CJam, 14 bytes

{{E*`mR}`mR}E*

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Explanation

Each character appears exactly twice, so the probabilities of the characters should all be the same.

{           e# Repeat this block 14 times.
  {E*`mR}   e# Push this (nonsensical) block.
  `         e# Stringify it, giving the string "{E*`mR}", which contains each of the
            e# seven characters once.
  mR        e# Select one of the characters at random.
}E*

Jelly, 13 bytes

“;⁾vṾWṁ$X€”vṾ

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How it works

“;⁾vṾWṁ$X€”vṾ  Main link. No arguments.

“;⁾vṾWṁ$X€”    Set the argument and return value to the string s := ';⁾vṾWṁ$X€'.
            Ṿ  Uneval; construct a string representation of s.
               This yields r := '“;⁾vṾWṁ$X€”'.
           v   Dyadic eval; evaluate s with argument r.


 ;⁾vṾWṁ$X€     Evaluated link (s). Argument: r

  ⁾vṾ          Yield 'vṾ'.
 ;             Concatenate r with 'vṾ'.
               This yields t := '“;⁾vṾWṁ$X€”vṾ', i.e., the original source code.
       $       Combine the previous two links into a monadic chain.
     W           Wrap; yield ['“;⁾vṾWṁ$X€”vṾ'].
      ṁ          Mold; repeat ['“;⁾vṾWṁ$X€”vṾ'] once for each charcter in t.
        X€     Random each; select a character, uniformly at random, of each
               of the 13 repetitions of t.

Japt, 22 bytes

"+Q ²£ZgMqB"+Q ²£ZgMqB

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How it works

"+Q ²£ZgMqB"+Q ²£ZgMqB  // Implicit: B = 11
"+Q ²£ZgMqB"            // Take this string.
            +Q          // Append a quotation mark.
               ²        // Double the result.
                £       // Replace each char in the result Z with
                 Zg     //  the char in Z at index
                   MqB  //   random integer in the range [0, 11).
                        // Implicit: output last expression

Pyth, 16 Bytes

smO+N"m1NOs+6"16

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Contains each char twice therefore the probability is the same as if each was only in there once.

smO+N"m1NOs+6"16     #
   +N"m1NOs+6"       # Add a quotation mark to the string: "m1NOs+6
  O                  # random choice from the string
 m            16     # do this 16 times.
s                    # join the list into a string

PHP, 71 140 110 124 140 120 bytes

for($i=2*$n=strlen($s='for($i=2*$n=strlen($s=.chr(39));$i--;)echo$s[rand(0,$n-1)];'.chr(39));$i--;)echo$s[rand(0,$n-1)];

run with php -d

1. creates a string containing the code without the quotation marks
2. and concatenates the quotation mark once using ord
   (same probability as if I would double the string and add two quotes);
3. then loops over twice the length of the string to get random characters from it.

Can possibly be golfed further, but my attempts on eval where futile so far.
I will probably not go deeper here.

Jelly, 44 bytes

I hope I have interpreted all the rules correctly (I'm not quite sure what the "carry payload" thing is in meta or if it's even relevant to this challenge).

“ẋ2;8220Ọ;8221ỌXµ44СḊ”ẋ2;8220Ọ;8221ỌXµ44СḊ

Test it out at TryItOnline

This constructs a string from which to choose characters. The initial string has all the character used except the open and close quotes. It then doubles that string and concatenates one of each of the open and close quotes from ordinals (hence the need to double the other characters). Lastly it repeatedly picks random characters from the composed string to the length of the program.

Pyke, 35 bytes

35F\";"\\+3322 5**F;H)s"\"2*+2* H)s

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To check: remove the final H and the resulting string contains the right number of each character (with the extra H)

This does NOT use a generalised quine or in fact a quine at all. It relies on being able to create a string containing all the characters in the source. It should be able to do it for any code but each character logarithmically increases code size. The only number of times a character is allowed in the source is 2 or 7

Python 2, 88 bytes

s='''from random import*; print "".join(choice(s) for c in "s='"+s+"';exec s")''';exec s

All actual merit in getting this far goes to mbomb007 - thanks for your help (and the pointer about backslashes)

Ruby, 81 bytes

s="s=%p;a=s%%s;81.times{$><<a[rand(a.size)]}";a=s%s;81.times{$><<a[rand(a.size)]}

I didn't realize that you had to randomly select every time; I thought a shuffle would do the trick. This can probably be golfed, but it's the shortest I could get it.

Standard Ruby quine with a few modifications so it prints out the shuffled string. I'm sad because it took like fifteen minutes to figure out the formatting quirks before I realized that I was subconsciously stealing it anyway.

