Equalising numbers

I'm sure some of the maths experts on here will provide a solution that absolutely rocks, but in the meantime I think a generic function to calculate each number in the series would be:

=ROUNDDOWN(AVERAGE(Numbers),0) +(MOD(AVERAGE(Numbers),1)>=(n/COUNT(Numbers)))

Where n is the position of the 'equalised' number in the equalised set.
Due to Double precision, 0.33333' does not necessarily equal 1/3, 0.66666' to 2/3 etc., so revise the function for Excel to this:

=ROUNDDOWN(AVERAGE(Numbers),0) +((1/1000+MOD(AVERAGE(Numbers),1))>=(n/COUNT(Numbers)))

Where the 1/1000 is a small fraction to ensure that the >= is correctly met. Obviously if your sets became much larger then this would fall short.

Examples up to 7 random numbers in a set in the attached example. Numeric order in the equalised set and negative averages not taken into consideration. For negative averages I think a function (which no doubt can be simplified) would be:

=TRUNC(AVERAGE(Numbers))+SIGN(AVERAGE(Numbers))*((1/1000+MOD(ABS(AVERAGE(Numbers)),1))>=(n/COUNT(Numbers)))