjustments made to compensate for gaps or vacant points. Look at the following diagrams:
 Diagram 6
A block of 10 checkers will always total 30 pips.
 Diagram 7
A block of 8 checkers will always total 20 pips.
 Diagram 8
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 1-point. 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.
 Diagram 9
With just the 1-point 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 (19&6) and one of a block of 10 (6 to 1). The opposites total 50 (2 x 25) and the block totals 30; therefore very quickly we can calculate 80 pips.
Now look again at Diagram 3. Quite clearly we have a block of ten that will have a base of 30, plus three vacant points @ 10 pips each = 60 total for the block; 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.
 Diagram 3a
Here we can now see an offset opposite. The single checkers on the 11- and 9-points form a triangle with the two checkers on the 15-point; this is akin to moving the two single checkers onto the 10-point and adding up the opposites to a total of 50 pips making a grand pip count of 60 (block) + 50 (opposites) = 110 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.
Look once again at Diagram 4 and see if you can work out the pip count the easy way using blocks and opposites ..... Couldn't do it? Sorry, I haven't explained how to account for gaps within your block, have I?
Well what do we already know? We have a nice easy opposite of 25 pips. We have a block of ten so they have a base of 30 pips. So what we don't know is what block adjustments we make for the vacant points 6 to 1 (ignore the checker on the 1-point, he's part of an opposite) and the gap on the 10-point. Well, the principle of counting vacant points and multiplying them by the number of checkers in a block is still relevant, but it is slightly altered. If you look closely at the block of ten it is in fact two blocks; one of four and one of six. There are six vacant points (ignore 1-point checker) for the block of six and block of four (ten in total) which is 6 x 10 = 60; and there is one vacant point (gap) between the two blocks for the block of four which is 4 x 1 = 4; so the total pip count is: 30 (base) + 64 (adjustments) = 94 (block total) + 25 (opposites) = 119 pips. Look at Diagram 10. Gaps galore. Vacant points below and gaps in between; how on earth do we do this one?
 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 8 |
4 |
| One gap of 8 |
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 ... Centre-point Block Counting Method.
Part 2
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