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  1. PO
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    <p>Just off the top of my head, what if you compare the % occurrence vs the % if all items had equal number of occurences</p> <p>In your example above<br> John, John, Jon, Jonny<br> 50% John<br> 25% Jon<br> 25% Jonny<br> 33.3% Normal? (I'm making up a word because I don't know what to call this. 3 items: 100%/3)<br> John's score = 50% - 33.3% = 16.7% </p> <p>John, John, Jon, Jon<br> 50% John<br> 50% Jon<br> 50% Normal (2 items, 100%/2)<br> John's score = 50% - 50% = 0% </p> <p>If you had [John, John, John, Jon, Jon] then John's score would be 60%-50% = 10% which is lower than the first case, but higher than the 2nd (hopefully that's the desired result, otherwise you'll need to clarify more what the desired results should be)</p> <p>In your first case [John, John, John, John, Jon] you'd get 80%-50% = 30%<br> For [John, John, John, John, Jon, Jonny] you'd get 66.6%-33.3% = 33.3%<br> That may or may not be what you want.</p> <p>Where the above might factor in more is if you had John*97+Jon+Jonny+Johnny, that would give you 97%-25% = 72%, but John*99+Jon would only give you a score of 99-50% = 49%</p> <p>You'd need to figure out how you want to handle the degenerate case of them all being the same, otherwise you'd get a score of 0% for that which is probably not what you want.</p> <p><strong>EDIT</strong> (okay I made lots of edits, but this one isn't just more examples :p)<br> To normalize the results, take the score as calculated above divide by the limit of max possible score given the number of different values. (Okay, that sounds way more complicated than it needs to, example time)</p> <p>Example:<br> [John, John, Jon, Jonny] 50% - 33.3% = 16.7%. That's the previous score, but with 3 items the upper limit of your score would be 100%-33.3% = 66.6%, so if we take that into account, we have 16.7/66.6 = 25%</p> <p>[John, John, Jon, Jon] gives (50-50) /50 = 0%<br> [John, John, John, Jon, Jon] gives (60-50) /50 = 20%<br> [John, John, John, John, Jon] gives (80-50)/50 = 60%<br> [John, John, John, John, Jon, Jonny] gives (66.6-33.3)/(66.6)= 50%<br> [John*97, Jon, Jonny, Johnny] gives (97-25)/75 = 96%<br> [John*99, Jon] gives (99-50)/50 = 98% </p>
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    1. POProbability algorithm: Finding probable correct item in a list (e.g John, John, Jon)
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    1. PTAnswer
    1. USDavy8
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    1. CONice thinking! I didn't have the number of unique items in the list. But I could calculate it. Until now I just had the frequency of the 2 most common elements and the list count.
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      1. USAndrew White
    2. COThe degenerate case of all elements being the same should return 100% (e.g. it is 100% likely the correct value). The example: John*97+Jon+Jonny+Johnny = 72% & John*99+Jon = 49% is not what I want. The likelihood of John being the correct item is higher in the second list. I think the answer is to combine the percentage the Mode value (had to look up Wiki statistics for that term) is of the total plus/multiplied by the percentage it is above the Normal.
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      1. USAndrew White
    3. CO@Andrew I meant to make the edit above sooner, but Stackoverflow went down for maintenance as I was typing it up, but I think that should give you what you need.
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      1. USDavy8
 

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