Differences

This shows you the differences between two versions of the page.

--- future_fabulators:formalised_decision_making [2013-10-17 19:42] – timbo
+++ future_fabulators:formalised_decision_making [2014-03-04 07:01] (current) – maja
@@ Line 1: / Line 1: @@
 ==== Formalised Decision Making ====
-In the process of creating scenarios, forecasts and otherwise moving onwards, we are left with a problem of decision making. On this page we look briefly at some ideas and thoughts.
+By Tim Boykett
+In the process of creating scenarios, forecasts and otherwise moving onwards, we are left with a problem of decision making and selecting a small amount of elements from larger clusters, for example in selecting two critical uncertainties in the 2x2 scenario matrix (see [[scenario methods]]). On this page we look briefly at some ideas and thoughts.
 Some language that we use here:
@@ Line 58: / Line 60: @@
 //(need to check the following calculations - done on paper in a shaky plane and then train!)//
-Some calculations show that the number of ways of doing this is (n-1)! sum(i=1,n-1) (1/i), where k! means the factorial of k. As the total number of permutations is n!, the proportion of permutations that have a 2-selection is (1/n)sum(i=1,n-1) (1/i). This means that with 6 factors, the likelihood of having a 2-selection is around 5/12, less than 50%. The likelihoof of a 1-selection is 1/6, so there is around 7/12 chance of having  a 1 or 2 selection. For 14 factors, the likelihood goes down to one quarter.
+Some calculations show that the number of ways of doing this is $(n-1)! \sum\limits_{i=1}^{n-1} (1/i)$, where k! means the factorial of k. As the total number of permutations is n!, the proportion of permutations that have a 2-selection is $1/n \sum_{i=1}^{n-1} 1/i$. This means that with 6 factors, the likelihood of having a 2-selection is around 5/12, less than 50%. The likelihoof of a 1-selection is 1/6, so there is around 7/12 chance of having  a 1 or 2 selection. For 14 factors, the likelihood goes down to one quarter.
 To Do: work out the corect formula for counting 3-selections. These would be xxxnyyyYzzzZ where Y is the largest factor not in xxxn and Z is the largest factor not in xxxnyyyY. My current conjecture is:
-(n-1)! sum(i=1,n-1)((1/i) sum(j=1,i-1)(1/j))
+$(n-1)! \sum\limits_{i=1}^{n-1} ( (1/i) \sum\limits_{j=1}^{i-1} (1/j))$
+but this needs some work to check it. (**note:** feel free to use inline $\LaTeX$ formatting)
-but this needs some work to check it.