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

Question

The following is an interview question.

Question: Given 4 players $A,B,C,D$ and a fair $50$-sided dice, assume that we do not allow repeated score (i.e. next player cannot get the same score as all previous players. Otherwise, the player roll again).

We group players with the highest and lowest scores together and second and third highest together. The winning team is the group that has the larger sum which will win the difference between team score.

For example, say $A,B$ and $C,D$ form 2 groups and $A=1,B=7,C=3,D=2,$ then $A,B$ groups wins with $8-5=3$ units.

What number should player $A$ hopes to get to maximize his expected winning?

I totally have no idea how to start this question at all.

Answer 1

If player $A$ scores $50$ and his team mate scores $1$, their team scores $51$. So the second and third highest players should not score more than $24 + 25 = 49$, thus allowing for $(25-1) \times (24-1) = 24 \times 23 = 552$ possibilities.

If player $A$ scores $50$ and his team mate scores $2$, their team scores $52$. So the second and third highest players should not score more than $25 + 26 = 49$, thus allowing for $25 \times 24 = 600$ possibilities.

Continuing in this way, if player $A$ scores $50$ and his teammate scores $46$, their team scores $50 + 46 = 96$, and then the second and third highest-scoring player can score at most $48 + 47 = 95$, thus allowing for $$47 \times 46$ possibilities.

Answer 2

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 $139940\cdot 6$.

110400
101476
93150
85441
78365
71935
66161
61050
56606
52830
49720
47271
45475
44321
43795
43880
44556
45800
47586
49885
52665
55891
59525
63526
67850
72450
77280
82295
87447
92685
97955
103200
108360
113372
118170
122685
126845
130575
133797
136430
138390
139590
139940
139347
137715
134945
130935
125580
118772
110400

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 $86710\cdot 6$.

0
-17296
-32430
-45494
-56580
-65780
-73186
-78890
-82984
-85560
-86710
-86526
-85100
-82524
-78890
-74290
-68816
-62560
-55614
-48070
-40020
-31556
-22770
-13754
-4600
4600
13754
22770
31556
40020
48070
55614
62560
68816
74290
78890
82524
85100
86526
86710
85560
82984
78890
73186
65780
56580
45494
32430
17296
0

Answer 3

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 $50$.