Rolling Quine Dice

Question

Create a program that outputs itself.

However, if the source code is repeated n times (meaning to concatenate a copy of the source code to the end n-1 times), then there should be 1/n probability outputting the original source code, a 1/n probability of outputting the source code repeated twice, a 1/n probability of outputting the source code repeated three times, ..., and a 1/n probability of outputting the source code n times.

For example, if your program is foobar, then it should always output exactly foobar. However, if you run foobarfoobarfoobarfoobar, then there should be a 1⁄4 chance each of outputting foobar, foobarfoobar, foobarfoobarfoobar and foobarfoobarfoobarfoobar.

• The distribution of each possible output should be equal
• In addition to standard I/O methods applying and standard loopholes forbidden, standard quine rules apply (can't access its own source, etc.)
• This is code golf so shortest answer in bytes wins

• wastl: Is pseudorandom good enough?
• JMigst: @wastl Yep, that's alright
• Veskah: Your definition is slightly off. If you repeat it 1 time (i.e. foobarfoobar) then it'd only print foobar
• Etheryte: @Veskah foobarfoobar is foobar repeated two times, not once.
• Veskah: @Nit You start with the source. If you use it again, you have only repeated it once but have two instances.

Answer 1

Perl 5, (削除) 82 (削除ここまで) 80 bytes

$s=q($s=q(%s);printf"$s\n+1#",$s for 0..rand);printf"$s\n+1#",$s for 0..rand
+1#

Try it online! or Test suite

Answer 2

Jelly, (削除) 24 (削除ここまで) 22 bytes

"Ṿ;ẋŻɼLX¤¤μ"Ṿ;ẋŻɼLX¤¤μ

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Try it x4!

"Ṿ;ẋŻɼLX¤¤μ"Ṿ;ẋŻɼLX¤¤μ
"Ṿ;ẋŻɼLX¤¤μ" String literal: Ṿ;ẋŻɼLX¤¤μ
 Ṿ Uneval. Return "Ṿ;ẋŻɼLX¤¤μ" (with quotes)
 ; Append the initial string. Yields the source code. 
 ɼ Apply the preceding link to the register and store the 
 result in the register.
 Ż Prepend 0. 
 Each time Żɼ is used when the source code is repeated
 the register's length increases by one. 
 We can't use ‘ because it closes string literals.
 L Length. Returns the number of times the source code has
 been repeated up till now.
 X Random element. If ɼ results in n, X returns a random 
 integer between 1 and n.
 ¤ Combines ŻɼLX as a nilad. 
 ẋ Repeat the source code a random number of times between
 1 and n.
 ¤ Close everything from the initial string literal as a
 nilad. 
 μ Start a new monadic chain. The register gets updated
 once for time the code is repeated but only the final
 repetition will result in output.

Answer 3

05AB1E, 32 bytes

"×ばつΩ"×ばつΩ

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Explanation

"×ばつΩ" # push this string
 34ç # push a "-character
 ì # prepend it to the string
 DJ # duplicate and join the copy to the string
 1⁄4 # increment the counter
 32ô # split the string into pieces of size 32
 н # take the first one
 ×ばつ # repeat it for each in [1 ... counter]
 Ω # pick one at random

• Shaggy: Could you add an explanation when you have a chance?
• Emigna: @Shaggy: Thanks for reminding me :)

Answer 4

Gol><>, 21 bytes

:QoaonC|P\Sx*F2ssS"
0

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Explanation:

:Q | If the top of stack is 0, skip to next |
 Top of stack is implicitly 0
 P Increment top of stack
 \ Redirect pointer down]
0:QoaonC|P\Sx*F2ssS" Increment for each copy of the source code
0:QoaonC|P\Sx*F2ssS"
 \ Redirect back to the right
 Sx* Multiply the number of copies of the source code by a random number between 0 and 1
 F Repeat that many times
 2ss Push a double quote
 S" Print the source code minus the quote, newline and 0
 :Q If top of stack is not 0
 oaonC Print the quote, a newline and a 0 to complete the source code and continue the loop
 \Sx* Error on the * for some reason

Answer 5

Alice, 35 bytes

"d3a*h-&;adddd12h&}Uh*t&w.odt,k@
 !

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Explanation

"

Like in many quines in 2D languages, this starts with a " that wraps around to itself and pushes the entire first line except the " itself.

d3a*h-&;

Adding one or more additional copies of the source code will place some implicit spaces at the end of the string literal. To make this actually a quine, we truncate the stack at 31 characters.

addd

Push a newline, then the stack height three times. The values pushed as the stack height are 32 (the space in the second line), 33 (the ! in the second line), and 34 (the initial ").

d

Push the stack height again, this time as the length of the original source code (35).

1

Initialize a counter at 1. This will count the number of times the source code is repeated.

2h&}

Turn right three times in place (i.e., turn left). Each additional repetition of the source code will contribute an h in the same column as this }, thus incrementing the counter. When the IP returns to the }, turn right again to continue in the same direction.

Uh

Take a uniform random number from 0 to n-1, then add 1 to get the number of times to output the original source.

*t&w

Multiply by the previously pushed stack height (code length), then repeat the following that many times by pushing a return address that many times minus one.

.o

Output the top of the stack without destroying it.

dt,

Move the bottom stack entry to the top.

k@

Repeat, then terminate after the loop is finished.

Answer 6

JavaScript (Node.js), 62 bytes

(f=a=>b=>b?f(a+.5):`(f=${f})(1)`.repeat(1+Math.random()*a))(1)

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• Nitrodon: Either I'm very unlucky, or this can't print more than two copies of the original source code.
• l4m2: @Nitrodon Thanks, fixed

Answer 7

Charcoal, 58 bytes

≔ ́θ ́⎚ ×ばつ ́⊕ ́‽ ́L ́⊞ ́O ́υ ́ω ́+ ́⪫ ́+ ́ ́ ́≔ ́θ ́ ́ ́ ́ ×ばつ⊕‽L⊞Oυω+⪫+ ́≔θ ́ ́θ

Try it online! No verbose version because the deverbosifier currently chokes on " ́". Mostly based on the Charcoal quine from Golf you a quine for great good!. Explanation:

≔ ́θ ́⎚ ×ばつ ́⊕ ́‽ ́L ́⊞ ́O ́υ ́ω ́+ ́⪫ ́+ ́ ́ ́≔ ́θ ́ ́ ́ ́ ́θθ

Assign the literal string ×ばつ⊕‽L⊞Oυω+⪫+ ́≔θ ́ ́θ to θ.

⎚

Clear any previous output, so that only the last output takes effect.

×ばつ⊕‽L⊞Oυω

Push the empty string to the predefined array. This makes the array length equal to the number of repetitions processed so far, so take its length, take a random number the implicit exclusive range, add 1, and repeat the following string that many times.

+⪫+ ́≔θ ́ ́θ

Prepend the literal string ≔ to θ, then insert literal ́s between each character, then suffix another copy of θ.

Answer 8

Python 3.8 (pre-release), (削除) 118 (削除ここまで) 116 bytes

for i in(1,c:=0):
 exec(s:="from random import*;c+=i or print('for i in(1,c:=0):\\n exec(s:=%r)#'%s*randint(1,c))")#

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