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future_fabulators:formalised_decision_making [2013-10-17 22:41] – nik | future_fabulators:formalised_decision_making [2014-02-11 07:54] – nik | ||
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==== Formalised Decision Making ==== | ==== Formalised Decision Making ==== | ||
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+ | 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 | + | Some calculations show that the number of ways of doing this is $(n-1)! |
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)! |
- | 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...) | ||