Top 10 contenders:

[![enter image description here][1]][1]

My approach using Ruby:

    num = [1,0,2,0,1,0,2,0,1,0,2,0,1,0,2,0,1]
    opr = [3,4,5,6]
    
    #3->+
    #4->-
    #5->/
    #6->*
    
    plc = [1,3,5,7,9,11,13,15]
    
    $i=0
    
    results = []
    express = []
    
    while $i < 100000 do
        opr = opr.shuffle
    
        tempRand=plc.shuffle.take(4)
    
        num[tempRand[0].to_i] = opr[0]
        num[tempRand[1].to_i] = opr[1]
        num[tempRand[2].to_i] = opr[2]
        num[tempRand[3].to_i] = opr[3]
    
        tempString = ''
    
        num.each do |e|
    
            if e.to_i == 0
                tempString << ''
            elsif e.to_i == 3
                tempString << '+'
            elsif e.to_i == 4
                tempString << '-'
            elsif e.to_i == 5
                tempString << '/'
            elsif e.to_i == 6
                tempString << '*'
            else
                tempString << e.to_s
            end
        end
    
        results << eval(tempString)
        express << tempString
    
        num[tempRand[0].to_i] = 0
        num[tempRand[1].to_i] = 0
        num[tempRand[2].to_i] = 0
        num[tempRand[3].to_i] = 0
    
        $i += 1
    end
    
    print_me = Hash[results.zip(express)]
    
    puts print_me.sort.uniq.reverse

<sub>(try it [here](https://codepad.remoteinterview.io/XEIHTRQDHB))</sub>

Basically just randomly swap the 0s (the blank spaces) with operators and then evaluate the result without 0s. It's not the best way to do it, but over a 100,000 iterations the max is 44944.


  [1]: https://i.sstatic.net/wptnA.png