Approximate Colorless Pip counting & strategy question
Hello all! Two part question here that I will try and make brief. I have always been terrible at math and now in my 40s I really don’t have any desire to improve all that much at it haha. One thing that has kept me away from playing OTB in clubs and at tournaments, has been my inability to count pips. I came across “approximate colorless counting” and I was curious who all has used this approach and how effective it has been? I understand it can be a few pips off here and there but it is by far the easiest method I have come across. I’m not looking to become a GM. I just want to feel more confident about decisions OTB and maybe play in a few tournaments.
My other question is basically about time management. For those that donÂ’t use a running pip count how do you know when itÂ’s best to use your time to count pips? Other than the obvious like you are being cubed by your opponent or you both have just broken contact. Will you take time to count pips in the early middle stages of the game when you are not sure if you should play boldly or safely?
Any responses are greatly appreciated! Hope Everyone is well!
4 Replies
Not sure about colorless pip counting, but you might like the half-crossover count. If you skip the last part where you get an exact count/difference, it's very easy and accurate enough if you don't need the precise numbers. Also it can work as a sanity check on an exact count.
I just try to make a quick assessment on whether the exact count is sufficiently likely to swing the decision before taking the time to count. Usually a rough estimate is good enough.
Another trick: if the positions are still symmetrical enough, you can often pretty easily get a rough idea of a relative pip count (I,e. I’m 10 pips behind or I’m 6 pips ahead, etc.) by “cancelling” you checkers and your opponent’s when they are on the same relative point. For example, you each have 5 checkers on your 6 point to start - if you both still have 3 or 4 checkers on this point, you can ignore them. Similarly with stacks on the midpoint or with any other made points. It can help in more developed spots too - if you each have 2 checkers on your 3-7 points, for example, you can ignore all ten checkers for each side and focus on the remaining 5.
That last example brings up another trick you might not have realized - pip counting a five prime is the easiest thing in the world. (Assuming no spares on any of the points). Take the middle point of the five prime and multiply by 10. In my example above the 3 through 7 points are the five prime. The middle point is the 5 point. These checkers then count to 5x10 = 50 pips. This trick can also work with other similar configurations; multiply the middle point by the number of checkers. It’s a bit more complicated for even numbers of points since there isn’t one single middle point. You have to average the two middle points. For instance 8 checkers covering the 6, 5, 4 and 3 points. The 4 and 5 are the two middle points. Average is 4.5. Multiply by 8 gives 36 pips.
It’s not as simple, but you can even use this logic when there are gaps in the prime. Suppose for example there are two checkers each on the 8,6,5, and 4 points. Note that if there also were two on the 7 point this would be a five prime with 6 as the midpoint. That would be 60 pips, so the actual configuration would be 14 pips less (missing two checkers on the 7 point), hence 46 pips.
I find that counting blocks like this can speed my counts immensely. There are certainly other methods out there, so find whatever works best for you. Good luck.
I second the recommendation to use the half crossover method and just skip the last step where you fine-tune. This is almost always within 3-5 pips of the true count (usually closer) and with practice you’ll recognize the clusters and can do it in a couple seconds per side.
For example, from the starting position I would just start from the back and go “12, 27, 30” (i.e., 2 x 6 half-crossovers + 5 x 3 half-crossovers + 3 x 1 half-crossover) to get an approximate pip count of 30 half-crossovers. Do the same for the opponent and you have the approximate different in pip counts (where one half crossover is 3 pips).
I personally prefer cluster counting and shifting.
