BrowserUk,
I have skim read the referenced thread and these are apparently random numbers which implies a normal distribution as many values above the mean as below the mean. Ignoring the requirement from the other thread about not having duplicates in any of the groups, I would apply a similar technique as I did in Re: Average Price Algorithm. Essentially:
- Determine the average of the entire list
- Start filling buckets of 6 picking the value in the list that brings the average of the group closest to the target average without going over the max for the group
As it turns out, the average * 6 = 851.71. This allows a little wiggle room. If there are left overs, you can spend some time swapping things around but also limit how many total iterations or time you spend swapping. I would code this up if I had time but I don't tonight. The code from the referenced node should make it clear the intent though.
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