How to test that a method generating random numbers does generate random numbers 
The links in comment by mob provide much detail on how difficult it is to
assert that a data source is random.
For your unit test, you may be better off using srand in the test setup
(making this technically a "white box" test), and relying on the fact that
behaviour of rand() is well known. This is normal practice as far as I
know, unless you want to test a PRNG or entropy source that you have
written.

How to generate 68 digit random numbers without colliding with previous generated numbers in time critical fashion? 
The easiest thing to do is generate random numbers and generating a new
random id if you get a duplicate. This works because with your figures the
probability of getting a duplicate is pretty small.
If that doesnt convince you, you can think of many schemes that guarantee
mathematically that the numbers will be unique and still look random, but
it gets complex.

java random numbers generator which generate twenty four numbers 
Here's the easiest way I can think of:
@Test
public void random24Numbers() {
String random = RandomStringUtils.random(24, false, true);
System.out.println(random);
}
This uses RandomStringUtils.random. The first parameter is the length, the
second says, "no letters". The third says, "give me numbers". Here's an
example output:
564266161898194666197908
Yes, it's a String, but I'm going to assume you know how to convert a
String into a number.

PHP generate two random numbers without repeating eachother neither previous numbers 
I am just shooting in the dark here, but if the numbers are in the same
call i.e $ticket_number1 and $ticket_number2 are same in one call to the
page, you can fix this by telling the function to reject an already chosen
number
function getTicket($exclude=1){
$count=1;
while($count){
$ticket_number = mt_rand(01990, 32000);
$query = "SELECT ticket1 FROM `tickets`
WHERE `ticket1` == '$ticket_number' or `ticket2` == '$ticket_number'";
$count = mysql_num_rows(mysql_query($query));
if(!$count)
$count=$ticket_number==$exclude;
}
return $ticket_number;
}
$ticket_number1 = getTicket();
$ticket_number2 = getTicket($ticket_number1);

Generate 30 Random Numbers from a given sum 
If you want to have 23 random numbers with sum of 2345, you can use this
code:
int sum = 2345;
int nums = 23;
int max = sum / nums;
Random rand = new Random();
int newNum = 0;
int[] ar = new int[23];
for (int i = 0; i < nums1; i++) {
newNum = rand.Next(max);
ar[i] = newNum;
sum= newNum;
max = sum / (numsi1);
}
ar[nums  1] = sum;
It will give you:

How do I generate a sum of random numbers? 
In pure Ruby
sum = [*0..70].sample(5).inject(&:+) until (140..220) === sum
if you use ActiveSupport then you can use #sum instead of inject(&:+).

PHP Generate x amount of random odd numbers within a range 
Generate x integer values over half the range, and for each value double it
and add 1.
ANSWERING REVISED QUESTION: 1) Generate a list of candidates in range,
shuffle them, and then take the first x. Or 2) generate values as per my
original recommendation, and reject and retry if the generated value is in
the list of already generated values.
The first will work better if x is a substantial fraction of the range, the
latter if x is small relative to the range.
ADDENDUM: Should have thought of this approach earlier, it's based on
conditional probability. I don't know php (I came at this from the
"random" tag), so I'll express it as pseudocode:
generate(x, upper_limit)
loop with index i from upper_limit downto 1 by 2
p_value = x / floor((i + 1) / 2)
if rand <= p_value

How would I generate a random sample in python with numbers between 1x? 
sample = random.sample(range(classNumber+1), classNumber+1)
Here's your problem. This sample makes no sense. First, you didn't set a
lower bound on the range, so it defaults to 0 instead of 1. Second, if you
just increase the lower bound, you'll be trying to draw more numbers from
the range than the range actually has. There's no part of your code that
says random.sample(range(1,16), classNumber). Is that what this line was
supposed to say?

