precision of Math.random()

Question

What is the precision of JavaScript's Math.random() function?

`Math.random()` is the correct function. – tcooc Jul 27 '10 at 14:10 — tcooc, Jul 27 '10 at 14:10
Right, that's what I meant. Thanks! – jpc826 Jul 27 '10 at 14:24 — jpc826, Jul 27 '10 at 14:24

themaarten · Answer 1 · 2015-01-15T10:17:40.303

10

It's browser/JavaScript engine dependent.

The maximum possible precision is 52 bits, because Math.random returns a double-precision floating-point between 0 (inclusive) and 1 (exclusive). This maximum corresponds to roughly 16 decimals, see Sly1024's answer.

In practice, many browsers give a smaller precision. This seems to be the current state:

Firefox: the full 52 random bits.
V8/Chrome: 32 bits precision (some years ago it used to be as low as 30 bits, see https://codereview.chromium.org/1599019)
Safari: 32 bits
IE: ?

Run the following piece of code several times and you'll see that the trailing 21 bits are consistently 0 in Chrome and Safari.

console.log((Math.random() * Math.pow(2,53)).toString(2));

edited Jan 15 '15 at 10:17

answered Jan 14 '15 at 13:45

themaarten

101
1
5

IE11: 52 random bits, although the code above gives results like `111000000101010100010110011011110111000111010101110.1` sometimes. – Andrew Morton Jan 15 '17 at 16:58
Chrome seems to have upgraded their precision, I can no longer observe the 21 consistently 0 bits. – Daniel Vestøl Aug 11 '19 at 16:25
Chrome (and in extension, all Blink-based browsers and V8-based products, such as Electron-apps and Node.js) updated the Math.random() precision to 128 bits in 2015. – Aron Cederholm Sep 04 '20 at 07:55

score 10 · Accepted Answer · edited Jul 27 '10 at 14:36

10

Math.random() generates a floating point number of 16 decimal places greater than or equal to zero and less than 1.

edited Jul 27 '10 at 14:36

James

106,638
30
159
173

answered Jul 27 '10 at 14:07

krs1

1,105
7
16

6

Javascript floating point numbers are 64-bit IEEE 754 values, so it's not *exactly* 16 decimal places. – Pointy Jul 27 '10 at 14:16
@volatile: If your precision is not a power of 10. – Brian Jul 27 '10 at 14:52
@Pointy This is true, I was imprecise in my answer. bump. – krs1 Jul 27 '10 at 15:04
Do you have any references for this precision? – Fabian Jakobs Oct 15 '11 at 17:16

score 3 · Answer 3 · edited Nov 15 '12 at 17:14

3

According to wikipedia : Double-precision floating-point format the fraction part is 52 bits, and we know that the number will be between 0 and 1 (I think not including 1), so the exponent is -1, that leaves us with 52 random bits.

The 52 random bits give you 52 bit precision, which in base 10 is about

52*log10(2) ~= 15.653559774527022151114422525674 digits.

edited Nov 15 '12 at 17:14

Kaalras

1,063
10
13

answered Nov 15 '12 at 16:52

Sly1024

359
3
2

precision of Math.random()

3 Answers3

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