Algorithm to distribute items "evenly"
I ran across this question while researching a similar problem: optimum additions of liquids to reduce stratification. It seems like my solution would be applicable to your situation, as well.

If you want to mix liquids A, B, and C in the proportion 30,20,10 (that is, 30 units of A, 20 units of B, and 10 units of C), you end up with stratification if you add all the A, then all the B, and then all the C. You're better off mixing smaller units. For example, do single-unit additions in the sequence [A,B,A,C,B,A]. That will prevent stratification altogether.

The way I found to do it is to treat it as a kind of merge, using a priority queue. If I create a structure to describe the additions:

MergeItem
    Item, Count, Frequency, Priority


The Frequency is expressed as "one every N". So A, which is added three out of six times, has a frequency of 2 (6/3).

And initialize a heap that initially contains:

(A, 3, 2, 2)
(B, 2, 3, 3)
(C, 1, 6, 6)


Now, I remove the first item from the heap and output it. Then reduce its count by 1 and increase Priority by Frequency and add it back to the heap. The resulting heap is:

(B, 2, 3, 0)
(A, 2, 2, 4)
(C, 1, 6, 6)


Next, remove B from the heap, output and update it, then add back to the heap:

(A, 2, 2, 4)
(C, 1, 6, 6)
(B, 1, 3, 6)


If I continue in that fashion, I get the desired mixture. I use a custom comparer to ensure that when equal Priority items are inserted into the heap, the one with the highest Frequency value (i.e. the least frequent) is ordered first.

I wrote a more complete description of the problem and its solution on my blog, and presented some working C# code that illustrates it. See Evenly distributing items in a list.

Update after comments

I do think my problem is similar to the OP's problem, and therefore that my solution is potentially useful. I apologize for not framing my answer more in the terms of the OP's question.

The first objection, that my solution is using A, B, and C rather than 0, 1, and 2, is easily remedied. It's simply a matter of nomenclature. I find it easier and less confusing to think about and say "two A's" rather than "two 1's". But for purposes of this discussion I have modified my outputs below to use the OP's nomenclature.

Of course my problem deals with the concept of distance. If you want to "spread things out evenly," distance is implied. But, again, it was my failing for not adequately showing how my problem is similar to the OP's problem.

I ran a few tests with the two examples that the OP provided. That is:

[1,1,2,2,3,3]  // which I converted to [0,0,1,1,2,2]
[0,0,0,0,1,1,1,2,2,3]


In my nomenclature those are expressed as [2,2,2] and [4,3,2,1], respectively. That is, in the last example, "4 items of type 0, 3 items of type 1, 2 items of type 2, and 1 item of type 3."

I ran my test program (as described immediately below), and have posted my results. Absent input from the OP, I can't say if my results are similar to, worse than, or better than his. Nor can I compare my results to anybody else's results because nobody else has posted any.

I can say, however, that the algorithm provides a good solution to my problem of eliminating stratification when mixing liquids. And it looks like it provides a reasonable solution to the OP's problem.

For the results shown below, I used the algorithm that I detailed in my blog entry, with the initial priority set to Frequency/2, and the heap comparer modified to favor the more frequent item. The modified code is shown here, with the modified lines commented.

private class HeapItem : IComparable&lt;HeapItem&gt;
{
    public int ItemIndex { get; private set; }
    public int Count { get; set; }
    public double Frequency { get; private set; }
    public double Priority { get; set; }

    public HeapItem(int itemIndex, int count, int totalItems)
    {
        ItemIndex = itemIndex;
        Count = count;
        Frequency = (double)totalItems / Count;
        // ** Modified the initial priority setting.
        Priority = Frequency/2;
    }

    public int CompareTo(HeapItem other)
    {
        if (other == null) return 1;
        var rslt = Priority.CompareTo(other.Priority);
        if (rslt == 0)
        {
            // ** Modified to favor the more frequent item.
            rslt = Frequency.CompareTo(other.Frequency);
        }
        return rslt;
    }
}


Running my test program with the OP's first example, I get:

Counts: 2,2,2
Sequence: 1,0,2,1,0,2
Distances for item type 0: 3,3
Stddev = 0
Distances for item type 1: 3,3
Stddev = 0
Distances for item type 2: 3,3
Stddev = 0


So my algorithm works for the trivial problem of all counts being equal.

