Blue duck, red duck, gray duck

Question

So, last week I posted a challenge to play Duck, Duck, Goose. This lead to a number of Minnesotans commenting about their regional 'Gray duck' variation.

So here's the rules:

Using this list of colours:

Red
Orange
Yellow
Green
Blue
Indigo
Violet
Gray

Write a program to follow these rules:

1. Select one of these colours, and prepend it to the word 'duck' and print the result to a new line.
2. If the colour was not 'Gray', repeat step 1.
3. If the colour was 'Gray', end your program.

Rules which must be followed:

• The program should not consistently print the same number of lines.
• It can begin on 'Gray duck', but should not do consistently.
• There should be exactly one duck on each line and no empty lines are output.
• There should be at least one space between a colour and a duck.
• White space before and after the significant output does not matter.
• The case of the output doesn't matter.
• Colours can be repeated.
• The output doesn't have to contain every colour every time, but it must be possible that your code will output every colour.
• No colours outside the above array can be included.
• Either grey or gray are acceptable in your answer.
• Colours should not consistently be in the same order.
• Aim for the shortest program. Smallest number of bytes wins.

Example output:

Green duck
Orange duck
Yellow duck
Indigo duck
Yellow duck 
Gray duck

Thanks to @Mike Hill for first alerting me to this variation.

Answer 1

05AB1E, 42 40 bytes

Saved 2 bytes thanks to Erik the Outgolfer

[“ëßigo°¯†¾›ÈŠÛˆ¨‡—°Íolet“#7ÝΩ©è'Ðœðý,®#

Try it online!

Answer 2

LuaLaTeX, 220 211 characters

command:

lualatex -interaction nonstopmode

Not the shortest, but the fanciest. Based on @skillmon's solution

\RequirePackage{tikzducks}\newcount\b\let~\def~\0{red}~\1{orange}~\2{yellow}~\3{green}~\4{blue}~\5{cyan}~\6{violet}~\7{gray}\loop\b\uniformdeviate8\tikz\duck[body=\csname\the\b\endcsname]; \ifnum\b<7\repeat\stop

Answer 3

6502 machine code (C64), 124 bytes

00 C0 AD 12 D0 85 02 A2 17 8E 18 D0 A5 02 F0 03 0A 90 02 49 1D 85 02 A8 CA 10
02 A2 2F BD 42 C0 D0 F6 88 D0 F3 86 FB E8 BD 42 C0 F0 06 20 16 E7 E8 D0 F5 AA
BD 73 C0 F0 06 20 16 E7 E8 D0 F5 A6 FB D0 C9 60 00 C7 52 45 59 00 D2 45 44 00
CF 52 41 4E 47 45 00 D9 45 4C 4C 4F 57 00 C7 52 45 45 4E 00 C2 4C 55 45 00 C9
4E 44 49 47 4F 00 D6 49 4F 4C 45 54 00 20 44 55 43 4B 0D 00

Online demo -- Usage: SYS49152.

---

Explanation (commented disassembly):

