具有较高概率给出低值的随机数生成器？

Question

如何生成伪随机数（最好在 Lua 中），其中生成器给出小数的概率更高？

在我的例子中，我想在游戏中给出一个随机分数，通常会获得较低的分数，但很少出现较高的分数。 我见过使用表格的加权随机数生成器，但它不符合我的计划。我只想指定最小值（0）和最大值（变量）并确保大多数数字保持在低位。

我确信这可以通过简单的数学运算来实现，但我不记得是哪一个了。喜欢过滤 math.random 的常规输出，不需要真正的随机生成器。

Answer 1

这可能不是您要找的，因为它不是平滑偏置的钟形曲线，但为什么不创建两个步骤呢？定义获得较低分数的概率，如果匹配，您的分数范围就是较低分数。否则，您的范围是从低范围的顶部到高范围的末端。

最终结果是您通常会获得低分，但有时您会获得高分。我敢打赌它看起来不错，而且非常简单。

你怎么看？

Answer 2

我只是转换标准随机函数的值，如下所示：

r1=math.random(0,255)
r2=math.exp(math.random(0,255))

你需要考虑你的界限，但你会有很多低值，很少有高值的东西。

Answer 3

试试math.floor(minscore+(maxscore-minscore)*math.random()^2)。调整功率以适合您所需的分布。

Answer 4

我发现Ihf's answer 非常有用，我为它创建了一个 C# 方法：

    private int GetRandomNumber(int max, int min, double probabilityPower = 2)
    {
        var randomizer = new Random();
        var randomDouble = randomizer.NextDouble();

        var result = Math.Floor(min + (max + 1 - min) * (Math.Pow(randomDouble, probabilityPower)));
        return (int) result;
    }

如果probabilityPower 大于 1，则较低的值将比较高的值更常见。 如果它在 0 到 1 之间，则较高的值将比较低的值更常见。 如果为 1，则结果为一般随机性。

示例（所有迭代次数为 100 万次，最小值 = 1，最大值 = 20）：

---

probabilityPower = 1.5

1: 135534 (13.5534%)
2: 76829 (7.6829%)
3: 68999 (6.8999%)
4: 60909 (6.0909%)
5: 54595 (5.4595%)
6: 53555 (5.3555%)
7: 48529 (4.8529%)
8: 44688 (4.4688%)
9: 43969 (4.3969%)
10: 44314 (4.4314%)
11: 40123 (4.0123%)
12: 39920 (3.992%)
13: 40466 (4.0466%)
14: 35821 (3.5821%)
15: 37862 (3.7862%)
16: 35222 (3.5222%)
17: 35902 (3.5902%)
18: 35202 (3.5202%)
19: 33961 (3.3961%)
20: 33600 (3.36%)

---

probabilityPower = 4

1: 471570 (47.157%)
2: 90114 (9.0114%)
3: 60333 (6.0333%)
4: 46574 (4.6574%)
5: 38905 (3.8905%)
6: 32379 (3.2379%)
7: 28309 (2.8309%)
8: 27906 (2.7906%)
9: 22389 (2.2389%)
10: 21524 (2.1524%)
11: 19444 (1.9444%)
12: 19688 (1.9688%)
13: 18398 (1.8398%)
14: 16870 (1.687%)
15: 15517 (1.5517%)
16: 15871 (1.5871%)
17: 14550 (1.455%)
18: 14635 (1.4635%)
19: 13399 (1.3399%)
20: 11625 (1.1625%)

---

probabilityPower = 1

1: 51534 (5.1534%)
2: 49239 (4.9239%)
3: 50955 (5.0955%)
4: 47992 (4.7992%)
5: 48971 (4.8971%)
6: 50456 (5.0456%)
7: 49282 (4.9282%)
8: 51344 (5.1344%)
9: 50841 (5.0841%)
10: 48548 (4.8548%)
11: 49294 (4.9294%)
12: 51795 (5.1795%)
13: 50583 (5.0583%)
14: 51020 (5.102%)
15: 51060 (5.106%)
16: 48632 (4.8632%)
17: 48568 (4.8568%)
18: 50039 (5.0039%)
19: 49863 (4.9863%)
20: 49984 (4.9984%)

---

probabilityPower = 0.5

1: 3899 (0.3899%)
2: 5818 (0.5818%)
3: 12808 (1.2808%)
4: 17880 (1.788%)
5: 23109 (2.3109%)
6: 26469 (2.6469%)
7: 33435 (3.3435%)
8: 35243 (3.5243%)
9: 42276 (4.2276%)
10: 47235 (4.7235%)
11: 52907 (5.2907%)
12: 58107 (5.8107%)
13: 63719 (6.3719%)
14: 66266 (6.6266%)
15: 72708 (7.2708%)
16: 79257 (7.9257%)
17: 81830 (8.183%)
18: 87243 (8.7243%)
19: 90958 (9.0958%)
20: 98833 (9.8833%)

---

probabilityPower = 0.4

1: 917 (0.0917%)
2: 3917 (0.3917%)
3: 3726 (0.3726%)
4: 10679 (1.0679%)
5: 13092 (1.3092%)
6: 17306 (1.7306%)
7: 22838 (2.2838%)
8: 29221 (2.9221%)
9: 35832 (3.5832%)
10: 38422 (3.8422%)
11: 47800 (4.78%)
12: 53431 (5.3431%)
13: 63791 (6.3791%)
14: 69460 (6.946%)
15: 75313 (7.5313%)
16: 86536 (8.6536%)
17: 95082 (9.5082%)
18: 103440 (10.344%)
19: 110203 (11.0203%)
20: 118994 (11.8994%)

Answer 5

function weighted_random (weights)
    local summ = 0
    for i, weight in pairs (weights) do
        summ = summ + weight
    end
    local value = math.random (summ)
    summ = 0
    for i, weight in pairs (weights) do
        summ = summ + weight
        if value <= summ then
            return i, weight
        end
    end
end

如何调用：

local elements = {"a", "b", "c", "d"} -- elements
local weights = {40, 24, 22, 14} -- weights of elements
local n = weighted_random (weights) -- returns 1, 2, 3 or 4
local element = elements[n] -- returns "a", "b", "c" or "d" 
-- with chances: 40/100, 24/100, 22/100, 14/100, as in weights