For the second problem that the OP posted, I got:

Counts: 4,3,2,1
Sequence: 0,1,2,0,1,3,0,2,1,0
Distances for item type 0: 3,3,3,1
Stddev = 0.866025403784439
Distances for item type 1: 3,4,3
Stddev = 0.471404520791032
Distances for item type 2: 5,5
Stddev = 0
Distances for item type 3: 10
Stddev = 0
Standard dev: 0.866025403784439,0.471404520791032,0,0


I don't see an obvious way to improve on that. It could be rearranged to make the distances for item 0 [2,3,2,3] or some other arrangement of 2 and 3, but that will change the deviations for items 1 and/or 2. I really don't know what "optimum" is in this situation. Is it better to have a larger deviation on the more frequent or on the less frequent items?

Lacking other problems from the OP, I used his descriptions to make up a few of my own. He said in his post:


  A typical list has ~50 items with ~15 different values in varied quantities.


So my two tests were:

[8,7,6,5,5,4,3,3,2,2,2,1,1,1,1]  // 51 items, 15 types
[12,6,5,4,4,3,3,3,2,2,2,1,1]     // 48 items, 13 types


And my results:

Counts: 8,7,6,5,5,4,3,3,2,2,2,1,1,1,1
Sequence: 0,1,2,3,4,5,7,6,0,1,2,8,9,10,4,3,0,1,5,2,0,1,3,4,6,7,14,11,13,12,0,2,5,1,0,3,4,2,8,10,9,1,0,7,6,5,3,4,2,1,0
Distances for item type 0: 8,8,4,10,4,8,8,1
Stddev = 2.82566363886433
Distances for item type 1: 8,8,4,12,8,8,3
Stddev = 2.76272565797339
Distances for item type 2: 8,9,12,6,11,5
Stddev = 2.5
Distances for item type 3: 12,7,13,11,8
Stddev = 2.31516738055804
Distances for item type 4: 10,9,13,11,8
Stddev = 1.72046505340853
Distances for item type 5: 13,14,13,11
Stddev = 1.08972473588517
Distances for item type 6: 17,20,14
Stddev = 2.44948974278318
Distances for item type 7: 19,18,14
Stddev = 2.16024689946929
Distances for item type 8: 27,24
Stddev = 1.5
Distances for item type 9: 28,23
Stddev = 2.5
Distances for item type 10: 26,25
Stddev = 0.5
Distances for item type 11: 51
Stddev = 0
Distances for item type 12: 51
Stddev = 0
Distances for item type 13: 51
Stddev = 0
Distances for item type 14: 51
Stddev = 0


And for the second example:

Counts: 12,6,5,4,4,3,3,3,2,2,2,1,1
Sequence: 0,1,2,0,3,4,7,5,6,0,1,8,9,10,0,2,0,3,4,1,0,2,6,7,5,12,11,0,1,0,3,4,2,0,1,10,8,9,0,7,5,6,0,
4,3,2,1,0
Distances for item type 0: 3,6,5,2,4,7,2,4,5,4,5,1
Stddev = 1.68325082306035
Distances for item type 1: 9,9,9,6,12,3
Stddev = 2.82842712474619
Distances for item type 2: 13,6,11,13,5
Stddev = 3.44093010681705
Distances for item type 3: 13,13,14,8
Stddev = 2.34520787991171
Distances for item type 4: 13,13,12,10
Stddev = 1.22474487139159
Distances for item type 5: 17,16,15
Stddev = 0.816496580927726
Distances for item type 6: 14,19,15
Stddev = 2.16024689946929
Distances for item type 7: 17,16,15
Stddev = 0.816496580927726
Distances for item type 8: 25,23
Stddev = 1
Distances for item type 9: 25,23
Stddev = 1
Distances for item type 10: 22,26
Stddev = 2
Distances for item type 11: 48
Stddev = 0
Distances for item type 12: 48
Stddev = 0