[Math] Understanding “randomness”
my answer to all random number questions is this
So I guess both methods are as random although my gutfeel would say that
rand() * rand() is less random because it would seed more zeroes. As soon as one
0, the total becomes
I can't get my head around this, which is more random?
rand() * rand()
I´m finding it a real brain teaser, could you help me out?
Intuitively I know that the mathematical answer will be that they are equally random, but I can't help but think that if you "run the random number algorithm" twice when you multiply the two together you'll create something more random than just doing it once.
Most rand() implementations have some period. I.e. after some enormous number of calls the sequence repeats. The sequence of outputs of
rand() * rand() repeats in half the time, so it is "less random" in that sense.
Also, without careful construction, performing arithmetic on random values tends to cause less randomness. A poster above cited "
rand() ..." (k times, say) which will in fact tend to k times the mean value of the range of values
rand() returns. (It's a random walk with steps symmetric about that mean.)
Assume for concreteness that your rand() function returns a uniformly distributed random real number in the range [0,1). (Yes, this example allows infinite precision. This won't change the outcome.) You didn't pick a particular language and different languages may do different things, but the following analysis holds with modifications for any non-perverse implementation of rand(). The product
rand() * rand() is also in the range [0,1) but is no longer uniformly distributed. In fact, the product is as likely to be in the interval [0,1/4) as in the interval [1/4,1). More multiplication will skew the result even further toward zero. This makes the outcome more predictable. In broad strokes, more predictable == less random.
Pretty much any sequence of operations on uniformly random input will be nonuniformly random, leading to increased predictability. With care, one can overcome this property, but then it would have been easier to generate a uniformly distributed random number in the range you actually wanted rather than wasting time with arithmetic.
Use a linear feedback shift register (LFSR) that implements a primitive polynomial.
The result will be a sequence of 2^n pseudo-random numbers, ie none repeating in the sequence where n is the number of bits in the LFSR .... resulting in a uniform distribution.
Use a "random" seed based on microsecs of your computer clock or maybe a subset of the md5 result on some continuously changing data in your file system.
For example, a 32-bit LFSR will generate 2^32 unique numbers in sequence (no 2 alike) starting with a given seed. The sequence will always be in the same order, but the starting point will be different (obviously) for a different seeds. So, if a possibly repeating sequence between seedings is not a problem, this might be a good choice.
I've used 128-bit LFSR's to generate random tests in hardware simulators using a seed which is the md5 results on continuously changing system data.
Some things about "randomness" are counter-intuitive.
Assuming flat distribution of
rand(), the following will get you non-flat distributions:
- high bias:
- bias peaking in the middle:
(rand(range) + rand(range))/2
range - sqrt(rand(range^2))
There are lots of other ways to create specific bias curves. I did a quick test of
rand() * rand() and it gets you a very non-linear distribution.
Multiplying numbers would end up in a smaller solution range depending on your computer architecture.
If the display of your computer shows 16 digits
rand() would be say 0.1234567890123
multiplied by a second
rand(), 0.1234567890123, would give 0.0152415 something
you'd definitely find fewer solutions if you'd repeat the experiment 10^14 times.
The obligatory xkcd ...
Consider you have a simple coin flip problem where even is considered heads and odd is considered tails. The logical implementation is:
rand() mod 2
Over a large enough distribution, the number of even numbers should equal the number of odd numbers.
Now consider a slight tweak:
rand() * rand() mod 2
If one of the results is even, then the entire result should be even. Consider the 4 possible outcomes (even * even = even, even * odd = even, odd * even = even, odd * odd = odd). Now, over a large enough distribution, the answer should be even 75% of the time.
I'd bet heads if I were you.
This comment is really more of an explanation of why you shouldn't implement a custom random function based on your method than a discussion on the mathematical properties of randomness.
OK, so I will try to add some value to complement others answers by saying that you are creating and using a random number generator.
