Random Golf of the Day #7: A distinctly random character

Question

About the Series

This is a guest entry for the Random Golf of the Day series.

First off, you may treat this like any other code golf challenge, and answer it without worrying about the series at all. However, there is a leaderboard across all challenges. You can find the leaderboard along with some more information about the series in the first post.

Input

No input is taken.

Output

A single letter of the alphabet (case irrelevant), with an optional trailing newline. Each letter must have non-zero probability of being chosen, and all 26 probabilities must be distinct. To remove all ambiguity: Distinct means that there must not be two probabilities that are equal to each other.

Scoring

This is code golf. Shortest code in bytes wins.

A valid entry is a full program or function that has zero probability of not terminating.

Alphabet

To avoid confusion, the particular alphabet to be used is the Latin alphabet:

Either

ABCDEFGHIJKLMNOPQRSTUVWXYZ

or

abcdefghijklmnopqrstuvwxyz

You may choose to output upper case or lower case. Alternatively, you may choose to output different cases on different runs if that helps. The probability for a given letter is the probability of that letter appearing in either case (upper or lower).

Explanation

As it won't be at all obvious from the output, please include a clear explanation of how you achieved the 26 distinct probabilities.

Leaderboard

(from here)

var QUESTION_ID=89621,OVERRIDE_USER=20283;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;

body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

The first post of the series also generates an overall leaderboard.

To make sure that your answers show up, please start every answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

(The language is not currently shown, but the snippet does require and parse it, and I may add a by-language leaderboard in the future.)

Answer 1

CJam, 10 bytes

CJam approach #3...

26mr)mr'A+

Try it online!

This creates a uniformly random number x between 1 and 26 and then uses that to create a uniformly random number between 0 and x-1 which is added to A. This biases results towards smaller characters.

Answer 2

Mathematica, 34 bytes

RandomChoice[Range@26->Alphabet[]]

RandomChoice[wlist -> elist] gives a random choice of elist weighted by wlist.

Answer 3

Ruby, 42 bytes

$><<(?A..?Z).flat_map{|c|[c]*c.ord}.sample

Quite straightforward: Generate 65 A's; 66 B's; ... 90 Z's, and randomly pick one of the letters generated.

Answer 4

LaTeX, 122 115 bytes

-7 bytes by replacing pgfmathparse{Hex(...)} with pgfmathHex{...}.

Or, if I'm allowed to skip the document class definition & setup, and just count the package import and functional code: 65 bytes.

pgfmathHex parses the expression and its (hexadecimal) result is then fed into \char which turns the code point into a unicode character. The expression itself is identical to many other answers here.

\documentclass{book}\usepackage{tikz}\begin{document}\pgfmathHex{65+sqrt(rnd)*26}\char"\pgfmathresult\char"\pgfmathresult\end{document}

Ungolfed (with cherry-picked for loop for page-filling output):

\documentclass{book}
\usepackage{tikz}
\begin{document}
\noindent
\foreach \n in {0,...,1513}
{
\pgfmathHex{65+sqrt(rnd)*26}\char"\pgfmathresult
}
\end{document} 

Output (w/ free page number :) ):

Answer 5

C#, 68 66 bytes

Thanks to Leaky Nun for fixing the chance distribution.

var r=new Random();char f(char c='A')=>c<'Z'&r.Next(3)>0?f(++c):c;

Surprisingly short for a C# answer.

It defines a function f that returns a random character between the given one and Z. It has a 1/3 chance of returning the current character, otherwise it recurses on the next. If the current character is Z, it always returns. Default parameter is A, so calling it without any argument satisfies the challenge.

The resulting chances follow 2^(n-1) / 3^n:

A: 1/3
B: 2/9
C: 4/27
D: 8/81
...
Y: 16777216/847288609443
Z: 67108864/2541865828329

Answer 6

x86_64 machine language for Linux, 15 19 17 16 bytes

L1:
48 0f c7 f0             rdrand %rax
f3 48 0f b8 c0          popcnt %rax,%rax
3c 1a                   cmp    $0x1a,%al
7d f3                   jge    L1
04 41                   add    $0x41,%al
c3                      retq

This requires support for the POPCNT and RDRAND instructions.

A uniform distributed random number is generated, the number of 1's in that number is counted, if that number is less than 26, a letter is returned. One will need to let the code run a long time before one sees a letter A.