         00 C0       .WORD $C000        ; load address
.C:c000  AD 12 D0    LDA $D012          ; current rasterline as seed
.C:c003  85 02       STA $02            ; to "random" value
.C:c005  A2 17       LDX #$17           ; cfg for upper/lower, also use as
.C:c007  8E 18 D0    STX $D018          ;    initial index into colors array
.C:c00a   .loop:
.C:c00a  A5 02       LDA $02            ; load current random val
.C:c00c  F0 03       BEQ .doEor         ; zero -> feedback
.C:c00e  0A          ASL A              ; shift left
.C:c00f  90 02       BCC .noEor         ; bit was shifted out -> no feedback
.C:c011   .doEor:
.C:c011  49 1D       EOR #$1D
.C:c013   .noEor:
.C:c013  85 02       STA $02            ; store new random val
.C:c015  A8          TAY                ; use as counter for next color string
.C:c016   .findloop:
.C:c016  CA          DEX                ; next char pos in colors (backwards)
.C:c017  10 02       BPL .xok           ; if negative ...
.C:c019  A2 2F       LDX #$2F           ;    load length of colors - 1
.C:c01b   .xok:
.C:c01b  BD 42 C0    LDA .colors,X      ; load character from colors
.C:c01e  D0 F6       BNE .findloop      ; not zero, try next character
.C:c020  88          DEY                ; decrement random counter
.C:c021  D0 F3       BNE .findloop      ; not zero, continue searching
.C:c023  86 FB       STX $FB            ; save character position
.C:c025  E8          INX                ; increment to start of color
.C:c026   .outloop:
.C:c026  BD 42 C0    LDA .colors,X      ; output loop for color string
.C:c029  F0 06       BEQ .duckout
.C:c02b  20 16 E7    JSR $E716
.C:c02e  E8          INX
.C:c02f  D0 F5       BNE .outloop
.C:c031   .duckout:
.C:c031  AA          TAX                ; A is now 0, use as char pos for duck
.C:c032   .duckoutloop:
.C:c032  BD 73 C0    LDA .duck,X        ; output loop for duck string
.C:c035  F0 06       BEQ .outdone
.C:c037  20 16 E7    JSR $E716
.C:c03a  E8          INX
.C:c03b  D0 F5       BNE .duckoutloop
.C:c03d   .outdone:
.C:c03d  A6 FB       LDX $FB            ; load saved character position
.C:c03f  D0 C9       BNE .loop          ; not zero -> continue main loop
.C:c041  60          RTS                ; zero was start of "Grey" -> done
.C:c042   .colors:
.C:c042  00 c7 52 45    .BYTE 0, "Gre"
.C:c046  59 00 d2 45    .BYTE "y", 0, "Re"
.C:c04a  44 00 cf 52    .BYTE "d", 0, "Or"
.C:c04e  41 4e 47 45    .BYTE "ange"
.C:c052  00 d9 45 4c    .BYTE 0, "Yel"
.C:c056  4c 4f 57 00    .BYTE "low", 0
.C:c05a  c7 52 45 45    .BYTE "Gree"
.C:c05e  4e 00 c2 4c    .BYTE "n", 0, "Bl"
.C:c062  55 45 00 c9    .BYTE "ue", 0, "I"
.C:c066  4e 44 49 47    .BYTE "ndig"
.C:c06a  4f 00 d6 49    .BYTE "o", 0, "Vi"
.C:c06e  4f 4c 45 54    .BYTE "olet"
.C:c072  00             .BYTE 0
.C:c073   .duck:
.C:c073  20 44 55 43    .BYTE " duc"
.C:c077  4b 0d 00       .BYTE "k", $d, 0

Answer 4

Perl 5, 79 bytes

say$_=(Grey,Red,Orange,Yellow,Green,Blue,Indigo,Violet)[rand 8]." duck"until/y/

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Answer 5

Taxi, 1995 bytes

Go to Heisenberg's:w 1 r 3 r 1 l.[a]Pickup a passenger going to Divide and Conquer.8 is waiting at Starchild Numerology.8 is waiting at Starchild Numerology.Go to Starchild Numerology:n 1 l 3 l 1 l 3 l.Pickup a passenger going to Divide and Conquer.Pickup a passenger going to Multiplication Station.Go to Divide and Conquer:w 1 r 3 r 1 r 2 r 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:e 1 l 1 l 2 l.Pickup a passenger going to What's The Difference.Pickup a passenger going to Trunkers.Go to Zoom Zoom:n.Go to Trunkers:w 3 l.Pickup a passenger going to What's The Difference.Go to What's The Difference:w 2 r 1 l.Pickup a passenger going to Multiplication Station.1 is waiting at Starchild Numerology.Go to Starchild Numerology:e 1 r 1 l 3 l.Pickup a passenger going to Addition Alley.Go to Multiplication Station:w 1 r 2 r 1 r 4 l.Pickup a passenger going to Addition Alley.Go to Addition Alley:n 2 l 1 r 3 l 1 l.Pickup a passenger going to The Underground.'Red duck\n' is waiting at Writer's Depot.'Orange duck\n' is waiting at Writer's Depot.'Yellow duck\n' is waiting at Writer's Depot.'Green duck\n' is waiting at Writer's Depot.'Blue duck\n' is waiting at Writer's Depot.'Indigo duck\n' is waiting at Writer's Depot.'Violet duck\n' is waiting at Writer's Depot.'Grey duck' is waiting at Writer's Depot.Go to Writer's Depot:n 1 l 1 l.[b]Pickup a passenger going to Narrow Path Park.Go to Narrow Path Park:n 3 r 1 l 1 r.Go to The Underground:e 1 r.Switch to plan "c" if no one is waiting.Pickup a passenger going to The Underground.Go to Writer's Depot:s 2 r 1 l 2 l.Switch to plan "b".[c]Go to Narrow Path Park:n 4 l.Pickup a passenger going to Post Office.Go to Post Office:e 1 r 4 r 1 l.Go to Writer's Depot:s 1 r 1 l 2 l.Switch to plan "a" if no one is waiting.[d]Pickup a passenger going to Sunny Skies Park.Go to Sunny Skies Park:n 2 r.Go to Writer's Depot:n 1 l.Switch to plan "e" if no one is waiting.Switch to plan "d".[e]Go to Heisenberg's:n 3 r 3 r.Switch to plan "a".