How to generate 100 random 3 digit numbers in java? 
Try for loop
for(int i=0;i<100;i++)
{
int pick = rand.nextInt(900) + 100;
System.out.println(pick);
}

Trying to generate a series of unique random numbers 
Instead of picking random integers, shuffle a list and pick the first two
items:
import random
choices = ['Candy', 'Steak', 'Vegetables']
random.shuffle(choices)
item1, item2 = choices[:2]
Because we shuffled a list of possible choices first, then picked the first
two, you can guarantee that item1 and item2 are never equal to one another.
Using random.shuffle() leaves the option open to do something with the
remaining choices; you only have 1 here, but in a larger set you can
continue to pick items that have so far not been picked:
choices = list(range(100))
random.shuffle(choices)
while choices:
if input('Do you want another random number? (Y/N)' ).lower() == 'n':
break
print(choices.pop())
would give you 100 random numbers without repeating.
If all you need is a

Generate random numbers from a three dimensional probability distribution 
If your spatial coordinate system is discrete, treat it as a univariate
generation problem to generate triplets.
If you're talking about a continuous distribution you probably want to use
conditional probability. In principle you should be able to derive the
marginal distribution of the X's, the conditional distribution of Y given
X, and the conditional distribution of Z given X & Y. Then generate X,
Y, and Z sequentially from their marginal and conditional distributions.
In practice this may be quite challenging.
Addendum
Perhaps the easiest scheme would be to generate a multidimensional normal
vector of length 3. This would give the highest density near the origin
and taper off symmetrically in all directions. You could displace it with
a mean vector if the density is highest

Generate more than one random numbers using native or builtin function 
try something like this
Create an ArrayList and add your random number and while adding check if
ArrayList already contains the number or not.
Eg:
ArrayList<Integer> numbers = new ArrayList<Integer>();
while (numbers.size()<=YOUR_MAX_SIZE)
{
int randomInteger = ( int )( Math.random() * 9999 );
if (!numbers.contains(randomInteger)) {
{
numbers.add(randomInteger);
}
}

Generate a specific amount of random numbers that add up to a defined value 
You will have no problem meeting any two out of your three constraints, but
all three might be a problem. As you note, the standard way to generate N
random numbers that add to a sum is to generate N1 random numbers in the
range of 0..sum, sort them, and take the differences. This is basically
treating your sum as a number line, choosing N1 random points, and your
numbers are the segments between the points.
But this might not be compatible with constraints on the numbers
themselves. For example, what if you want 10 numbers that add to 1000, but
each has to be less than 100? That won't work. Even if you have ranges
that are mathematically possible, forcing compliance with all the
constraints might mean sacrificing uniformity or other desirable
properties.
I suspect the only way to

How to generate random double numbers with high precision in C++? 
In a typical system, RAND_MAX is 2311 or something similar to that. So
your "precision" from using a method like:L
double r = rand()/RAND_MAX;
would be 1/(2<sup>31</sup)1  this should give you 89 digits
"precision" in the random number. Make sure you print with high enough
precision:
cout << r << endl;
will not do. This will work better:
cout << fixed << sprecision(15) << r << endl;
Of course, there are some systems out there with much smaller RAND_MAX, in
which case the results may be less "precise"  however, you should still
get digits down in the 912 range, just that they are more likely to be
"samey".

How do I generate random numbers from a column, without duplicates  SQL Server 
You could try using newid():
INSERT INTO #RandomUsers
Select TOP 10 userId From Profile WHERE userId <> @UserID
ORDER BY newid()
This way you get ten different users that are not the one you passed in
parameter. But I don't know how you can get really good randomness in SQL
Server.
Edit:
I think you can also use tablesample() for this kind of issue, but I don't
really know how to use this.
By the way, if a hole appears in the id sequence (like user 34 is deleted),
you could still pick 34 with your method, while methods taking a sample
from the table directly will still work.

How do I generate random numbers in an array that add up to a defined total? 
A good way to achieve uniformity is, for example, to fill up a = 15 units
into an 8 element array:
Put 1 in each element in the array as this is your requirement, you have
now 7 values left to distribute
Roll a random number between 0 and the max index of the array, and add 1 to
that element, and subtract 1 from 7. Do this until 7 goes down to zero.
In this way, you'll meet your minimum conditions by having each element
have minimum value 1. Then you distribute the remaining totals in a
completely random way.