To test, try something like

#include<stdio.h>
#define TEST "\x48\xf\xc7\xf0\xf3\x48\xf\xb8\xc0\x3c\x1a\x7d\xf3\4\x41\xc3"
int main(){
  int hist[26]={0};
  for(int i=0;i<10000000;i++){
    hist[ ((int(*)())TEST)() - 'A' ]++;
  }
  for(int i=0;i<26;i++){
    printf("%c %d\n", 'A'+i, hist[i] );
  }
}

Sample output

A 0
B 0
C 0
D 0
E 0
F 0
G 0
H 0
I 0
J 0
K 0
L 8
M 32
N 137
O 511
P 1639
Q 5188
R 14475
S 37539
T 91670
U 205638
V 431381
W 842259
X 1536776
Y 2626524
Z 4206223

The analytical expression for the probability of each letter can be derived from the binomial distribution. The letter A is assigned index k=0, B is assigned k=1 and so on.

        /  \
       | 64 |
       | k  |
        \  /
p(k)=------------
      25
      --- /  \
      \  | 64 |
      /  | i  |
      --- \  /
      i=0

 p(A)~1.0483e-18
 p(B)~6.7093e-17
 p(C)~2.1134e-15
 p(D)~4.3678e-14
 p(E)~6.6608e-13
 p(F)~7.9930e-12
 p(G)~7.8598e-11
 p(H)~6.5124e-10
 p(I)~4.6401e-09
 p(J)~2.8872e-08
 p(K)~1.5879e-07
 p(L)~7.7953e-07
 p(M)~3.4429e-06
 p(N)~1.3772e-05
 p(O)~5.0169e-05
 p(P)~1.6723e-04
 p(Q)~5.1214e-04
 p(R)~1.4460e-03
 p(S)~3.7758e-03
 p(T)~9.1413e-03
 p(U)~2.0568e-02
 p(V)~4.3095e-02
 p(W)~8.4231e-02
 p(X)~1.5381e-01
 p(Y)~2.6276e-01
 p(Z)~4.2042e-01

Answer 7

Matlab, 24 18 bytes

char(floor(26*rand^2+65))

it looks like using floor(x) isn't necessary as char also takes non-integer inputs

char(25*rand^2+65)

not using ['' x] instead of char(x) so I won't get warning: implicit conversion from numeric to char for "purer" output.

rand yields a uniformly distributed random number in the interval (0,1) but rand^2 isn't uniformy distributed anymore, the probability density function follows , see here.

with (b-a)*rand+a one can shift the interval of the distribution from (0,1) to (a,b), this also works with rand^2.

because i use floor i need to stretch the interval to (65,91) so i don't lose "Z"

It might no be the shortest answer but i like the approach with using a uniform random distribution to get a non-uniform random distribution.

Below is the count for each generated numbers after 20000 iterations (using my first version of my answer).

Answer 8

MATL, 12 bytes

1Y2toY"tnYr)

Try it online!

Explanation

1Y2          % Pushes the alphabet A...Z
   to        % Make a numeric copy [65 ... 90]
     Y"      % Run length decoding: 65 A's, 66 B's, etc. Let's call this [S]
        nYr  % Random number between 1 and length of [S], let's call this [i]
       t   ) % Index into a copy of [S] at the [i]'th position. Implicit display.

The selection from [S] is according to a uniform distribution, but since each letter in the alphabet occurs a distinct number of times in [S], the resulting distribution is also distinct for each letter in the alphabet.

Answer 9

SmileBASIC, 26 22 bytes

I think it's unfair that most of these are biased towards earlier letters.

Here's one that prints Z the most:

?CHR$(90-RND(RND(27)))

Answer 10

Phooey, 26 bytes

~?675@(&>@+1@<--@@^)>+63$c

Try it online!

The demo contains a simple infinite loop which resets the tape (this program expects a clean tape and doesn't clean up lul)

Similar to a lot of other answers, this generates a number from \$0-675\$ and uses roughly \$\lfloor \sqrt{n}\rfloor\$ to select the letter with varying probability.

Unfortunately, AFAICT, there is no way of doing a square root automatically since the pow operator uses integers. Maybe there is a fixed point method, but idk yet.

However, the naïve subtraction method works and is probably shorter than any other solution. It's only going to be running a few iterations, so the inefficiency isn't that bad.

int64_t isqrt(int64_t n)
{
    int64_t d = 0;
    while (n >= 0) {
        n -= d;
        d += 1;
        n -= d;
    }
    return d - 1;
}

My implementation actually has an off-by-one problem with my loop condition check. I use \$x^x\$ (@^) instead of \$2^x\$ (@&2^) which is the "proper" (at least, in Phooey) way to check for negatives, and when \$x = -1\$, it will report a false negative (or is it a false positive? 🤨), as \${-1}^{-1} = {-1}, {-1} \ne 0\$.