Try it online!

I think it's worth noting that 47% of this code is just picking a random integer from 1 to 8.
Also, Taxi is so verbose that it is way shorter to hard code the duck\n after each color rather than concatenating it later.
Here's the un-golfed version:

Go to Heisenberg's: west 1st right 3rd right 1st left.

[Pick up a random INT 1-8 going to The Underground]
[a]
Pickup a passenger going to Divide and Conquer.
8 is waiting at Starchild Numerology.
8 is waiting at Starchild Numerology.
Go to Starchild Numerology: north 1st left 3rd left 1st left 3rd left.
Pickup a passenger going to Divide and Conquer.
Pickup a passenger going to Multiplication Station.
Go to Divide and Conquer: west 1st right 3rd right 1st right 2nd right 1st right.
Pickup a passenger going to Cyclone.
Go to Cyclone: east 1st left 1st left 2nd left.
Pickup a passenger going to What's The Difference.
Pickup a passenger going to Trunkers.
Go to Zoom Zoom: north.
Go to Trunkers: west 3rd left.
Pickup a passenger going to What's The Difference.
Go to What's The Difference: west 2nd right 1st left.
Pickup a passenger going to Multiplication Station.
1 is waiting at Starchild Numerology.
Go to Starchild Numerology: east 1st right 1st left 3rd left.
Pickup a passenger going to Addition Alley.
Go to Multiplication Station: west 1st right 2nd right 1st right 4th left.
Pickup a passenger going to Addition Alley.
Go to Addition Alley: north 2nd left 1st right 3rd left 1st left.
Pickup a passenger going to The Underground.

[Use the random INT to select a color]
'Red duck\n' is waiting at Writer's Depot.
'Orange duck\n' is waiting at Writer's Depot.
'Yellow duck\n' is waiting at Writer's Depot.
'Green duck\n' is waiting at Writer's Depot.
'Blue duck\n' is waiting at Writer's Depot.
'Indigo duck\n' is waiting at Writer's Depot.
'Violet duck\n' is waiting at Writer's Depot.
'Grey duck' is waiting at Writer's Depot.
Go to Writer's Depot: north 1st left 1st left.
[b]
Pickup a passenger going to Narrow Path Park.
Go to Narrow Path Park: north 3rd right 1st left 1st right.
Go to The Underground: east 1st right.
Switch to plan "c" if no one is waiting.
Pickup a passenger going to The Underground.
Go to Writer's Depot: south 2nd right 1st left 2nd left.
Switch to plan "b".

[Output the selected color]
[c]
Go to Narrow Path Park: north 4th left.
Pickup a passenger going to Post Office.
Go to Post Office: east 1st right 4th right 1st left.

[If the color was grey, exit by error]
Go to Writer's Depot: south 1st right 1st left 2nd left.
Switch to plan "a" if no one is waiting.

[Get rid of the rest of the colors]
[You could throw them off a bridge but you won't get paid]
[d]
Pickup a passenger going to Sunny Skies Park.
Go to Sunny Skies Park: north 2nd right.
Go to Writer's Depot: north 1st left.
Switch to plan "e" if no one is waiting.
Switch to plan "d".

[Start over from the beginning]
[e]
Go to Heisenberg's: north 3rd right 3rd right.
Switch to plan "a".

