What is the probability that a natural number is a sum of two squares?

A simple numerical experiment shall confirm the answer given by zhoraster: the latter is only deviant from mine by a factor $2$. We define an initial segment of the naturals with length $N$ and count all sums of two squares in that segment. The accompanying program is in Pascal:

program kwadraat;

procedure test(N : integer);
type
  data = record
  b : boolean; i,j : integer;
end;
var
  i,j,k,t : integer;
  rij : array of data;
begin
  t := 0;
  SetLength(rij,N);
  for k := 0 to N-1 do
  begin
    rij[k].b := false;
  end;
  i := 0;
  while true do
  begin
    if sqr(i) > N-1 then Break;
    j := i;
    while true do
    begin
      k := sqr(i)+sqr(j);
      if k > N-1 then Break;
    { Writeln(k,' = ',i,'^2 + ',j,'^2'); }
      rij[k].i := i; rij[k].j := j;
      rij[k].b := true;
      j := j + 1;
      t := t + 1;
    end;
    i := i + 1;
  end;
  if N < 100 then
  for k := 0 to N-1 do
  begin
    if rij[k].b then
      Writeln(k,' = ',rij[k].i,'^2 + ',rij[k].j,'^2');
  end;
  Writeln(t/N,' ->',Pi/8);
end;

begin
  test(10);
  test(1000);
  test(100000);
  test(10000000);
end.

Output (details for $N=10$ only):

0 = 0^2 + 0^2
1 = 0^2 + 1^2
2 = 1^2 + 1^2
4 = 0^2 + 2^2
5 = 1^2 + 2^2
8 = 2^2 + 2^2
9 = 0^2 + 3^2
 7.00000000000000E-0001 -> 3.92699081698724E-0001
 4.19000000000000E-0001 -> 3.92699081698724E-0001
 3.95420000000000E-0001 -> 3.92699081698724E-0001
 3.92969900000000E-0001 -> 3.92699081698724E-0001

Note that the results converge to $\;\pi/8$ , quite in agreement with the argument given by zhoraster, provided though that $i^2 + j^2$ and $j^2 + i^2$ give a double count of $\,\pi/4\,$ which must be halved.

EDIT. Question & Answer is related to : Double Think about Numerosity .

BONUS. In one of the comments with the answer by zhoraster, VividD has been asking for a variant of the original question for the sum of three/four squares. Minor modification of the above program gives the following output for the three squares case. It is seen that some numbers can be written as a sum of three squares in more than one way. Therefore two cases shall be distinguished: with or without these duplicates. Details again for $N=10$ :

0 = 0^2 + 0^2 + 0^2
1 = 0^2 + 0^2 + 1^2
4 = 0^2 + 0^2 + 2^2
9 = 0^2 + 0^2 + 3^2
2 = 0^2 + 1^2 + 1^2
5 = 0^2 + 1^2 + 2^2
8 = 0^2 + 2^2 + 2^2
3 = 1^2 + 1^2 + 1^2
6 = 1^2 + 1^2 + 2^2
9 = 1^2 + 2^2 + 2^2
    with duplicates = 10/10
 without duplicates = 9/10
    with duplicates = 3254/1000
 without duplicates = 835/1000
    with duplicates = 2807201/100000
 without duplicates = 83336/100000
    with duplicates = 87741031/1000000
 without duplicates = 833336/1000000

If the duplicates are counted, then the results are seen to diverge $\to \infty$ .
It is conjectured that, without the duplicates, the results converge to : $5/6$ (Michael Lugo).
Any takers to prove the latter statement?