HOWEVER, that doesn't affect the output's randomness. It just means that Z will appear once instead of A and we need \$d - 2\$ instead of \$d - 1\$.

With all values from \$0-675\$ tested, this is the frequency:

A 3   B 5
C 7   D 9
E 11  F 13
G 15  H 17
I 19  J 21
K 23  L 25
M 27  N 29
O 31  P 33
Q 35  R 37
S 39  T 41
U 43  V 45
W 47  X 49
Y 51  Z 1

How it works:

~?675@(&>@+1@<--@@^)>+63$c
~?675                       # Generate a random integer between 0 and 675 inclusive
     @                      # Push the initial n to the stack
                            # Begin our naïve semi-sqrt loop
      (            )        # while n >= 0
       &                    #    Pop n from the stack
        >                   #    Move left to d
         @                  #    Push d
          +1                #    Increment d
            @               #    Push d + 1
             <              #    Move back to n
              --            #    Subtract d and d + 1, popping them
                            #    Use a modified method I found in my Brexit post to check for < 0
                @           #    Save a copy of n
                 @^         #    Calculate n^n using double<->int64 conversion
                            #    The current cell will contain zero if n < -1, or nonzero otherwise.
                            #    Use that as our loop condition.
                    >       # Move to d
                     +63    # d += 'A' - 2
                        $c  # print as char

Answer 11

Vyxal, 5 bytes

ka¦∑℅

Try it Online!

same idea as the Pyth answer.

Answer 12

Zsh, 26 bytes

eval shuf -en1 {{a..z}..z}

Try it online! Includes a basic frequency analysis.

Explanation:

• {a..z} - construct the string a b ... z
• {{a..z}..z} - construct the string {a..z} {b..z} ... {z..z}
• evaluate that string - executes shuf -en1 a b ... z b c ... z c d ... z ... y z z
• shuf -en1 - picks one of those letters at random. In the pool, z is repeated 26 times, y 25 times, etc., which creates the distinct probabilities.

Answer 13

05AB1E, 4 bytes

AƶJΩ

Try it online! Beats all other answers.

AƶJΩ  # full program
   Ω  # random character from...
  J   # joined...
A     # "abcdefghijklmnopqrstuvwxyz"...
 ƶ    # with each character repeated 1-indexed index of character in string times
      # implicit output

AƶJ generates the string 'abbcccddddeeeeeffffffggggggghhhhhhhhiiiiiiiiijjjjjjjjjjkkkkkkkkkkkllllllllllllmmmmmmmmmmmmmnnnnnnnnnnnnnnoooooooooooooooppppppppppppppppqqqqqqqqqqqqqqqqqrrrrrrrrrrrrrrrrrrsssssssssssssssssssttttttttttttttttttttuuuuuuuuuuuuuuuuuuuuuvvvvvvvvvvvvvvvvvvvvvvwwwwwwwwwwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxyyyyyyyyyyyyyyyyyyyyyyyyyzzzzzzzzzzzzzzzzzzzzzzzzzz', so the chances are as follows:

a: 0.285%
b: 0.57%
c: 0.855%
d: 1.14%
e: 1.425%
f: 1.709%
g: 1.994%
h: 2.279%
i: 2.564%
j: 2.849%
k: 3.134%
l: 3.419%
m: 3.704%
n: 3.989%
o: 4.274%
p: 4.558%
q: 4.843%
r: 5.128%
s: 5.413%
t: 5.698%
u: 5.983%
v: 6.268%
w: 6.553%
x: 6.838%
y: 7.123%
z: 7.407%

Alternative, 4 bytes

AηJΩ

Try it online!