Answer 6

Ruby, 93 91 90 89 87 86 85 bytes

^{Thanks to Dom Hastings for saving 2 bytes, Kirill L. 1 byte and Asone Tuhid 1 byte!}

puts %w(Red Orange Yellow Green Blue Indigo Violet Grey)[$.=rand(8)]+" duck"while$.<7

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Answer 7

Java (OpenJDK 9), 133 bytes

v->{for(int x=9;x>0;)System.out.println("Grey,Red,Orange,Yellow,Green,Blue,Indigo,Violet".split(",")[x+=Math.random()*8-x]+" duck");}

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Explanations

v->{                              // Void-accepting void lambda function
  for(int x=9;x>0;)               //  Loop until x is zero
    System.out.println(           //   Print...
        "Grey,Red,Orange,         //       colors, "Grey" first
         Yellow,Green,Blue,       //          more colors
         Indigo,Violet"           //          more colors
        .split(",")               //       as an array
          [x+=Math.random()*8-x]  //       pick one randomly, use implicit double to int cast with "x+=<double>-x" trick
        +" duck");                //      Oh, and append " duck" to the color.
}

Answer 8

Operation Flashpoint scripting language, 133 bytes

f={s="";v=s;while{v!="grey"}do{v=["Red","Orange","Yellow","Green","Blue","Indigo","Violet","Grey"]select random 7;s=s+v+" duck\n"};s}

Call with:

hint call f

Example output:

At first I somehow misread the challenge as if the goal was to just output a varying amount of lines, not necessarily ending at the "Grey duck" -line. Following that incorrect interpretation produced a slightly more interesting piece of code:

f={s="";c=[1];c set[random 9,0];{s=s+(["Red","Orange","Yellow","Green","Blue","Indigo","Violet","Grey"]select random 7)+" duck\n"}count c;s}

Answer 9

pdfTeX, 231 220 219 209 207 bytes

\newcount\b\let~\def~\0{Red}~\1{Orange}~\2{Yellow}~\3{Green}~\4{Blue}~\5{Indigo}~\6{Violet}~\7{Gray}~\9{ }\newlinechar`z\loop\b\pdfuniformdeviate8\message{z\csname\the\b\endcsname\9duck}\ifnum\b<7\repeat\bye

LuaTeX, 216 206 204 bytes

\newcount\b\let~\def~\0{Red}~\1{Orange}~\2{Yellow}~\3{Green}~\4{Blue}~\5{Indigo}~\6{Violet}~\7{Gray}~\9{ }\newlinechar`z\loop\b\uniformdeviate8\message{z\csname\the\b\endcsname\9duck}\ifnum\b<7\repeat\bye

Answer 10

Brachylog, 68 bytes

"Grey","
Red
Orange
Yellow
Green
Blue
Indigo
Violet"ṇṛS," duck"ẉS¬?↰

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Answer 11

Ruby, 84 81 bytes

Thanks to Dom Hastings for -3 bytes.

puts$_=%w[Red Orange Yellow Green Blue Indigo Violet Grey].sample+" duck"until/y/

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Answer 12

PHP, 89 bytes

do echo[Grey,Red,Orange,Yellow,Green,Blue,Indigo,Violet][$i=rand()%8]," Duck
";while($i);

Run with -nr or try it online.

Answer 13

><>, 107 bytes

x<>" duck"a>
x<^"deR"
x<^"egnarO"
x<^"wolleY"
x<^"neerG"
x<^"eulB"
x<^"ogidnI"
x<^"teloiV"
x"Grey duck"r>o|

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Answer 14

Octave, 114 112 bytes

do disp([strsplit('Red Orange Yellow Green Blue Indigo Violet'){i=randi(7)},' duck'])until i>6
disp('Grey duck')

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There are a bunch of different options that are all between 112 and 118 bytes... Some initialize an index in the start and decrements it by a random number for each loop, and waits until it's 0. Others use printf instead of disp to avoid some brackets and so on.

Answer 15

PHP, 133 125 111 108 97 92 bytes

<?for(;$b=[Red,Orange,Yellow,Green,Blue,Indigo,Violet][rand(0,7)];)echo"$b duck
"?>Grey duck

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-8 bytes thanks to @Olivier Grégoire

-3 bytes thanks to @manatwork

-11 bytes thanks to @Dom Hastings

Answer 16

bash, 96 bytes

a=(Grey Red Orange Yellow Green Blue Indigo Violet);for((i=1;i;));{ echo ${a[i=RANDOM%8]} duck;}

Thanks to @DigitalTrauma.

Answer 17

Bash + GNU utilities, 72

shuf -e {Red,Orange,Yellow,Green,Blue,Indigo,Violet,Grey}\ Duck|sed /y/q

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Answer 18

Retina, 69 68 bytes

Thanks to Leo for saving 1 byte.