Generate a random number from a density object (or more broadly from a set of numbers) 
If all you need is to draw values from your existing pool of numbers, then
sample is the way to go.
If you want to draw from the presumed underlying distribution, then use
density , and fit that to your presumed distribution to get the necessary
coefficients (mean, sd, etc.), and use the appropriate R distribution
function.
Beyond that, I'd take a look at Chapter7.3 ("rejection method") of
Numerical Recipes in C for ways to "selectively" sample according to any
distribution. The code is simple enough to be easily translated into R .
My bet is someone already has done so and will post a better answer than
this.

How to generate scaleindependent random floating point numbers? 
You are correct that your approach doesn't return some numbers. For
example, there is no floatingpoint number between 1.0 and
1.0000000000000002, but 10**1.0000000000000002 is 10.000000000000005, and
there are two numbers between 10.0 and 10.000000000000005:
10.000000000000002 and 10.000000000000004. Those two numbers will never be
returned by your algorithm.
But you can cheat and use Decimal to exponentiate with greater precision:
>>> float(10 ** Decimal('1'))
10.0
>>> float(10 ** Decimal('1.0000000000000001'))
10.000000000000002
>>> float(10 ** Decimal('1.00000000000000015'))
10.000000000000004
>>> float(10 ** Decimal('1.0000000000000002'))
10.000000000000005
So, arbitrary needs to generate random Decimal exponents of sufficient
precision and use

Very nonrandom factor of Math.random when filling an array with random numbers 
If the length is even, you add all the numbers once and subtract all of
them once – of course, the result is 0.
Example with length==6:
/*i==0*/ sum += rand[0] + rand[5];
/*i==1*/ sum = rand[1] + rand[4];
/*i==2*/ sum += rand[2] + rand[3];
/*i==3*/ sum = rand[3] + rand[2];
/*i==4*/ sum += rand[4] + rand[1];
/*i==5*/ sum = rand[5] + rand[0];
Did you try any uneven lengths?

Given 4 numbers of array of 1 to 10 elements. Find 3 numbers whose sum can generate all the four numbers? 
Since the solution contains only 3 numbers which are in 1..10 range, brute
force
is an effective algorithm here (at most 1000 possibilities to check even in
naive implementation). So C# code could be
public static int[] BruteForce(int[] data) {
HashSet<int> hs = new HashSet<int>();
for (int x = 1; x <= 10; ++x)
for (int y = x; y <= 10; ++y)
for (int z = y; z <= 10; ++z) {
hs.Clear();
for (int i = 0; i < data.Length; ++i)
hs.Add(data[i]);
hs.Remove(x);
hs.Remove(y);
hs.Remove(z);
hs.Remove(x + y);
hs.Remove(x + z);
hs.Remove(y + z);
hs.Remove(x + y + z);
if (hs.Count <= 0)
return new int[] { x, y, z }; // < Solution
}
return new in

Java arraylist. I need to pass random numbers to an array list and pass them to another class to check those numbers against another set of numbers 
You can always transform the random int to a string by using
Integer.toString() before inserting into your array list.
You can convert the String back to int using Integer.parseInt()
E.g.
for (int i = 0 ; i < 3 ; i++)
{
locations.add(Integer.toString((int)(Math.random() * 5));
}

Adding the sum of all the random numbers in a random number generator 
You'll want to put this inside of the for loop:
sum = sum + n;
(or sum += n;)
Then after the loop, the average will be the sum divided by 100.
Also you can take out the min and max part, unless that is needed for
something else.

How many random numbers can I get with .nextInt() from Random class in Java? 
StackOverflowError is what you get when your recursion runs too deep.
Solution: convert it to an iterative version.
Way to go from recursion to iteration

Generate Random Number Using Random Series 
$series = array(
'BT4YX***1',
'BT4YX**X1',
'BT4Y**X*1',
'BT4YXXXX1',
);
$result = preg_replace_callback(
'/[X*]/',
function ($matches) {
switch($matches[0]) {
case '*' : return range(0,9)[array_rand(range(0,9))];
case 'X' : return range('A','E')[array_rand(range('A','E'))];
}
},
$series
);
var_dump($result);
EDIT
Requires PHP >= 5.4.0; but if you refactored the array dereferencing, you
could use it with 5.3.0 or above, e.g.
$series = array(
'BT4YX***1',
'BT4YX**X1',
'BT4Y**X*1',
'BT4YXXXX1',
);
$result = preg_replace_callback(
'/[X*]/',
function ($matches) {
switch($matches[0]) {
case '*' :
$range = range(0,9);

