1/N probability

Question

Because there are not enough simple code-golf challenges:

Create an optionally unnamed program or function that, given (by any means) an integer 1 ≤ N ≤ 10000, outputs your language's True value with a pseudo-random probability of 1/N, False otherwise.

Please note that the requirement for naming has been removed. Feel free to edit answers and scores accordingly.

Some languages use 1 (or -1) and 0 for True and False, that is fine too.

Example:

Example input tests:

4 -> True
4 -> False
4 -> False
4 -> False
4 -> False
4 -> True
4 -> False
4 -> False

I.e. given 4; it returns True with a 25% chance and False with a 75% chance.

Answer 1

MediaWiki templates with ParserFunctions, 48 bytes

{{#ifexpr:1>{{#time:U}} mod {{{n}}}|true|false}}

Answer 2

Pyth, 3 bytes

!OQ

Try it online

Simple inversion of random choice from 0 to input

Amusingly in Pyth it is not possible to make a function that does this without $ because Pyth functions are automatically memoized.

Answer 3

CJam, 5 bytes

Gotta be quick with these ones...

rimr!

Test it here.

Explanation

ri e# Read input and convert to integer N.
mr e# Get a uniformly random value in [0 1 ... N-1].
!  e# Logical NOT, turns 0 into 1 and everything else into 0.

Answer 4

TI-BASIC, 4 bytes using one byte tokens

not(int(Ansrand

Determines if the integer part of the input times a random number in [0,1) is zero. Ansrand<1 also works.

Answer 5

MATL, 5 bytes

Three different versions of this one, all length 5.

iYr1=

which takes an input (i), generates a random integer between 1 and that number (Yr), and sees if it it is equal to 1 (1=). Alternatively,

li/r>

make a 1 (l, a workaround because there is a bug with doing 1i at the moment), take an input (i), divide to get 1/N (/), make a random number between 0 and 1 (r), and see if the random number is smaller than 1/N. Or,

ir*1<

take and input (i), and multiply by a random number between 0 and 1 (r*), and see if the result is smaller than 1 (1<).

In Matlab, not MATL, you can do this anonymous function

@(n)n*rand<1

for 12 bytes, which is used by doing ans(5), for example.

Answer 6

JavaScript ES6, 15 bytes

-5 bytes thanks to Downgoat.

x=>1>new Date%x

Based off (uses) of this answer's technique.

Answer 7

Julia, 17 16 15 bytes

n->2>rand(1:n)

This is a function that generates a random integer between 1 and n and tests whether it's less than 2. There will be a 1/n chance of this happening, and thus a 1/n chance of returning true.

Saved 1 byte thanks to Thomas Kwa!

Answer 8

Microscript II, 3 bytes

NR!

Reads an integer n, generates a random integer between 0 and n-1 (inclusive), then applies a boolean negation to that value.

Answer 9

Candy, 2 bytes

Hn

H stands for Heisen-double

n stands for not

The 'n' is passed with the -i flag as numeric input. Values left on the stack are printed on exit.

"Long" form:

rand   # number between 0 and pop()
not    # cast to int, invert non-zero to zero, and zero to one

Answer 10

Seriously, 3 bytes

,JY

0 is falsey and 1 is truthy. Try it online

Explanation:

,JY
,    get input
 J   push a random integer in range(0, input) ([0, ..., input-1])
  Y  logical not: push 0 if truthy else 1  

Answer 11

R, 30 22 bytes

code

cat(runif(1)<1/scan())          #new
f=function(N)cat(runif(1)<1/N)  #old

It generates a number from a uniform distribution (0 to 1) and should evaluate to true 1/n of the times.

Answer 12

Japt, 6 bytes

1>U*Mr

Try it online!

Mr is equivalent to JS's Math.random. The rest is pretty obvious. I could probably add a number function that generates a random float between 0 and the number. When this happens, two bytes will be saved:

1>Ur    // Doesn't currently work

Alternate version:

1>Ð %U

Ð is equivalent to new Date(, and the Date object, when asked to convert to a number, becomes the current timestamp in milliseconds. Thus, this is entirely random, unless it is run multiple times per ms.

Answer 13

Marbelous, 21 bytes

}0    # takes one input n
--    # decrements n
??    # random value from range 0..n (inclusive)
=0?0  # push right if not equal to 0, fall through otherwise | convert to zero
++    # increment | no-op
{0//  # output | push left

I've taken 0 to be falsey and 1 to be truthy, though there is no real reason for that seeing as Marbelous doesn't really have an if. More Marbelousy would be output to {0 for true and {> for false. This would look like this:

}0
--
??
=0{>
{0

But I'm not sure that's valid.

Answer 14

APL, 6 3 bytes

+=?

This is a function train that takes an integer and returns 1 or 0 (APL's true/false). We generate a random integer from 1 to the input using ?, then check whether the input is equal to that integer. That results in a 1/input chance of true.

Saved 3 bytes thanks to Thomas Kwa!

Answer 15

PlatyPar, 3 bytes

#?!

#? gets a random number [0,n) where n is input. ! returns true if the number before it is 0, else it returns false.