.^/y/{K`Red¶Orange¶Yellow¶Green¶Blue¶Indigo¶Violet¶Grey
" duck¶">?G`

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Explanation

./y/^{K`Red¶Orange¶Yellow¶Green¶Blue¶Indigo¶Violet¶Grey

. suppresses implicit output at the end of the program (otherwise, we'd get two grey ducks). /y/^{ wraps the entire program in a loop which continues as long as the working string doesn't contain a y. The rest of that line sets the working string to a linefeed-separated list of all the colours.

" duck¶">G?`

We grep a random line from the working string (and therefore a random colour). And we print the result with a trailing duck and a linefeed.

Answer 19

JavaScript, 104 bytes

f=_=>`grey,red,orange,yellow,green,blue,indigo,violet`.split`,`[n=Math.random()*8|0]+` duck
${n?f():``}`

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Answer 20

Python 3, 130, 128, 127, 126, 125 bytes

from random import*
d,c=1,'Grey Red Orange Yellow Green Blue Indigo Violet'.split()
while d!=c[0]:d=choice(c);print(d,'duck')

-1 by @ElPedro!
-1 by me
-1 by @Bubbler!

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Answer 21

Python 2, 138 133 120 117 116 bytes

import os
while id:id=ord(os.urandom(1))%8;print"Grey Red Orange Yellow Green Blue Indigo Violet".split()[id],'duck'

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Much better with some ideas from @EriktheOutgolfer. Thanks!

-3 more with thanks to @ovs

-1 with thanks to @Rod for a cool new trick learned :-)

Answer 22

AWK, 114 bytes

{srand();for(split("Red9Orange9Yellow9Green9Blue9Indigo9Violet9Grey",A,9);r<8;print A[r]" duck")r=int(8*rand()+1)}

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Explanation

{srand();                  # Seed rand to obtain different sequence each run
for(
     split("Red9Orange9Yellow9Green9Blue9Indigo9Violet9Grey",
            A,9);          # Split on 9 to avoid using '"'s
     r<8;
     print A[r]" duck")    # Print the colored ducks
     r=int(8*rand()+1)     # AWK uses 1-indexing not 0-indexing when splitting strings into arrays
}

Note that this requires "some" input. The input can be empty. To avoid the need for input prepend the first line with BEGIN

Answer 23

PowerShell, 94 bytes

for(;$r-ne'Grey'){$r=-split"Red
Orange
Yellow
Green
Blue
Indigo
Violet
Grey"|Random;"$r Duck"}

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Loops until $r is equal to Grey. Inside the loop, -splits the literal string on newlines, chooses a Random one thereof, then prints out the color plus Duck (technically, it's left on the pipeline and pipeline cleanup on the next loop iteration causes a Write-Output to happen). Note that it's theoretically possible for Grey to never be chosen, and the loop to continue infinitely, but this almost never (in the probability sense) will happen.

Answer 24

R, 101 bytes

cat(paste(c(sample(scan(,""),rexp(1),T),"gray"),"duck\n"))
Red
Orange
Yellow
Green
Blue
Indigo
Violet

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Heavily inspired by @user2390246's answer to the related challenge. We need two sources of randomness: Changing the order of the colors and sampling the non-gray duck colors. The sample will take a random sample of random size given by an exponential distribution with rate parameter 1, truncated to an integer. Using an exponential distrubtion unfortunately means that there is a probability of exp(-8) or around 0.0003354 that the sample will be at least 8, so we have to sample with replace=T.

Answer 25

MATL, 68 64 bytes

`'DYCIXMSQ(qm#Q$4{#is,Gh1(=lAjUSId;&'F2Y232hZaYb8YrX)' duck'h7Mq

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Explanation

`                         % Do...while
  'DYCI···Id;&'           %   Push this string (to be decompressed by base conversion)
  F                       %   Push false
  2Y2                     %   Push string 'abc...xyz'
  32                      %   Push 32 (ASCII for space)
  h                       %   Concatenate horizontally. Gives 'abc...xyz '
  Za                      %   Base-convert from alphabet of all printable ASCII
                          %   characters except single quote (represented by input
                          %   false) to alphabet 'abc...xyz '. Produces the string
                          %   'grey red ··· violet'
  Yb                      %   Split on space. Gives a cell array of strings
  8Yr                     %   Random integer from 1 to 8, say k
  X)                      %   Get the content of the k-th cell
  ' duck'                 %   Push this string
  h                       %   Concatenate horizontally
  7M                      %   Push k again
  q                       %   Subtract 1
                          % Implicit end. Run a new iteration if top of the stack
                          % is non-zero
                          % Implicit display

