Best Answer Pickett, 10 December 2013 - 04:58 PM

As a result, that part of the equation can effectively be ignored...since it's constant (we'll just factor it in at the end to get the final values)

This leaves us with ONE_GROUP * INVERSE_OF_ANOTHER_GROUP...

In order to find the maximum, you would always want the INVERSE to the when the group was of length 1...because that means you would always have NUM_MARKERS/1 as one of the multipliers. Which means you just want to choose the largest group as your first one.

Therefore, to find the MAXIMUM value of n markers, given m groups, it would be

GROUP 1: (n - m + 1) / n

GROUP 2: 1 / n

...

GROUP m: 1 / n

Your maximum value would be (n - m + 1) / m...which in your case (n=23, m=3), would be 7.

To find the minimum, you would just do the exact opposite (choose the largest group as your reciprocal, and a group of ONE as your base...which is the same grouping as the maximum...which means your minimum value would be 1 / (m * (n-m+1))...which in your case above would simply be 1/63 = 0.015873...

So basically, you can always use the grouping I have listed above to get both the maximum and minimum values...