Using more recent features that were implemented (but unfortunately for me not committed) before this question was asked I can get it down to 2 with ~! Try it online!

Answer 16

Java, 43 bytes

boolean b(int a){return a*Math.random()<1;}

Answer 17

C, 24 bytes

f(n){return!(rand()%n);}

Answer 18

><>, 27 + 3 for -v = 30 bytes

Here is a not-uniform-at-all solution where I mod N the sum of 15876 random picks of 0 or 1 :

0"~":*>:?vr%0=n;
1-$1+$^-1x

N must be input on the stack with -v flag, output is 0 for falsey and 1 for truthy.

A much smarter and uniform solution that work for 1/2^N instead :

4{:?!v1-}:">"$2p:"x"$3p:"^"$4p1+:">"$3p1+!
   ^1<
0n;
1n;>
 

For an input 3 you've got 1/8 chances of getting 1 and 7/8 of getting 0.

Explanation :

I append as much x as needed on the 4th line and surround them with directions so there is only two ways out of the x: either the falsey output or the next x. If all x go in the right direction, the last one will route to the truthy output.

For example for N=5, the final codespace is the following :

4{:?!v1-}:">"$2p:"x"$3p:"^"$4p1+:">"$3p1+!
   ^1<
0n; > > > > >
1n;>x>x>x>x>x>
    ^ ^ ^ ^ ^

Answer 19

Mathematica, 18 16 bytes

#RandomReal[]<1&

Basic solution. The unnamed Function creates a random number in [0, 1), multiplies it by its argument, and checks if it is still less than 1.

Answer 20

Python, 42 bytes

import random
lambda n:1>random.random()*n

Edit: Removed the time.time() answer because of the distribution.

Answer 21

TeaScript, 3 bytes

!N×

Try it here.

Explanation

 N  maps to Math.rand which is a utility function that returns an integer
    between `arg1` and `arg2` or `0` and `arg1` if only one argument is
    provided.
  × is expanded to `(x)`, where `x` is initialised with the value provided
    in the input boxes; × represents byte '\xd7'
!   negate the result, 0 results in true, anything else false

Answer 22

Fuzzy Octo Guacamole, 10 bytes

^-!_[0]1.|

Explanation:

^-!_[0]1.|

^          # Get input.
 -         # Decrement, so we can append <ToS> zeros and a 1 to the stack.
  !        # Set loop counter.
   _       # Pop, since we are done with the input.
    [      # Start loop
     0     # Push 0
      ]    # End for loop. We have pushed input-1 0s to the stack.
       1   # Push a single 1 to the stack.
        .  # Switch stacks
         | # Pick a random item from the inactive stack, which has n-1 falsy items and 1 truthy item, so the truthy probability is 1/n.
           # (implicit output)

Answer 23

Ruby, 14

Function

->n{rand(n)<1}

Program is slightly longer

p rand(gets.to_i)<1

Answer 24

Perl 6,  10  8 bytes

!(^*).pick
#  ^- The * is the argument

This code creates a Range from 0 up-to but excluding the input *. It then picks one at random and the ! returns True when it receives a 0.

1>*.rand
# ^- The * is the argument

This takes the input * and multiplies it by a random Num from 0..^1 then returns True if it was smaller than 1.

# store it in a lexical code variable for ease of use
my &code = 1>*.rand;

die "never dies here" unless code 1;

for ^8 { say code 4 }

False
True
False
False
False
True
False
False

Answer 25

Prolog (SWI), 24 bytes

Code:

p(N):-X is 1/N,maybe(X).

maybe(+P) is a function which succeeds with probability P and fails with probability 1-P

Example:

p(4).
false

p(4).
false

p(4).
true

Answer 26

PowerShell, 25 Bytes

!(Random -ma($args[0]--))

The Get-Random function when given a -Maximum parameter n returns a value from the range [0,n). We leverage that by subtracting 1 from our input $args[0], so we're properly zero-indexed, and get a random value. Precisely 1/nth of the time, this value will be 0, so when we Boolean-not it with ! it will return True. The other times will return False.

Answer 27

J, 3 bytes

0=?

This is a monadic fork that takes an argument on the right. Similarly to APL, ? generates a random integer; however, J arrays are zero-based. So we compare to 0 instead of to the input.

Answer 28

Minkolang 0.14, 7 bytes

1nH1=N.

Try it here.

Explanation

1          Pushes 1
 n         Takes number from input
  H        Pops b,a and pushes a random integer between a and b, inclusive
   1=      1 if equal to 1, 0 otherwise
     N.    Output as number and stop.

Answer 29

PHP, 22 bytes

<?=2>rand(1,$argv[1]);

Reads n from command line, like:

$ php probability.php 4

Outputs (false is cast to an empty string in PHP) or 1 (in case of true).

Answer 30

C#, 56 45 bytes

Thanks to, pinkfloydx33 it's 45 now.

bool b(int n){return new Random().Next(n)<1;}

Old 56 bytes

Generates random positive integer bigger or equal to 0 and smaller than n and checks if it's smaller than 1 and return comparison result.

bool a(int n){Random r=new Random();return r.Next(n)<1;}