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future_fabulators:formalised_decision_making [2013-10-17 22:41] nikfuture_fabulators:formalised_decision_making [2014-02-11 07:54] nik
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 ==== Formalised Decision Making ==== ==== Formalised Decision Making ====
 +
 +By Tim Boykett
  
 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. 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.
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 //(need to check the following calculations - done on paper in a shaky plane and then train!)// //(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 $1/n \sum_{i=1}^{n-11/i$. 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: 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.+but this needs some work to check it. (**note:** feel free to use inline $\LaTeX$ formatting)
  
-(note: you can use $\LaTeXe$ if it helps...) 
  
  • future_fabulators/formalised_decision_making.txt
  • Last modified: 2014-03-04 07:01
  • by maja