Answer 26

Java (JDK 10), 287 bytes

Random r=new Random();int i;String c;do{i=r.nextInt(8);switch(i){case 0:c="Red";break;case 1:c="Orange";break;case 2:c="Yellow";break;case 3:c="Green";break;case 4:c="Blue";break;case 5:c="Indigo";break;case 6:c="Violet";break;default:c="Gray";}System.out.println(c+" duck");}while(i!=7)

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My very first codegolf! Obviously not competitive, just happy to have learned enough Java (currently in CS200) to be able to participate.

Answer 27

Japt, 54 53 52 bytes

Could probably be shorter but my tool for finding the optimal permutation of the string to compress is crapping out on this, for some reason.

_èSiy}a@P±` Ýõ
`i`gÎ9h½uehéHÃÎÁe⸩ºÄgohgyËB`qh ö

Test it

1 byte saved thanks to Oliver

Answer 28

APL (Dyalog Unicode), 86 bytes

{×⍵:∇r⊣⎕←'Grey' 'Red' 'Orange' 'Yellow' 'Green' 'Blue' 'Indigo' 'Violet'[r←?8]'duck'}1

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Assumes ⎕IO←0.

Answer 29

Kotlin, 122 bytes

while({x:Any->println("$x duck");x!="Grey"}("Red,Orange,Yellow,Green,Blue,Indigo,Violet,Grey".split(",").shuffled()[0])){}

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Answer 30

T-SQL, 211 178 bytes

SELECT c+' Duck'FROM(SELECT TOP(CAST(RAND()*9AS INT))c FROM(VALUES('Red'),('Orange'),('Yellow'),('Green'),('Blue'),('Indigo'),('Violet'))v(c)ORDER BY NEWID()UNION SELECT'Grey'c)q

SQLFiddle

Answer 31

T-SQL, 195 185 184 bytes

DECLARE @ VARCHAR(9)=''WHILE @<>'Grey'BEGIN SELECT @=c FROM(VALUES('Red'),('Orange'),('Yellow'),('Green'),('Blue'),('Indigo'),('Violet'),('Grey'))v(c)ORDER BY NEWID()PRINT @+' Duck'END

Answer 32

Python 2, 98 bytes

Look ma no imports!

v=0
while 1:x=id(v)%97%8;print"GVIYORGBrinererleodladeuyliln ee egog n  towe"[x::8],"duck";v=1/x,v

(Prints extra spaces between the colours and duck as allowed in the question)

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A pretty poor pseudo-random number generator seeded with the object id of 0 (but it seems to fit the spec) which repeatedly yields an integer, x, in [0,7] which is used to slice a list of characters from that index in steps of 8 to get the colour name which is printed along with duck as a tuple, forcing a space in-between. When x becomes zero Grey is printed and the evaluation of the next input to the modulo based random number generator errors attempting to divide by zero (v=1/x,v tries to make a new tuple with the first element 1/x = 1/0)

---

Same way is 100 in Python 3 with

v=0
while 1:x=id(v)%17%8;print("GVIYORGBrinererleodladeuyliln ee egog n  towe"[x::8],"duck");v=1/x,v

Answer 33

MS-SQL, 158 bytes

DECLARE @ VARCHAR(6)a:SELECT @=value FROM STRING_SPLIT('Red,Orange,Yellow,Green,Blue,Indigo,Violet,Grey',',')ORDER BY NEWID()PRINT @+' duck'IF @<>'Grey'GOTO a

Based largely on Razvan's excellent answer, but using the STRING_SPLIT function that is specific to MS-SQL 2016 and later. Also uses a GOTO instead of a WHILE loop.

Formatted:

DECLARE @ VARCHAR(6)
a:
    SELECT @=value FROM 
        STRING_SPLIT('Red,Orange,Yellow,Green,Blue,Indigo,Violet,Grey',',')
        ORDER BY NEWID()
    PRINT @+' duck'
IF @<>'Grey'GOTO a

Answer 34

x86_64 machine code, 120 bytes.

I got inspired by the 6502 machine code and implemented the same in x86_64 machine code.

