THE PIP-COUNT OF MONTE CRISTO
Part 2
CENTRE-POINT BLOCK COUNTING METHOD
As the name implies, blocks are counted using the central point of the block (odd blocks are easiest but even ones can be calculated). Look at the next diagram:
 Diagram 11
Firstly we need to identify the central point, in this easy example it is quite obviously the 4-point. 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.
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 this:
 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 quicker using the block 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.
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).
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
| Point # |
White men on pt |
Black men on pt |
Black men - White men |
col. 4 x col. 1 |
Black running total of pip count |
| one |
2 |
2 |
2 - 2 = 0 |
0 x 1 = 0 |
0 (even) |
| two |
3 |
2 |
3 - 2 = +1 |
1 x 2 = +2 |
+2 |
| three |
0 |
2 |
0 - 2 = -2 |
-2 x 3 = -6 |
+2 - 6 = -4 |
| four |
0 |
2 |
0 - 2 = -2 |
-2 x 4 = -8 |
-4 - 8 = -12 |
| five |
0 |
2 |
0 - 2 = -2 |
-2 x 5 = -10 |
-12 - 10 = -22 |
| six |
4 |
2 |
4 - 2 = +2 |
2 x 6 = +12 |
-22 + 12 = -10 |
| seven |
3 |
0 |
3 - 0 = +3 |
3 x 7 = +21 |
-10 + 21 = +11 |
| eight |
1 |
0 |
1 - 0 = +1 |
1 x 8 = +8 |
+11 + 8 = +19 |
| thirteen |
2 |
3 |
2 - 3 = -1 |
-1 x 13 = -13 |
+19 - 13 = +6 |
The final total indicates that Black is +6, or 6 pips behind in the race against White.
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.
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.
 Diagram 14
2+2-1+3+1=7
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 hoe 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:
 Step One: Diagram 15
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.
 Step Two: Diagram 16
5. Starting in your own board, add the amount indicated for each checker occupying that point; thus in Diagram 16, 126 + 26 (4 x 2 + 5 x 2 + 4 x 1 + 2 x 2) = 152 total pip count.
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.
 Diagram 17
Count for Black:
Step One: 15 - 9 = 6 + 2 = 8 x 6 = 48
Step Two: 48 + 42 = 90 pip count for Black
Count for White:
Step One: 15 - 9 = 6 + 2 = 8 = 48
Step Two: 48 + 30 = 78 pip 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 ... the Thorp Count
Part 3
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