## Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

Assuming one needs to have a function returning false or true on each
call in pseudo-random order.and using JavaScript native Math.random()
method as the basis of the pseudo-randomness. Say the variants of such
function are:

var n = Math.round(Math.random());
return n ? true : false;
}

var n = Math.floor(Math.random()*2);
return (n==2) ? true : false;
}

Leaving obvious practical testing by platforms aside:

Is there are theoretical considerations that pseudo-randomness
(predictability) of either of above will be better or worse than the
other one? JavaScript Kit site claims that the second bits first:
http://www.yqcomputer.com/
but they don't disclose the underlaying reasoning.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

Why not { return Math.random() >= 0.5; } ?
Jorge.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

I have no idea. The linked source at
http://www.yqcomputer.com/
claims this: "Some of you may be curious as to why Math.floor(),
instead of Math.round(), is used in the above code. While both
successfully round off its containing parameter to an integer within
the designated range, Math.floor does so more "evenly", so the
resulting integer isn't lopsided towards either end of the number
spectrum. In other words, a more random number is returned using
Math.floor()."

It may be some actual behavior or an author's fantasy - no arguments
are given on the page. From the Math.round and Math.floor methods
descriptions:

https://developer.mozilla.org/En/Core_JavaScript_1.5_Reference/Objects/Math/Round
Returns the value of a number rounded to the nearest integer.

https://developer.mozilla.org/En/Core_JavaScript_1.5_Reference/Objects/Math/Floor
Returns the largest integer less than or equal to a number.

I am failing to grasp the exact difference between of them. I only
assume that the only place of rounding inequality results could be in
"border cases" like 0.5xxxxxx etc. so with .5 being the first
fractional digit. Respectively if such inequality really exists then
there must be something with pseudo-random generation in whole or in
how it is implemented in Math.random() that would suggest 0.5xxxxx or
0.5 generation being lesser random than other results. Or it is just
another urban legend.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

Another direction to look for is that the computer pseudo-random
generation operates in non-closed properly set [0,1[ with the upper
border not included so the result can be 0 but never 1. That shifts
the predictability pattern down toward 0. This is why actually why
Shannon's Clairvoyant can quickly tell for any sequence is it's truly
random or pseudo-random. btw Wiki's http://www.yqcomputer.com/
claim that "No efficient algorithm can distinguish (with significant
advantage) between a function chosen randomly from the PRF family" is
a complete b.s. but I am too lazy to argue with the entire ACM. Let
them believe what they want to believe.

So it might be that Math.floor somehow "re-balance" the outcome making
it lesser predictable. I just don't see how could it be. I am really
puzzled... Maybe I should get all JavaScript stuff out and post it as
a purely math question to sci.math

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

Posted at sci.math as a mathematical problem:
http://www.yqcomputer.com/

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

<snip>
<snip>

You leave practical testing aside far to often in your posted code. Any
reasonable testing (or your just understanding the methods/operations
ever returns false. Thus, it fails to satisfy your "returning false or
true on each call in pseudo-random order".

Yes, but mostly because the obvious bugs in the second prevent it from
doing anything useful at all.

The subject of the comments on that page is the choice of the use of
Math.floor over Math.round (where Math.round is commonly used in example
javascript random number generators found on the Internet).

Math.random returns a (pseudo-random) number that is in the range zero
to less than one (i.e. anything non-negative that is smaller than one).
If you multiply that number by an integer you will get a result that is
in the range from zero to less than that number. If two were taken as an
example of such a number (i.e. - (Math.random * 2) - the result would be
a number in the range zero to less than two.

If you apply - Math.round - to all the numbers in the range zero to less
than two the range zero to <0.5 (a quarter of the total range) would
result in zero, between 0.5 and <1.5 (half the total range) would result
in one, and 1.5 to <2 would result in two. So, using Math.round, a
random number in the 0 to <2 range has a 25% chance of coming our zero,
a 50% chance of coming out one and a 25% chance of coming out 2. This is
not an even distribution. (Extending this to multiplying by any positive
integer; it is the values at the two extremes of output that end up
being half as likely in the output as any numbers in between, however if
that integer were one then there would be no numbers in between the
extremes of the range and so then the distribution between zero and one
would be equal.)

If you apply - Math.floor - to numbers in the range 0 to <2 then the
range 0 to <1 (half the original range) result in 0 and the range 1 to
<2 (the other half of the original range) result in 1; a 50/50
distribution.

Richard.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

Given that there has been a prevalence of examples of javascript (so-
called) random number generators posted to the web that did use
Math.round and did then produce a non-even distribution of numbers in
their output, a fair number of readers of that article may well be

This bit isn't actually true as if you did a direct substitute of -
Math.round - for - Math.floor - then the range of the output would
increase by one.

"Lopsided" is probably inappropriate as well, as the output
distribution following a substitution of - Math.random - for -
Math.floor - is still symmetrical.