I think the string shuffling can be shortened but I don't know how; I might also be able to finagle the formatting into being shorter once I put some thought into it. Help would be appreciated.

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Python 3, 134 132 bytes

I use every character in my source code within the string the correct number of times, then multiply the string by two to include itself. The program prints a random character from that string for each character in the code (the length is hard-coded.)

from random import*
for i in[0]*134:print(choice("""from random import*
for i in[0]*134:print(choice(""*""2""),end='')"""*2),end='')

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I avoided backslashes like the plague. As soon as the code contains \n or \", you have a problem, because the string doesn't contain backslashes yet, so you have to add those also, but in a separate string multiplied by a higher number, because it takes two backslashes to represent one (\\).

Example output:

i(tc*"3]i o''r=,,,h34t"r
ri"](fco t)niap)t "it2nc0o  npoi3'"nto(*4 i(ido' r*4f"oi]d rm ,i"eif)m"d
m emi
dfr n*p 3*(i""r1d"dr menc hio'

I gotta say, it reminds me of FlogScript.

PowerShell v2+, 175 bytes

$d='$d={0}{1}{0}{2}-join(0..174|%{3}[char[]]($d-f [char]39,$d,"`n",[char]123,[char]125)|Random{4})'
-join(0..174|%{[char[]]($d-f [char]39,$d,"`n",[char]123,[char]125)|Random})

Quines in PowerShell suck, because the string replacement delimiters {} also denote loops and whatnot, so you need to use a bunch of chars in the -f operator, which bloats the code.

Similar-ish to my Quine on Every Line answer. Basically we loop from 0 to 174 and each iteration re-calculate the quine $d, cast it as a char-array, and pull out a Random element chosen uniformly from the input. By definition, this gives probability (occurrences in source) / (length of source) as required. Those characters are encapsulated in parens and -joined together back into a string.

Example

PS C:\Tools\Scripts\golfing> .\faux-souce-code.ps1
}}[${hr[`ini}}] [5i,=[]0,j2($=n4(dm]jh]jc]]7
}..j"rnj9|fn,4r]{9]('["jdh0}$rd,-f,a.c"}{h1 ]5d,),0n5|nh(]73a9da4aRi[5}a}430}}rd$,$r)-hhr%or79-R.R-`'r'aa|=1f0][|[{7}do1]$ja0 rd{h

(Yes, that's a newline in the output -- when a string containing a newline is char-array'd, the `n is treated as a character, since a char-array is just an array of byte codes, so it also has a 1/175th chance of being selected.)

Dyalog APL, 20 bytes

f←{(,⎕CR'f')[?⍴⍨20]}

f←{...} define f as

(,⎕CR'f') listified (,) Character (table) Representation (⎕CR) of f ('f')

[?⍴⍨20] indexed with ([...]) random-up-to (?) repeat-itself-times (⍴⍨) of twenty

Let us run it (with a dummy argument) a few times:

      f⍬
)0'0](⍨(([],}⎕))⎕'f2
      f⍬
{'['}f[←R[)2←?}⍨]}C2
      f⍬
,,⍨←?2}⍴?'⍨}C,'{⎕(C0

Fine, but is the distribution correct? Let us run it on 10,000 dummy arguments and see how many times each character occurs:

      {⍺ , 1E¯4× ⍴⍵}⌸ ∊ f¨ ⍳1E4
C 0.9952
⎕ 0.9996
' 1.9777
f 2.004 
← 1.0018
⍨ 1.0173
0 1.0213
] 1.0049
[ 0.9988
2 0.9943
{ 0.9895
) 1.0093
R 1.0054
, 1.0029
? 0.9943
} 0.9861
⍴ 1.0032
( 0.9944

Clearly, f and ' occur twice as often as the other characters, just like in the original source code.

How did we do it?

{⍺ , 1E¯4× ⍴⍵}⌸ ∊ f¨ ⍳1E4`

⍳1E4 generates the first 10,000 integers

f¨ runs f on each of those numbers

∊ flattens all the pseudo-quines into a single 200,000-character string

⌸ is a higher-order function which for each unique character in the right side data, feeds the left-side function the unique element as left-argument and the indices where that character occurs as right-argument. The left-side function is

{⍺ , 1E¯4× ⍴⍵}

⍺ left-argument, i.e. the unique character

, followed by

1E¯4× 1×10⁻⁴ times

⍴⍵ the shape of the right-argument (the occurrence indices), i.e.how many times it occurs

Finally, ⌸ puts it all together in a table.

C, 60 bytes

l,i;f(char*s){l=i=strlen(s);while(i--)putchar(s[rand()%l]);}

While for counting characters my solution needed 86:

c[256];i;f(char*s){i=256;while(*s)c[*s++]++;while(--i)while(c[i]?c[i]--:0)putchar(i);}