AηJΩ  # full program
   Ω  # random character from...
  J   # joined...
 η    # prefixes of...
A     # "abcdefghijklmnopqrstuvwxyz"
      # implicit output

AηJ generates the string 'aababcabcdabcdeabcdefabcdefgabcdefghabcdefghiabcdefghijabcdefghijkabcdefghijklabcdefghijklmabcdefghijklmnabcdefghijklmnoabcdefghijklmnopabcdefghijklmnopqabcdefghijklmnopqrabcdefghijklmnopqrsabcdefghijklmnopqrstabcdefghijklmnopqrstuabcdefghijklmnopqrstuvabcdefghijklmnopqrstuvwabcdefghijklmnopqrstuvwxabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyz', so the chances are as follows:

a: 7.407%
b: 7.123%
c: 6.838%
d: 6.553%
e: 6.268%
f: 5.983%
g: 5.698%
h: 5.413%
i: 5.128%
j: 4.843%
k: 4.558%
l: 4.274%
m: 3.989%
n: 3.704%
o: 3.419%
p: 3.134%
q: 2.849%
r: 2.564%
s: 2.279%
t: 1.994%
u: 1.709%
v: 1.425%
w: 1.14%
x: 0.855%
y: 0.57%
z: 0.285%

Answer 14

Clojure, 61 bytes

#(char(- 90(count(take-while(fn[a](<(rand)0.5))(range 26)))))

Takes each successive element from the range of 26 elements with probability 1/2. So the chance that one element is taken is 1/2, 2 elements - 1/4 and so on. After that subtract from 90 number of elements taken and turn it into char.

p(Z) = 1/2, p(Y) = 1/4, p(X) = 1/8 ... p(A) = 1/2^(26)

See it online: https://ideone.com/Cg15FY

Answer 15

Julia, 33 bytes

f()=['A':'Z'][ceil(26*√rand())]

Applies the square root to RNG.

Graph showing distribution: Generated with:

using Gadfly
plot(x=ceil(26*√rand(10000)),Geom.histogram)

Answer 16

ForceLang, 64 bytes

Noncompeting, uses language features (the string.char method) that postdate the question.

set s math.sqrt 676.mult random.rand()
io.write string.char 65+s

Answer 17

Actually, 7 bytes

ú#;╚♀MJ

Try it online!

This answer uses the same approach as Dennis's Jelly answer.

Explanation:

ú#;╚♀MJ
ú#       lowercase English alphabet, as a list (this is to dodge a bug with the shuffle command)
  ;╚     duplicate, shuffle
    ♀M   pairwise maximum
      J  random element

Answer 18

perl, 21 bytes

say+(A..Z)[rand$$%27]

Needs -M5.010 or -E to run :

perl -E 'say+(A..Z)[rand$$%27]'

How it works : (A..Z) is an array. $$ is process ID, which introduces some randomness. $$%27 is a number between 0 and 26 (inclusive, and roughly evenly spread). Calling rand on that produces a number between 0 and 26 (inclusive at 0, exclusive at 26), but biased towards smaller numbers. We then use that number for an array lookup, which we print.

Opportunities for improvement :

$$ is random, but not cleanly random. We know that the chance of getting particular numbers differs, but not by much, and not in ways that are easy to predict cleanly. Therefore, the possibility of getting different each letter is probably not identical, but it's very close. If (and only if) we accept that probability{ $$%26 == x } is different for all x, then we could replace "rand$$%27" with "$$%26".

Alternately, we could replace "rand$$%27" with (for example) "26*sin$$", and save one character. This gives a much less regular spread than simply "$$%26":

 perl -E 'for(1..32768){say+(A..Z)[26*sin$_]}' | sort | uniq -c
    832 A
   3269 B
   1648 C
   1360 D
   1183 E
   1133 F
   1043 G
   1071 H
    959 I
    978 J
    985 K
    939 L
    926 M
    950 N
    924 O
    939 P
    985 Q
    977 R
    959 S
   1072 T
   1043 U
   1135 V
   1182 W
   1360 X
   1647 Y
   3269 Z

But if you check, you'll see that it's not perfect. There are three pairs of numbers that have the same probability as each other: K&Q, I&S, L&P.

(If you use actual PIDs, then depending on what processes are actually running, and depending on the highest allowable pid on your machine, you might get more clashes, or less.)

"15*sin$$" works better, only 2 clashes - but that is still 2 clashes. (Any number between 13 and 26 will work, in the sense of producing all 26 letters, because sin(x) spans -1 to +1 and perl treats negative number array look-ups as starting from the array end).

There might be some way of getting all 26 numbers, with different probabilities, using less (or no more) than the 8 characters that "rand$$%27" needs. But if so, it isn't coming to me.

Answer 19

Clojure, 62 bytes

(loop[i 65](if(or(> i 89)(neg?(rand)))(char i)(recur(inc i))))

This is somewhat stupid shortcut. The probability of exact zero double - the only (rand) to not break on condition (pos? x) - is around 1/(2^62) [source].