Random number generators are devices (in a very general sense) that have multiple characteristics which can be modified to fit a purpose. Some of them (from me) are:
- Entropy: as in Shannon Entropy
- Distribution: statistical distribution (poisson, normal, etc.)
- Type: what is the source of the numbers (algorithm, natural event, combination of, etc.) and algorithm applied.
- Efficiency: rapidity or complexity of execution.
- Patterns: periodicity, sequences, runs, etc.
- and probably more...
In most answers here, distribution is the main point of interest, but by mix and matching functions and parameters, you create new ways of generating random numbers which will have different characteristics for some of which the evaluation may not be obvious at first glance.
The accepted answer is quite lovely, but there's another way to answer your question. PachydermPuncher's answer already takes this alternative approach, and I'm just going to expand it out a little.
The easiest way to think about information theory is in terms of the smallest unit of information, a single bit.
In the C standard library,
rand() returns an integer in the range 0 to
RAND_MAX, a limit that may be defined differently depending on the platform. Suppose
RAND_MAX happens to be defined as
2^n - 1 where
n is some integer (this happens to be the case in Microsoft's implementation, where
n is 15). Then we would say that a good implementation would return
n bits of information.
rand() constructs random numbers by flipping a coin to find the value of one bit, and then repeating until it has a batch of 15 bits. Then the bits are independent (the value of any one bit does not influence the likelihood of other bits in the same batch have a certain value). So each bit considered independently is like a random number between 0 and 1 inclusive, and is "evenly distributed" over that range (as likely to be 0 as 1).
The independence of the bits ensures that the numbers represented by batches of bits will also be evenly distributed over their range. This is intuitively obvious: if there are 15 bits, the allowed range is zero to
2^15 - 1 = 32767. Every number in that range is a unique pattern of bits, such as:
and if the bits are independent then no pattern is more likely to occur than any other pattern. So all possible numbers in the range are equally likely. And so the reverse is true: if
rand() produces evenly distributed integers, then those numbers are made of independent bits.
So think of
rand() as a production line for making bits, which just happens to serve them up in batches of arbitrary size. If you don't like the size, break the batches up into individual bits, and then put them back together in whatever quantities you like (though if you need a particular range that is not a power of 2, you need to shrink your numbers, and by far the easiest way to do that is to convert to floating point).
Returning to your original suggestion, suppose you want to go from batches of 15 to batches of 30, ask
rand() for the first number, bit-shift it by 15 places, then add another
rand() to it. That is a way to combine two calls to
rand() without disturbing an even distribution. It works simply because there is no overlap between the locations where you place the bits of information.
This is very different to "stretching" the range of
rand() by multiplying by a constant. For example, if you wanted to double the range of
rand() you could multiply by two - but now you'd only ever get even numbers, and never odd numbers! That's not exactly a smooth distribution and might be a serious problem depending on the application, e.g. a roulette-like game supposedly allowing odd/even bets. (By thinking in terms of bits, you'd avoid that mistake intuitively, because you'd realise that multiplying by two is the same as shifting the bits to the left (greater significance) by one place and filling in the gap with zero. So obviously the amount of information is the same - it just moved a little.)
Such gaps in number ranges can't be griped about in floating point number applications, because floating point ranges inherently have gaps in them that simply cannot be represented at all: an infinite number of missing real numbers exist in the gap between each two representable floating point numbers! So we just have to learn to live with gaps anyway.
As others have warned, intuition is risky in this area, especially because mathematicians can't resist the allure of real numbers, which are horribly confusing things full of gnarly infinities and apparent paradoxes.
But at least if you think it terms of bits, your intuition might get you a little further. Bits are really easy - even computers can understand them.
It's not exactly obvious, but
rand() is typically more random than
rand()*rand(). What's important is that this isn't actually very important for most uses.
But firstly, they produce different distributions. This is not a problem if that is what you want, but it does matter. If you need a particular distribution, then ignore the whole “which is more random” question. So why is
rand() more random?