0x603000: 0x47    0x72    0x61    0x79    0x20    0x00    0x00    0x00
0x603008: 0x52    0x65    0x64    0x20    0x00    0x00    0x00    0x00
0x603010: 0x4f    0x72    0x61    0x6e    0x67    0x65    0x20    0x00
0x603018: 0x59    0x65    0x6c    0x6c    0x6f    0x77    0x20    0x00
0x603020: 0x47    0x72    0x65    0x65    0x6e    0x20    0x00    0x00
0x603028: 0x42    0x6c    0x75    0x65    0x20    0x00    0x00    0x00
0x603030: 0x49    0x6e    0x64    0x69    0x67    0x6f    0x20    0x00
0x603038: 0x56    0x69    0x6f    0x6c    0x65    0x74    0x20    0x00
0x603040: 0x64    0x75    0x63    0x6b    0x0a    0x00    0x0f    0x31
0x603048: 0xc1    0xe8    0x03    0x33    0xdb    0x83    0xc3    0x07
0x603050: 0x23    0xd8    0x8b    0xfb    0xc1    0xe7    0x03    0x48
0x603058: 0x81    0xc7    0x00    0x30    0x60    0x00    0xe8    0x73
0x603060: 0xd5    0xdf    0xff    0x48    0xc7    0xc7    0x40    0x30
0x603068: 0x60    0x00    0xe8    0x67    0xd5    0xdf    0xff    0x85
0x603070: 0xdb    0x0f    0x85    0xcf    0xff    0xff    0xff    0xc3

Or the disassembly:

   0x603046:  rdtsc  
   0x603048:  shr    eax,0x3
   0x60304b:  xor    ebx,ebx
   0x60304d:  add    ebx,0x7
   0x603050:  and    ebx,eax
   0x603052:  mov    edi,ebx
   0x603054:  shl    edi,0x3
   0x603057:  add    rdi,0x603000
   0x60305e:  call   0x4005d6 <printf>
   0x603063:  mov    rdi,0x603040
   0x60306a:  call   0x4005d6 <printf>
   0x60306f:  test   ebx,ebx
   0x603071:  jne    0x603046
   0x603077:  ret     

The first 70 bytes is a string table that looks like:

Gray\0\0\0\0
Red\0\0\0\0\0
Yellow\0\0
...
 duck\0

There is certainly room for optimizations and it might not be the shortest implementation in terms of bytes but it probably competes for least amount of bytes executed :)

Answer 35

Red, 96 91 bytes

until[n: random 8 print[pick[Red Orange Yellow Green Blue Indigo Violet Grey]n"duck"]n = 8]

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Answer 36

APL+WIN, 106 bytes

⊃('red' 'orange' 'yellow' 'green' 'blue' 'indigo' 'violet' 'grey')[(↑(m=8)/⍳⍴m)↑m←1+8|n←1E3?1E3],¨⊂' duck'

Generates a vector of random integers between 1 and 8 and takes the series up to and including the first occurance of an 8 as indices into the colour vector and then concatenates the duck.

Answer 37

C (gcc), 138 133 123 122 bytes

Saved a byte thanks to @ceilingcat

*c[]={"Grey","Red","Orange","Yellow","Green","Blue","Violet"};main(i){for(;printf("%s duck\n",c[i=time(sleep(1))%6])*i;);}

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Answer 38

PHP, 166, 148, 142 98 Bytes

Down to 98 bytes thanks to Shaggy and Titus.

I tried a recursive solution.

function f(){echo[grey,red,orange,yellow,green,blue,indigo,violet][$n=rand()&7]." duck 
 ";$n&&f();}

Explanation

function f(){
  echo[grey,red,orange,yellow,green,blue,indigo,violet][$n=rand()&7]
  //a line break goes here, thats why it skips a line after duck
  //saves one byte over \n
        ." duck
         ";
  $n&&f();
}

For real it does not need much, the function will stop when $n == false or as php asumes $n == 0 as a false then it leaves the method.

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Don't really know how to "golf code", but i found this forum really, really fascinating.

