如何在随机生成器中获得 50/50 的机会

Question

I am trying to get a 50/50 chance of get either 1 or 2 in a random generator.

For example:

Random random = new Random();
int num = random.nextInt(2)+1;

This code will output either a 1 or 2.

Let's say I run it in a loop:

for ( int i = 0; i < 100; i++ ) {
    int num = random.nextInt(2)+1 ;
}

How can I make the generator make an equal number for 1 and 2 in this case?

So I want this loop to generate 50 times of number 1 and 50 times of number 2.

Answer 1

一种方法：用五十个 1 和五十个 2 填充ArrayList<Integer> ，然后在其上调用Collection.shuffle(...) 。

Answer 2

50/50 is quite easy with Random.nextBoolean()

private final Random random = new Random();

private int next() {
  if (random.nextBoolean()) {
    return 1;
  } else {
    return 2;
  }
}

Test Run:

final ListMultimap<Integer, Integer> histogram = LinkedListMultimap.create(2);
for (int i = 0; i < 10000; i++) {
    nal Integer result = Integer.valueOf(next());
  histogram.put(result, result);
}
for (final Integer key : histogram.keySet()) {
  System.out.println(key + ": " + histogram.get(key).size());
}

Result:

1: 5056
2: 4944

Answer 3

You can't achieve this with random . If you need exactly 50 1s and 50 2s, you should try something like this:

int[] array = new int[100];
for (int i = 0; i < 50; ++i)
 array[i] = 1;
for (int i = 50; i < 100; ++i)
 array[i] = 2;

shuffle(array); // implement shuffling algorithm or use an already existing one

Answer 4

EDIT: I understand that if you are looking to accomplish exactly 50-50 results, then my answer was not accurate. You should use a pre-filled collection, since it is impossible to achive that using any kind of randomness. This considered, my answer is still valid for the title of the question, so, this is it:

Well, you do not need the rnd generator to do this. Comming from javascript, I would go with a single liner:

return Math.random() > 0.5 ? 1: 2;

Explanation: Math.random() returns a number between 0(inclusive) and 1(exclusive), so, we just examine weather is larger than 0.5 (middle value). In theory there is a 50% change that does.

For a more generic use, you can just replace 1:2 to true:false

Answer 5

You can adjust the probability along the way so that the probability of getting a one decreases as you get more ones. This way you don't always have a 50% chance of getting a one, but you can get the result you expected (exactly 50 ones):

int onesLeft = 50;

for(int i=0;i<100;i++) {
  int totalLeft = 100 - i;
  // we need a probability of onesLeft out of (totalLeft)
  int r = random.nextInt(totalLeft);
  int num;
  if(r < onesLeft) {
    num = 1;
    onesLeft --;
  } else {
    num = 2;
  }
}

This has an advantage over shuffling because it generates numbers incrementally so it desn't need memory to store the numbers.

Answer 6

You have already successfully created a random generator that returns 1 or 2 with equal probability.

As (many) other's have mentioned, your next request, to force an exact 50/50 distributions in 100 trials, does not fall in line with random number generation. As shown in https://math.stackexchange.com/questions/12348/probability-of-getting-50-heads-from-tossing-a-coin-100-times , the realistic expectation of that occurring is only around 8%. So even while you might expect 50 of each, that exact outcome is actually rather rare.

The Law of Large Numbers states that you should close in on expected value as your number of trials increases.

So for your actual question: How can I make the generator make an equal number for 1 and 2 in this case?

The best (humorous) answer I can come up with is: "Run it in an infinite loop."