Randomly select a character, plinko-style

Question

Let's see how good your language of choice is at selective randomness.

Given 4 characters, A, B, C, and D, or a string of 4 characters ABCD as input, output one of the characters with the following probabilities:

• A should have a 1/8 (12.5%) chance to be chosen
• B should have a 3/8 (37.5%) chance to be chosen
• C should have a 2/8 (25%) chance to be chosen
• D should have a 2/8 (25%) chance to be chosen

This is in-line with the following Plinko machine layout:

   ^
  ^ ^
 ^ ^ ^
A B \ /
     ^
    C D

Your answer must make a genuine attempt at respecting the probabilities described. A proper explanation of how probabilities are computed in your answer (and why they respect the specs, disregarding pseudo-randomness and big numbers problems) is sufficient.

Scoring

This is code-golf so fewest bytes in each language wins!

Answer 1

Java (OpenJDK 8), 40 38 bytes

s->s[5551>>2*(int)(Math.random()*8)&3]

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5551 in base 10 is 01112233 in base 4. So let's randomly pick one of those base-4 digits using bit shift and selection, then pick the n-th character from the originating char[].

Answer 2

Mathematica, 25 bytes

RandomChoice@{##,##2,#2}&

input

["A", "B", "C", "D"]

-23 bytes from @MarkS!

Answer 3

braingasm, 17 bytes

4[,>]<4r<#z[2r>].

Works like this:

4[,>]<             Input 4 bytes from stdin and stay in cell 3 (0-indexed).
        <          Move left ...
      4r           ... randomly between 0 (inclusive) and 4 (exclusive) times.
         #z[   ]   If we're at cell 0,
              >      move right ...
            2r       ... with 50% probability.
                .  Print the byte value of the cell.

Given the input ABCD, each character has 1/4 = 2/8 chance of being printed because of 4r, but if we selected A, we have 1/2 chance of printing B instead, giving A (1/4)/2 = 1/8 chance and B 1/4 + 1/8 = 3/8 chance.

Answer 4

J, 16 bytes

(1?8:){1 3 2 2&#

how?

• 1 3 2 2&# - copy the input elementwise, ie, 1 copy of A, 3 copies of B, etc, yielding the list ABBBCCDD
• 1?8: - choose 1 element at random from the list 0 1 2 3 4 5 6 7
• { - "From", ie, choose from ABBBCCDD the random index generated by 1?8:

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Answer 5

C, 106 bytes

int f(){int r=(int)random()%8;if(!r--)return 'A';if(--r-->0)if(r<2)return 'C';else return 'B';return 'D';}

Answer 6

q/kdb+, 15 13 bytes

Solution:

1?a[0],7#1_a:

Example:

q)1?a[0],7#1_a:"ABCD"
,"B"

Explanation:

This is pretty much a q version of the jelly solution. Create the list "ABCDBCDB" and then pick one item from it at random

Bonus:

Another q solution in 15 bytes, same style as the APL one:

1?(,/)1 3 2 2#'

A k version of this solution is 13 bytes:

1?,/1 3 2 2#'

Answer 7

CJam, 11 bytes

XZYY]q.*smR

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XZYY]    e# add [1,3,2,2] to stack
q        e# add input string to stack
.*       e# element-wise multiply the two arrays. char * int in CJam = repeat character x times
s        e# flatten result to string
mR       e# pick random char from string

Answer 8

05AB1E, 9 bytes

•20åÝ•.Rè

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•20åÝ•   # Push 33221110.
     .R  # Randomly selected index.
       è # Character at that index.

Answer 9

Octave, 30 28 bytes

@(S)S([1:4 2 2:4])(randi(8))

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Previous answer:

@(S)S('ABBBCCDD'(randi(8))-64)

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Input is taken as a string of 4 characters.

Explanation:

Instead of [1 2 2 2 3 3 4 4 ] we can write 'ABBBCCDD'-64 and randomly select one of indexes and use the index to extract the desired character from the string.

Answer 10

JavaScript (ES6), 25 bytes / 30 bytes

In both following versions, we're looking for two positive integers m and n such that n MOD (m + i) MOD 4 gives us results in [0...3] at the expected frequencies for i in [0...7].

---

Version 1, 25 bytes

If a time-dependent formula is allowed, we can just do:

s=>s[11%(new Date%8+2)%4]

This one is using n = 11 (smallest possible value of n) and m = 2.

i | 2 + i | 11 % (2 + i) | 11 % (2 + i) % 4
--+-------+--------------+-----------------
0 |     2 |            1 |                1
1 |     3 |            2 |                2
2 |     4 |            3 |                3
3 |     5 |            1 |                1
4 |     6 |            5 |                1
5 |     7 |            4 |                0
6 |     8 |            3 |                3
7 |     9 |            2 |                2

Demo

let f =

s=>s[11%(new Date%8+2)%4]

for(stat = {A:0, B:0, C:0, D:0}, i = 0; i < 80000; i++) {
  stat[f("ABCD")]++;
}
console.log(stat);

---

Version 2, 30 bytes

With the native JavaScript PRNG -- which generates a floating-point number in [0...1) -- we'd rather do:

s=>s[21%(Math.random()*8|8)%4]

Here, we use n = 21 and m = 8 because it allows us to do the addition and isolate the integer part at the same time with Math.random()*8|8.

i | 8 + i | 21 % (8 + i) | 21 % (8 + i) % 4
--+-------+--------------+-----------------
0 |     8 |            5 |                1
1 |     9 |            3 |                3
2 |    10 |            1 |                1
3 |    11 |           10 |                2
4 |    12 |            9 |                1
5 |    13 |            8 |                0
6 |    14 |            7 |                3
7 |    15 |            6 |                2

Demo

let f =

s=>s[21%(Math.random()*8|8)%4]

for(stat = {A:0, B:0, C:0, D:0}, i = 0; i < 80000; i++) {
  stat[f("ABCD")]++;
}
console.log(stat);

Answer 11

PowerShell, 64 bytes

$args|%{($_-replace '(^.)(.)',"`$1`$2`$2`$2$'")[(Get-Random)%8]}

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It replaces the string abcd with abbbcdcd using regex witchcraft and then randomizes which of the eight characters it chooses. The regex is expensive; it requires a backtick per grouping, so there may be a better method using recursive lookups.

Here's a quick explanation of the regex:

For input abcd the full match is ab, grouping $1 is 'a', grouping $2 is 'b', and $' (note quote, not backtick) returns everything after the full match. Then, the pattern $1$2$2$2$' is substituted for the full match: abcd goes to $1$2$2$2$' + cd = abbbcdcd

Answer 12

Bash, 52 bytes

f(){ c=$1$2$2$2$3$3$4$4;echo ${c:$(($RANDOM%8)):1};}

Usage exemple:

$ f a b c d
b
$ f a b c d
d
$ f a b c d
b
$ f a b c d
a

Answer 13

TI-Basic, 25 bytes

sub(Ans,randInt(1,8),1
If int(2rand(Ans="A
"B

Answer 14

Perl 5, 26 bytes

say((A..D,B..D,B)[rand 8])

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Answer 15

SmileBASIC, 31 bytes

LINPUT S$N=RND(8)?S$[N+!N*2>>1]

Pick a number from 0 to 7, add 2 if it's 0, then divide by 2.

Answer 16

Thunno 2, 7 bytes

«¦/d×Jɼ

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Explanation

«¦/d×Jɼ  # Implicit input
«¦/      # Push 1322
   d     # Cast to digits
    ×    # Multiply elementwise
     J   # Join into a string
      ɼ  # Choose a random character
         # Implicit output