Adding random numbers 1-1000000 in an array?

Question

I have looked up how to add 0-99 to an array. But, my assignment was 1-1000000. I just keep getting really large numbers and no small numbers. Is it just because the chance of getting large numbers is a lot higher? I just wanted to make sure I was doing it right. Thanks in advance for any help!

int a[]= new int[50];
for(int i = 0; i < 50; i++) {
    a[i] = (int)(Math.random() * 10000000);
}

This is the normal case. If the numbers all have equal distribution, the probability of getting a number > 1000000 is 0.9, while a number <= 1000000 is 0.1. Going one step further, a number <= 100000 has a probability of 0.01. And that's still a pretty large number. A number below 1000 has a probability of 1/10000. I've just noticed you use 1mio as upper-bound in the question itself, but 10mio in the code. My values are based on an upper bound of 10mio. — , Feb 09 '17 at 03:12
I mean, there are never any 1-5 digit numbers. Just numbers with 6+ digits. I might just acting stupid, because there are so many more 6+ digits numbers. I just wanted to make sure my code was right. Thanks! — MostPalone, Feb 09 '17 at 03:13
Are you asking how to bias results in favor of smaller numbers, or are you just confirming your code is correct? — shmosel, Feb 09 '17 at 03:16

score 0 · Answer 1 · answered Feb 09 '17 at 03:12

0

Well if you hoped to get a number below 1000, there's a 1/10000 chance, which translates to 0.01%. Your code is fine, but with these settings getting a low number is quite unlikely

answered Feb 09 '17 at 03:12

Cedric Martens

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score 0 · Accepted Answer · answered Feb 09 '17 at 03:13

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Depends what do you call by small numbers.

For example, getting number lower than 1000 means that it would need to fit to lower 0.1% of the interval as 1000 it's just 0.1% of all the numbers from 1 million.

Additionally, note that this way you'll get numbers between 0 and 999999 (inclusive). To get 1-1000000 you need to add 1:

a[i] = (int)(Math.random() * 10000000)+1;

answered Feb 09 '17 at 03:13

Zbynek Vyskovsky - kvr000

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Thanks for your help! I was just over-thinking it. – MostPalone Feb 09 '17 at 03:16

score 0 · Answer 3 · answered Feb 09 '17 at 03:41

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Yes, large numbers will dominate with that. That's because half the numbers are over 500,000. Similarly, 9/10 of the numbers are over 100,000. So 9/10 of the numbers will have six digits.

answered Feb 09 '17 at 03:41

Chai T. Rex

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score 0 · Answer 4 · edited May 23 '17 at 10:29

[Somewhat orthogonal solution to your question]

One of the options you can consider is to shuffle the array after adding the numbers based on your expected distribution, which can be all numbers added once/certain numbers added multiple times/certain prime numbers multiple times, etc.

I personally used it to cover some cases for a game development.

Here is an answer which can help you shuffle: Credits

 List<Integer> solution = new ArrayList<>();
    for (int i = 1; i <= 6; i++) {
        solution.add(i);
    }
 Collections.shuffle(solution);

Adding random numbers 1-1000000 in an array?

4 Answers4