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    <h2>Lua, 895</h2> <p>OK, correct now. Can even score 1 as ace.</p> <pre><code>t=table p=ipairs j=arg x={}r={}o={}b=0 function f()local e,s,n,m=0,0,{}for i,v in p(j)do s=s+v e=e+(v+0==1 and 1 or 0)n[i]=v end t.sort(n)for i,v in p(n)do if o[i]and v~=o[i]then m=v&gt;o[i]break end end while e&gt;0 and s&lt;12 do s=s+10 e=e-1 end if s&lt;=21 and s&gt;b or(s==b and(#j&gt;#r or(#j==#r and m)))then b=s for i=1,#r&gt;#j and #r or #j do r[i]=j[i]o[i]=j[i]end t.sort(o)end if s&gt;0 then for i=1,#j do t.insert(x,t.remove(j,i))f()t.insert(j,i,table.remove(x))end end end f()t={" __________ ","| |","|__________|","| /\\ |","| /\\ |","| /\\ |","| /\\ /\\ |","| \\/ /\\ \\/ |","|_\\/____\\/_|"}c={"12224D223","125E226F3","125E4D6F3","127G227G3","127G4D7G3","127G7G7G3","1278HG7G3","1278H8HG3","17G78HG79","178HG78H9"}for i=1,9 do for j,x in p(r)do v=c[tonumber(x)]:sub(i,i):byte()io.write(v&lt;64 and t[v-48]or(t[v-64]:gsub("[/\\]",{["/"]="\\",["\\"]="/"})))end io.write("\n")end </code></pre> <p>Sample output:</p> <pre><code>&gt;lua card.lua 8 9 10 7 __________ __________ | /\ /\ || /\ /\ | | \/ \/ || \/ /\ \/ | | /\ /\ || /\ \/ /\ | | \/ /\ \/ || \/ \/ | | /\ \/ /\ || /\ /\ | | \/ \/ || \/ /\ \/ | | /\ /\ || /\ \/ /\ | |_\/____\/_||_\/____\/_| &gt;lua card.lua 1 4 5 4 7 4 8 __________ __________ __________ __________ __________ | || || || || | | || /\ /\ || /\ /\ || /\ /\ || /\ /\ | | || \/ \/ || \/ \/ || \/ /\ \/ || \/ \/ | | /\ || /\ || || /\ \/ /\ || | | \/ || \/ || || \/ \/ || | | || /\ /\ || /\ /\ || /\ /\ || /\ /\ | | || \/ \/ || \/ \/ || \/ \/ || \/ \/ | |__________||__________||__________||__________||__________| &gt;lua card.lua 1 9 5 5 __________ __________ __________ | || || | | || /\ /\ || /\ /\ | | || \/ \/ || \/ \/ | | /\ || /\ || /\ | | \/ || \/ || \/ | | || /\ /\ || /\ /\ | | || \/ \/ || \/ \/ | |__________||__________||__________| </code></pre>
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    1. POCode Golf: Diamond Blackjack
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    1. COThis doesn't answer the question? It says to output the optimised hand, not just the hand. So it should only print cards 10 and 9
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    2. COI'm agreeing with 9 10, 8+9+10=27...
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    3. COI was thinking in hexadecimal... `0x8 + 0x9 + 0x10 = 0x21`
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