随机选择指定概率分布范围内的数字

Question

我想从3-10或4-6等范围中选择一个随机数.应该选择数字,使得数字越低,选择的机会就越多.我下面的代码只选择具有相同概率的每个数字.

private int bonusPoints;
private double randomBonusPoints = Math.Random() * 100;

bonusPoints = (int)randomBonusPoints;

如何从分布中选择P(3,4,5)=85%, P(6,7,8)=10%, P(9,10)=5%？

Answer 1

如果您的发行版可能会发生变化,那么您不希望以Kevin的回答方式对您的发行版进行硬编码,您可以使用NavigableMap该ceilingEntry方法.这允许您为选择选项指定权重.然后,您可以生成从0到权重总和的随机数.然后根据您的加权分布统计输出.

创建地图后,选择条目的代码非常短:

Random rand = new Random();
...
double rnd = rand.nextDouble() * totalWeight;
int elem = map.ceilingEntry(rnd).getValue();

要创建任意分布,您可以执行以下操作:

int[] options = new int[]{3,4,5,6,7,8,9,10};
double[] weights = new double[]{ 0.85/3d, 0.85/3d, 0.85/3d,
                                 0.10/3d, 0.10/3d, 0.10/3d,
                                 0.05/2d, 0.05/2d };

NavigableMap<Double, Integer> map = new TreeMap<Double, Integer>();
double totalWeight = 0d;

for (int i = 0; i < weights.length; i++) {
    totalWeight += weights[i];
    map.put(totalWeight, options[i]);
}

测试这10000多个选项给出了以下权重:

03: 28.99% 
04: 28.10% 
05: 28.06% 
06: 3.27% 
07: 3.62% 
08: 3.08% 
09: 2.40% 
10: 2.48%  

测试代码:

//  select from the weighted elements
Random rand = new Random();
HashMap<Integer, Double> freqs = new HashMap<Integer, Double>();
int iterations = 10000;
for(int i = 0; i < iterations; i++) {
    double rnd = rand.nextDouble() * totalWeight;
    int elem = map.ceilingEntry(rnd).getValue();
    freqs.put(elem, (freqs.containsKey(elem) ? freqs.get(elem) : 0) + (1d/iterations));
}
Map<Integer, Double> sortedFreqs = new TreeMap<Integer, Double>(freqs);

for(Map.Entry<Integer,Double> entry : sortedFreqs.entrySet()) {
    System.out.printf("%02d: %.2f%% %n", entry.getKey(), entry.getValue() * 100d);
}