Expected winning of a player with highest lowest and second highest lowest grouping

The triple integral for the continuous case comes to $$\frac18(1-x)^4 + x^3(1-x) + \frac18x^4$$ Its local maximum is near $x=0.86$. The three terms are for $x=a$, $x=b \text{ or }c$ and $x=d$.
Alex's numbers follow one quartic for $n=1$ to $25$, and a different quartic for $n=26$ to $50$.

The question can be easily answered by a short computer program. Below are provided a graph and a table of the total gain (divided by $6$) for each choice $a_1$ of the first player. The best choice is $a_1=43$ with the total gain $139940\cdot 6$.

enter image description here

The program (in Pascal):

program p3542562;
const
 n=50;
var
 OFi:Text;
 a1,a2,a3,a4,gain:LongInt;
 w:array[1..n]of LongInt;
begin
assign(OFi,'3542562.txt');
Rewrite(OFi);
for a1:=1 to n do begin
 w[a1]:=0;
 for a2:=1 to n-2 do for a3:=a2+1 to n-1 do for a4:=a3+1 to n do begin
  if (a1<>a2) and (a1<>a3) and (a1<>a4) then begin
   if a1>a4 then gain:=a1+a2-a3-a4 else
   if a1<a2 then gain:=a1+a4-a2-a3 else
    gain:=a1+a3-a2-a4;
   if gain>0 then w[a1]:=w[a1]+gain;
  end;
 end;
 writeln(OFi,w[a1]);
end;
Close(OFi);
end.

Below are provided a graph and a table of the total gain (divided by $6$) for each choice $a_1$ of the first player *for the case when the winning team gain is paid by the losing team. The best choice now is $a_1=40$ with the total gain $86710\cdot 6$.

enter image description here

Expected winning of a player with highest lowest and second highest lowest grouping

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