Most likely an overlay hurried explanation of a common fault in
javascript authoring, with a couple of mistakes getting in the way of
making the point.

Then you are the author of:-

<URL: http://www.yqcomputer.com/

- so we know the subtleties of rounding in javascript don't have to
get that subtle before you cannot comprehend them.

<snip>

Don't waste you time in assuming. The best you will do is invent an
elaborate fantasy explanation. Just wait for someone to tell you the
answer, and then avoid ever trying to put it into your own words.

Richard.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

n May 1, 3:16m, "Richard Cornford" < XXXX@XXXXX.COM >
wrote:

Yeah... My mind was distracted a lot by the binary trees observations,
sorry.

Right. Similar answer from sci.math :

<quote>

Let X be uniformly distributed in [0,1). Then floor(X*11) takes
values
0,1,...,10 each with probability 1/11 ... that is what the page means
by "even". However round(X*10) takes value 0 with probability 1/20,
values 1,...,9 each with probability 1/10 and value 10 with
probability
1/20. Not "even" according to the page.
</quote>

So the question is then
a) if it is possible to use Math.floor for a pseudo-random input to
get binary output (1/0, true/false)
b) if so than will it be more even probability for output than for
Math.round(Math.random()) or (Math.random >= 0.5)

.... or Math.floor(Math.random()*N) benefits appear only for ternary
and wider ranges "0 or 1 or 2", "0 or 1 or 2 or 3" etc. ?

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

Not so. A probability theory outcome: see my answer to Richard
Cornford

First of all that was about IEEE-754 FP-DP rounding and calculation
vs. top level methods - not about the probability theory and how does
it apply to JavaScript Math methods. And yes, I couldn't grasp how
Math.floor would be more "even probability-friendly" than Math.round
until I got explanations of that. I don't see you grasping it on the
quoted at the top was wrong.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

The thousandth excuse for making the same mistake isn't any more
convincing than the second.

<snip>
<snip>
<snip>

Unsurprisingly.

Strange question; of course it is possible.

Properly implemented, there should be no difference. A bias towards the
numbers that are not at the ends of the range of output does not apply
when all of the possible outputs are at the ends of their range.

The issue only applies to rages with more than two values in them. But
still, Jorge's remains the better implementation for javascript (FPU
handled math operation over a method call), and it uses neither
Math.round nor Math.floor.

Richard.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

On May 1, 4:02m, "Richard Cornford" < XXXX@XXXXX.COM >

Then would it be properly to state that in order to have the least
compromised pseudo-random sequence of integers from a set of two
elements one should use
return (Math.random() >= 0.5) ? _this : _that;
and for all sets with more than two elements one should use
return Math.floor( n * Math.random() );
where the range is [0, n-1]

Would it be appropriate to correct this in the FAQ
http://www.yqcomputer.com/ #randomNumber
and maybe add a short math explanation note based on the sci.math post
so not leaving reader to wonder why the hey floor() and what's wrong
with round() ?

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

VK < XXXX@XXXXX.COM > writes:

As stated elsewhere, this should read
return (n == 1) ? true : false;
or, preferably,
return n == 1;

No, they are (obviously?) exactly identical. They map exactly the same
results of Math.random() to true and false respectively.
In both cases, a value in the range [0..0.5[ is mapped to false
and a value in the range [0.5..1[ is mapped to true.

I guess their point is that to generate an integer in the range [0..n[,
Math.floor(Math.random() * n)
is better, in general, than
Math.round(Math.random() * (n - 1))
... which is pretty old news (not that people still don't bungle it
regularly, but it's embarrasing every time it happens).

The funny thing is that for n = 2, the unevenness of using the Math.round

--
Lasse Reichstein Holst Nielsen
'Javascript frameworks is a disruptive technology'

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

Lasse Reichstein Nielsen wrote on 01 mei 2010 in comp.lang.javascript:

or:

return !n-1;

--
Evertjan.
The Netherlands.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

> > eturn (n == 1) ? true : false; >>>> or, preferably, >>>> eturn n == 1;> >> > or:> >> > return !n-1;

return !n;
Jorge.

### Math.random() and Math.round(Math.random()) and Math.floor(Math.random()*2)

VK :

Of course not. It is quite difficult to define "random" and "pseudo-
random" in a precise and satisfactory way, but an essential part of
any reasonable definition is that there is no quick way to tell the
difference given the sequences. What you call "Shannon's Clairvoyant"
(I don't know it by that name from any other source than you) is a
rather clever illustration of the shortcomings of *humans* as sources
of random or pseudo-random sequences, nothing more. Computers are
much better, and easily defeat your "clairvoyant".

No, it is a sensible definition.

Perhaps some lingering remains of sanity, as well, like when you
chickened out of a bet you proposed yourself on the subject. Some
part of you may dimly realise that some people actually know more