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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| future_fabulators:formalised_decision_making [2014-02-11 07:54] – nik | future_fabulators:formalised_decision_making [2014-03-04 07:01] (current) – maja | ||
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| By Tim Boykett | 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 |
| Some language that we use here: | Some language that we use here: | ||
| Line 60: | Line 60: | ||
| //(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 $(n-1)! \sum\limits_{i=1}^{n-1} (1/i)$, 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 | + | 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 |
| 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: | ||
nik