The core of why
rand() is more random (under the assumption that it is producing floating-point random numbers with the range [0..1], which is very common) is that when you multiply two FP numbers together with lots of information in the mantissa, you get some loss of information off the end; there's just not enough bit in an IEEE double-precision float to hold all the information that was in two IEEE double-precision floats uniformly randomly selected from [0..1], and those extra bits of information are lost. Of course, it doesn't matter that much since you (probably) weren't going to use that information, but the loss is real. It also doesn't really matter which distribution you produce (i.e., which operation you use to do the combination). Each of those random numbers has (at best) 52 bits of random information – that's how much an IEEE double can hold – and if you combine two or more into one, you're still limited to having at most 52 bits of random information.
Most uses of random numbers don't use even close to as much randomness as is actually available in the random source. Get a good PRNG and don't worry too much about it. (The level of “goodness” depends on what you're doing with it; you have to be careful when doing Monte Carlo simulation or cryptography, but otherwise you can probably use the standard PRNG as that's usually much quicker.)
Oversimplification to illustrate a point.
Assume your random function only outputs
random() is one of
random()*random() is one of
You can clearly see that the chances to get a
0 in the second case are in no way equal to those to get a
When I first posted this answer I wanted to keep it as short as possible so that a person reading it will understand from a glance the difference between
random()*random(), but I can't keep myself from answering the original ad litteram question:
Which is more random?
(random()+1)/2 or any other combination that doesn't lead to a fixed result have the same source of entropy (or the same initial state in the case of pseudorandom generators), the answer would be that they are equally random (The difference is in their distribution). A perfect example we can look at is the game of Craps. The number you get would be
random(1,6)+random(1,6) and we all know that getting 7 has the highest chance, but that doesn't mean the outcome of rolling two dice is more or less random than the outcome of rolling one.
As others already pointed out, this question is hard to answer since everyone of us has his own picture of randomness in his head.
That is why, I would highly recommend you to take some time and read through this site to get a better idea of randomness:
To get back to the real question. There is no more or less random in this term:
both only appears random!
In both cases - just rand() or rand() * rand() - the situation is the same: After a few billion of numbers the sequence will repeat(!). It appears random to the observer, because he does not know the whole sequence, but the computer has no true random source - so he can not produce randomness either.
e.g.: Is the weather random? We do not have enough sensors or knowledge to determine if weather is random or not.
There is no such thing as more random. It is either random or not. Random means "hard to predict". It does not mean non-deterministic. Both random() and random() * random() are equally random if random() is random. Distribution is irrelevant as far as randomness goes. If a non-uniform distribution occurs, it just means that some values are more likely than others; they are still unpredictable.
Since pseudo-randomness is involved, the numbers are very much deterministic. However, pseudo-randomness is often sufficient in probability models and simulations. It is pretty well known that making a pseudo-random number generator complicated only makes it difficult to analyze. It is unlikely to improve randomness; it often causes it to fail statistical tests.
The desired properties of the random numbers are important: repeatability and reproducibility, statistical randomness, (usually) uniformly distributed, and a large period are a few.
Concerning transformations on random numbers: As someone said, the sum of two or more uniformly distributed results in a normal distribution. This is the additive central limit theorem. It applies regardless of the source distribution as long as all distributions are independent and identical. The multiplicative central limit theorem says the product of two or more independent and indentically distributed random variables is lognormal. The graph someone else created looks exponential, but it is really lognormal. So random() * random() is lognormally distributed (although it may not be independent since numbers are pulled from the same stream). This may be desirable in some applications. However, it is usually better to generate one random number and transform it to a lognormally-distributed number. Random() * random() may be difficult to analyze.
For more information, consult my book at www.performorama.org. The book is under construction, but the relevant material is there. Note that chapter and section numbers may change over time. Chapter 8 (probability theory) -- sections 8.3.1 and 8.3.3, chapter 10 (random numbers).