Answer 39

SOGL V0.12, 40 39 38 bytes

]"sω2a⁄Ξ┌┼0⁾    a║EΣ?GΘƨ+}‘θJ;7ψ:≥w⁽OX@oo

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Answer 40

Pyth, 72 bytes^(SBCS)

K" Duck"#=kOc." y9Fª¹Ú¹«·.Û»õx¾jL[`ýõ?;ß¼<|9ü"\ +kKIqk"Gray"B

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Python 3 translation:

import random
K=" Duck"
while True:
    k=random.choice("Red Orange Yellow Green Blue Indigo Violet Gray".split())
    print(k+K)
    if k=="Gray":
        break

Answer 41

Swift, 161 143 bytes

import Foundation;var x=9;while x>0{x=time(nil)%7;print(["Grey","Red","Orange","Yellow","Green","Blue","Indigo","Violet"][x]+" duck");sleep(1)}

Thanks to @cleblanc, whom I got the idea of the shorter code from.

Explanation

import Foundation                   //we need Foundation for time

var x = 9

while x > 0 {
    x = time(nil)%7
    print(["Grey",
           "Red",
           "Orange",
           "Yellow",
           "Green",
           "Blue",
           "Indigo",
           "Violet"][x] + " duck") //subscripting color array, appending "duck"
            sleep(1)
}

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Answer 42

Python 3, 127 bytes

from random import *;x=''
while'Grey'!=x:x=choice('Red,Orange,Yellow,Green,Blue,Indigo,Violet,Grey'.split(','));print(x,'duck')

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Answer 43

Perl 6/Rakudo 78 bytes

say $_=<Grey Red Orange Yellow Green Blue Indigo Violet>.roll~" duck"until /y/

Stealing directly from Dom Hastings' Perl 5 answer

Perl 6 wins slightly with .roll

Answer 44

SmileBASIC, 134 116 bytes

DIM C$[8]C=RND(8)COPY C$,@L@L?C$[C];" duck
EXEC!C
DATA"Gray","Red","Orange","Yellow","Green","Blue","Indigo","Violet

Answer 45

Python 3, 124 bytes

import random
b=1
while b>0:b=random.randint(0,7);print('Gray Red Orange Yellow Green Blue Indigo Violet'.split()[b],'duck')

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I almost always accompany my posts with explanations, but this one seems simple enough that I feel confident going without one...

Answer 46

F#, 171 bytes

let r=System.Random()
let mutable i=r.Next(8)
while i<>7 do
 printfn"%s duck"["Red";"Orange";"Yellow";"Green";"Blue";"Indigo";"Violet"].[i]
 i<-r.Next(8)
printf"Grey duck"

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That flipping "mutable" keyword! I thought that F#'s lack of a do/while loop (it only has while/do) would hurt me, but on retrospect it worked out, since "Grey duck" is always printed without a line separator (printfn versus printf), so there's no trailing empty line.

Answer 47

Visual Basic for Applications, 110 102 bytes

Edit: saved eight more bytes by using split() instead of array()

while rnd>.3:?split("Red,Orange,Yellow,Green,Blue,Indigo,Violet",",")(rnd*6)+" duck":wend:?"Gray duck"

Type in the immediate window.

Answer 48

TI-BASIC, 80 bytes

Repeat not(I
int(8rand→I
Disp sub("GRAY  RED   ORANGEYELLOWGREEN BLUE  INDIGOVIOLET",6I+1,6)+" DUCK
End

Answer 49

Bash + Coreutils, 100 Bytes

B=`shuf -en1 Red Orange Yellow Green Blue Indigo Violet Gray`;echo $B duck;[[ $B = Gray ]]||bash $0

Answer 50

ink, 78 87 bytes

-(e){RANDOM(0,1):{~Red|Orange|Yellow|Green|Blue|Indigo|Violet} duck|Gray duck->END}
->e

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Explanation

-(e){RANDOM(0,1):{~Red|Orange|Yellow|Green|Blue|Indigo|Violet} duck|Gray duck->END}  // With a 50% chance of each, either pick a colour that hasn't been picked already (unless all others have also been picked) and print it followed by "duck", or print "Gray duck" and terminate.
->e   // if we haven't terminated we get here and go back to the line above

+9 bytes: Apparently shuffles in ink will not repeat an option until all options have been visited. This means the previous version of my program was not valid, since it would always encounter "gray" before it could encounter any other colour for the second time, and thus broke the "Colours can be repeated" rule. The new answer will print each non-gray colour once until all have been printed, then shuffle again and restart - except every time it's about to pick a colour there's a 50% chance it instead prints gray and terminates.

Answer 51

C (gcc), 104 bytes

f(n){for(;printf("%.6s duck\n","Gray\0 OrangeYellowIndigoVioletGreen\0Blue\0 Red"+(n=time(0)%7)*6),n;);}

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