So chance for letter char(65+N) is around (1/(2^62))^N, slightly higher for Z (because it is last).

With (rand-int 2) - 6 more bytes - it becomes testable.

Answer 20

Perl 6, 30 bytes

{('a'..'z'Zxx 1..*).flat.pick}

Explanation:

# bare block lambda
{
  (

    'a' .. 'z' # the alphabet
    Z[xx]      # zipped ｢Z｣ using the list repeat operator ｢xx｣
    1 .. *     # with 1 to 26

    # ((a)(b b)(c c c)(d d d d)(e e e e e)...

  )
  .flat # flatten it from a list of lists into a single list
  .pick # pick an element from the list
}

Answer 21

Pip, 8 bytes

(zr*r*h)

z is a built-in for the lowercase alphabet, h for 100. r is a special variable: each time it is evaluated, it generates a random number with uniform distribution, 0 <= r < 1. Thus, r*r*h squares the uniform distribution and scales it a hundredfold (saving a character over using 26). The result is used to index into z, auto-truncated and with index wrapping.

Try it here (frequency chart of 100,000 runs, takes about 30 seconds).

Answer 22

Lua, 100 99 bytes

s="A"math.randomseed(os.time())for i=66,90 do
s=.6<math.random()and string.char(i)or s end
print(s)

Not really competitive, but still an interesting algorithm I think.

Answer 23

IA-32 machine code, 12 bytes

Hexdump:

0F C7 F0 F6 E4 88 E0 D4 1A 04 61 C3   

Source code:

rdrand eax
mul ah
mov al, ah
aam 26
add al, 'a'
ret

Or, as a C expression (assuming r is a uniformly generated random number of type uint32_t):

(r & 0xff) * ((r >> 8) & 0xff) / 0x100 % 26 + 'a'

Running this expression on values in the range 0...65535 gives the probabilities of the letters:

z 1968
y 2012
x 2068
w 2076
v 2134
u 2140
t 2194
s 2209
r 2233
q 2292
p 2335
o 2358
n 2410
m 2437
l 2475
k 2545
j 2571
i 2654
h 2696
g 2738
f 2818
e 2922
d 2955
c 3098
b 3241
a 3957

Answer 24

Dyalog APL, 14 bytes

See this meta post for information about the code page.

(?351)⊃⎕A/⍨⍳26

Gives instant results. Works by selecting an evenly distributed random character among "ABBCCCDDDD..."

(?351) RandInt(1,351)
⊃picks among
⎕A "ABC...Z"
/⍨ element-by-element replicated
⍳26 {1, 2, 3, ..., 26} times

TryAPL online!

Answer 25

MathGolf, 6 bytes

▄mÅî*w

Try it online.

Explanation:

▄       # Push the lowercase alphabet
 m      # Map over each letter
  Å     # using the following two characters as inner code-block:
   î    #  Push the 1-based map-index
    *   #  Repeat the current character that many times
     w  # After the map: pop and push a random character from this string
        # (after which the entire stack joined together is output implicitly as result)

Same probability as most other answers:

• a: \$\frac{1}{351}\$
• b: \$\frac{2}{351}\$
• ...
• y: \$\frac{25}{351}\$
• z: \$\frac{26}{351}\$

Answer 26

Javascript, 93 91 bytes

Yes there is a shorter one, but that does it differently. Also I spent half an hour on this so I'm not giving up.

q=>'abcdefghijklmnopqrstuvwxyz'.split``.map((i,k)=>i.repeat(k)).join``[Math.random()*351|0]

Explained:

q=> // define function, meaningless arg saves a byte
'abcdefghijklmnopqrstuvwxyz'.split`` // get each alphabet character in array
.map((i,k)=>i.repeat(k)) //replace each with it repeated index times
.join``[Math.random()*351|0] //turn to string and get random character. 

Probability:

a - 1/351
b - 2/351
c - 3/351

Etcetera.

Note: according to this comment, I don't have to use a function declaration if it isn't recursive.

Answer 27

Python 3, 58 bytes

from random import*;print(chr(randint(65,randint(65,91))))

it is like that perl answer.

Answer 28

k4, 15 bytes

*1?a|-26?a:.Q.A

A port of @Dennis's Jelly answer.

Answer 29

JavaScript ES6, 42 Bytes, modified from user81655's solution:

_=>(0|10+26*Math.random()**2).toString(36)

Answer 30

Javascript, 49 bytes

f=()=>String.fromCharCode(65+Math.random()**2*26)