Probability calculation

[taalyn]

So, I thought I made some progress:

Given:

  A balanced caern (equal number of motes of each color)
  m: number of motes of each color, number of target motes
  a: number of adjacent motes (=2m)
  c: number of motes in caern (=7m+3)
  h: hand, number of motes drawn from caern
 
Then:

 X: number of possible ways to draw h motes from caern of c motes:            
            c!/(c-h)!h!
 number of ways to get x target motes: same formula - m!/(m-x)!x!
 Y: number of ways to get x target and y adjacent motes:
             m!a!/(m-x)!(a-y)!x!y!

So, to get probabilities of successes:
        sum the #s of ways for possible combinations (i.e. add the # of ways in which you can get 5 success with a hand of 4: ways for 5 asdjacent motes (A), 3A and 1 target (T), 1A2T).
        this sum, divided by number of possible ways of drawing 4 motes, should give the probability of drawing exactly 5 successes with a hand of 4.

  Or so I thought.

  Something's still wrong with the calculations - except for hands 1 or 2,
everything is apparently not possible. What am I doing wrong?

  Aidan

[taalyn]

so, I just did a brute force crunch - just ran 10 million draws, and calculated odds that way. It was much easier.

 Aidan

[bladamson]

hahahaha, yeah, it is most of the time. :)

Still useful to know at least the basis of the theory from the textbook though; helps you design and fix the mechanic, even if it's still easier to brute force the exact figures.