Concept of Math.floor(Math.random() * 5 + 1), what is the true range and why?

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南方客
南方客 2021-01-05 15:39

By multiplying the random number (which is between 0 and 1) by 5, we make it a random number between 0 and 5 (for example, 3.1841). Math.floor() rounds this number down to a

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  • 2021-01-05 16:10

    15.8.2.14 Math.random from the ES5 spec,

    Returns a Number value with positive sign, greater than or equal to 0 but less than 1, chosen randomly or pseudo randomly with approximately uniform distribution over that range, using an implementation-dependent algorithm or strategy. This function takes no arguments.

    So,

    x = Math.random(); // 0 ≤ x < 1
    y = x * 5;         // 0 ≤ y < 5
    z = y + 1;         // 1 ≤ z < 6
    i = Math.floor(z); // 1 ≤ i ≤ 5, i ∈ ℤ, ℤ integers
    

    Which means

    i ∈ {1, 2, 3, 4, 5}
    
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  • 2021-01-05 16:12

    Math.Random() returns a number between 0 and 1, excluding 1.

    So when you multiply it with 5, you get a number between 0 and 5 but not 5.

    Math.floor() on this number rounds down to a whole number.

    So numbers you will get are either 0, 1, 2, 3 or 4.

    Adding 1 to this range gives you a number in [1, 2, 3, 4, 5].

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  • 2021-01-05 16:26

    From the Mozilla Developer Networks' documentation on Math.random():

    The Math.random() function returns a floating-point, pseudo-random number in the range [0, 1) that is, from 0 (inclusive) up to but not including 1 (exclusive).

    Here are two example randomly generated numbers:

    Math.random() // 0.011153860716149211
    Math.random() // 0.9729151880834252
    

    Because of this, when we multiply our randomly generated number by another number, it will range from 0 to a maximum of 1 lower than the number being multiplied by (as Math.floor() simply removes the decimal places rather than rounding the number (that is to say, 0.999 becomes 0 when processed with Math.floor(), not 1)).

    Math.floor(0.011153860716149211 * 5) // 0
    Math.floor(0.9729151880834252 * 5)   // 4
    

    Adding one simply offsets this to the value you're after:

    Math.floor(0.011153860716149211 * 5) + 1 // 1
    Math.floor(0.9729151880834252 * 5) + 1   // 5
    
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  • 2021-01-05 16:28

    Note that:

    • 0 <= Math.random() **<** 1
    • Math.floor(x.yz) = x

    And therefore, the number given is a integer in the interval:

    x = Math.floor((0..0.999999999) * 5 + 1)
    x = (0..4) + 1
    
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