Find random numbers in a given range with certain possible numbers excluded 
If you have no constraints at all, i guess this is the easiest way: create
an array containing the valid values, a[0]...a[m] . Return
a[rand(0,...,m)].
If you don't want to create an auxiliary array, but you can count the
number of exceptions e and of elements n in the original range, you can
simply generate a random number r=rand(0 ... ne), and then find the valid
element with a counter that doesn't tick on exceptions, and stops when it's
equal to r.

How to specify upper and lower limits when using numpy.random.normal 
It sounds like you want a truncated normal distribution.
Using scipy, you could use scipy.stats.truncnorm to generate random
variates from such a distribution:
import matplotlib.pyplot as plt
import scipy.stats as stats
lower, upper = 3.5, 6
mu, sigma = 5, 0.7
X = stats.truncnorm(
(lower  mu) / sigma, (upper  mu) / sigma, loc=mu, scale=sigma)
N = stats.norm(loc=mu, scale=sigma)
fig, ax = plt.subplots(2, sharex=True)
ax[0].hist(X.rvs(10000), normed=True)
ax[1].hist(N.rvs(10000), normed=True)
plt.show()
The top figure shows the truncated normal distribution, the lower figure
shows the normal distribution with the same mean mu and standard deviation
sigma.

SPSS: generating random data on a normal distribution 
You are using the old form of the normal RNG. In that function, the
parameter is sigma, not the mean.
Use RV.NORMAL(mean, std deviation).
The RV.xxx notation can be used for all the random number generators,
including RV.UNIFORM(min, max).

Random() not generating new random numbers 
You haven't shown your RandomizePosition() method, but it is likely that
you are calling randomise with the same time seed in quick succession, i.e.
creating the Random object inside your RandomizePosition() method.
To stop this happening create the Random object outside your
RandomizePosition() method.
Update: you have now shown the code for RandomizePosition() and that is
indeed the problem.
As mentioned by @dlev, the reason you see the expected behavior when you
step through the code is that you are causing enough time to elapse for
your newly created Random() object to use a new seed (since by default it
is based on the current time) .

How to generate a random number from 1 to 100 only once? 
check out this for more detail
class NonRepeatedPRNG {
private final Random rnd = new Random();
private final Set<Integer> set = new HashSet<>();
public int nextInt() {
for (;;) {
final int r = rnd.nextInt();
if (set.add(r)) return r;
}
}
}

PHP generate random usernames 
Place the random number generator and nickname building variable at the
start of your loop. Also increase the range of numbers that rand is allowed
to return, as it will always return 2 at the moment
$usernames = array();
do {
// inside generate random number,
// build nickname,
// query to see if its taken,
// and if NOT taken, add to the usernames array
} while(count($usernames) < 3);

Generate random names in sql 
Store the 5 random names for male in one table and the 5 random names for
female in another table. Select a random integer between 1 and 5 and cross
reference to male or female table using an inner join.

Generate a random DB insert with PHP 
This takes multiple statements to do one row... and I may have
misunderstood your requirements, but here goes:
EDIT: Modified to be one statement and actually work with RAND()
INSERT INTO matrix
SELECT
NULL,
1,  NO idea what round_id should be...
IF( 0 in (one8, two8), 8, (SELECT element_id
FROM matrix_elements ORDER BY RAND() LIMIT
1)),
IF( 1 in (one8, two8), 8, (SELECT element_id
FROM matrix_elements ORDER BY RAND() LIMIT
1)),
IF( 2 in (one8, two8), 8, (SELECT element_id
FROM matrix_elements ORDER BY RAND() LIMIT
1)),
IF( 3 in (one8, two8), 8, (SELECT element_id
FROM matrix_elements ORDER BY RAND() LIMIT
1)),
IF( 4 in (one8, two8), 8, (SELECT element_i

generate a random number between 1 and 10 in c 
You need a different seed at every execution.
You can start to call at the beginning of your program:
srand(time(NULL));
Note that % 10 yields a result from 0 to 9 and not from 1 to 10: just add 1
to your % expression to get 1 to 10.

Generate random uint 
The simplest approach would probably be to use two calls: one for 30 bits
and one for the final two. An earlier version of this answer assumed that
Random.Next() had an inclusive upper bound of int.MaxValue, but it turns
out it's exclusive  so we can only get 30 uniform bits.
uint thirtyBits = (uint) random.Next(1 << 30);
uint twoBits = (uint) random.Next(1 << 2);
uint fullRange = (thirtyBits << 2)  twoBits;
(You could take it in two 16bit values of course, as an alternative... or
various options inbetween.)
Alternatively, you could use NextBytes to fill a 4byte array, then use
BitConverter.ToUInt32.

How to generate a Random number? 
This is how you generate a random number:
Random r = new Random();
int random = r.nextInt(100); // returns a value between 0 (incl) and 100
(excl)
Random r = new Random();
int random = r.nextInt(100) + 20; // returns a value between 20 (incl) and
120 (excl)
Concerning your issue that the numbers should be unique, you could probably
save each generated number somehow. Then, when drawing a new Random number
check the saved numbers if the new random number has already been
generated. If so, redraw.
Java Random doc:
http://docs.oracle.com/javase/6/docs/api/java/util/Random.html

Trying to Generate a Random password 
$alphabet=$NULL;For ($a=65;$a –le 90;$a++) {$alphabet+=,[char][byte]$a }
GETTemppassword –length 10 –sourcedata $alphabet
NewADUser $cn SamAccountName $sAMAccountName GivenName $givenName
Surname $sn DisplayName $displayName UserPrincipalName
$userPrincipalName accountPassword (ConvertToSecureString AsPlainText
$alphabet Force) PasswordNeverExpires $true Path
"OU=Service,OU=Accounts,DC=rjfdev,DC=com"
You aren't capturing the value returned from GetTemppassword. You're
passing $alphabet as the new password for the user account. Try this
(untested):
$alphabet=$NULL;For ($a=65;$a –le 90;$a++) {$alphabet+=,[char][byte]$a }
$TempPassword = GETTemppassword –length 10 –sourcedata $alphabet
NewADUser $cn SamAccountName $sAMAccountName GivenName $givenName Sur

vb.net generate random number and string 
The first problem to solve is the random number generator. You should use
only one instance and not multiple instances that gives back the same
sequence if called in short time distance. Then it is
difficult to say what 'something' means in your requirements, but supposing
you are fine with rougly 70% numbers and 30% a mix of numbers and strings
then you call the random generator to decide for a sequence of only numbers
or a mixed one. Based on the output of the random selection of the sequence
build from the appropriate string
' A global unique random generator'
Dim rng As Random = New Random
Sub Main
Console.WriteLine(rand())
Console.WriteLine(rand())
Console.WriteLine(rand())
Console.WriteLine(rand())
End Sub
Public Function rand() As String
Dim sb As New StringBu

PHP generate a random minus or plus percentage of a given value 
A bit of basic mathematics
$number = 1000;
$below = 20;
$above = 20;
$random = mt_rand(
(integer) $number  ($number * (abs($below) / 100)),
(integer) $number + ($number * ($above / 100))
);

Can you use 'rand' to generate random operators? 
well you can put operators in an array and then call array_rand to retrive
it
$ops = array('+','','*','/');
$rand_key = array_rand($ops);
$operator = $ops[$rand_key];

Generate random BLOB in MySQL 
I happened to require something similar lately. I therefore created a UDF
for generating random BLOBs. You can find it on my github.com account:
https://github.com/adrpar/mysql_udf_randomBlob
After compilation using cmake and following the post compilation procedures
as described in the README file you can use the UDF like this:
SELECT randomBlob( sizeOfBLOBInBytes);
e.g. for a 128 byte BLOB:
SELECT randomBlob( 128 );
or combine it with your INSERT statement.
