Php 以 % 的机会随机化

Question

我需要一些我真的不知道怎么做的东西。

这是一个例子：

• 用户 1 = 3
• 用户 2 = 1
• 用户 3 = 6

我想要的是，用户 1 有 30% 的机会，用户 2 有 10%，用户 3 有 60%。但这些数字也可以是 0.01 而不是 6。现在想要的是它是随机的，但也有 % 的机会。我真的不知道该怎么解释。然后它会为用户 3 抽出一个 60% 的机会，为用户 2 提供 10% 的机会和为用户 1 提供 30% 的机会的数字。但是，最后一个可以尽可能长。这个怎么做？对不起，我解释得不好。

谢谢！

Answer 1

您想要的可以描述为“加权概率分布”或技术上的“discrete distribution”和“Categorical distribution”：

分类分布是一种离散概率分布，它描述了一个随机变量的可能结果，该随机变量可以处理 K 个可能的基本事件之一，并分别指定每个基本事件的概率。

-- 维基百科

假设您有一个random variable with uniform distribution in the range 0 to 1，您可以使用Inverse Method 构建您想要的任何发行版。

您的第一步是规范化分布。这意味着要确保曲线下方的面积等于 1（即您的体重总和不超过 100%）。对于离散分布，这意味着确保权重之和等于 1。 【相当于取值作为向量，计算同方向的单位向量】，取每个值除以值之和即可。

因此，你从这里开始：

(original)
user 1 = 3
user 2 = 1
user 3 = 6

到这里：

sum = 3 + 1 + 6 = 10

(normalized)
user 1 = 3 / 10 = 0.3
user 2 = 1 / 10 = 0.1
user 3 = 6 / 10 = 0.6

接下来，获取累积分布。也就是说，对于每个值，您不想要它的（标准化）权重，而是它的权重加上所有以前的权重。

因此，你从这里出发

(normalized)
user 1 = 0.3
user 2 = 0.1
user 3 = 0.6

到这里：

(cumulative)
user 1 = 0.3
user 2 = 0.1 + 0.3 = 0.4
user 3 = 0.6 + 0.3 + 0.1 = 1

最后，你得到你的random variable with uniform distribution in the range 0 to 1 并检查它低于哪个值：

$r = (float)rand()/(float)getrandmax();

if ($r <= 0.3) return "user 1"; // user 1 = 0.3
else if ($r <= 0.4) return "user 2"; // user 2 = 0.4
else return "user 3"; // user 3 = 1

注意：范围包括在内，因为PHP is weird。

好的，一口气，在（丑陋的）PHP中：

$p = ['user 1' => 3, 'user 2' => 1, 'user 3' => 6];
$s = array_sum($p);
$n = array_map(function($i) use ($p, $s){return $i/$s;},$p);
$a = []; $t = 0;
foreach($n as $k => $i) {$t += $i; $a[$k] = $t;}
$r = (float)rand()/(float)getrandmax();
foreach($a as $k => $i) { if ($r <= $i) return $k; }

Try online.

---

让我们在 MySQL 中重新实现因为原因。

首先我们需要一个带有输入的表格，例如：

SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`

然后我们将值相加

SELECT sum(chance) FROM (SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`) input

然后我们归一化

SELECT id, chance / sum FROM (SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`) input CROSS JOIN (SELECT sum(chance) as sum FROM (SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`) input) s

然后我们累积

SELECT id, chance / sum as sum, (@tmp := @tmp + chance / sum) as csum FROM (SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`) input CROSS JOIN (SELECT sum(chance) as sum FROM (SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`) input) s CROSS JOIN (SELECT @tmp := 0) cheat

然后我们选择

SELECT id from (
SELECT id, chance / sum as sum, (@tmp := @tmp + chance / sum) as csum FROM (SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`) input CROSS JOIN (SELECT sum(chance) as sum FROM (SELECT 'user 1' AS `id`, 3 AS `chance`
UNION
SELECT 'user 2' AS `id`, 1 AS `chance`
UNION
SELECT 'user 3' AS `id`, 6 AS `chance`) input) s CROSS JOIN (SELECT @tmp := 0) cheat) a
CROSS JOIN (SELECT RAND() as r) random
WHERE csum > r
LIMIT 1

Try online.

Answer 2

编辑：此答案基于标题“PHP....”而不是标签“mysql”

每个用户有 n 个票。在您的情况下，您有 10 张票，因此需要 0 到 9 之间的随机数。一个不雅的解决方案可能是：

<?php
$tickets = array();
$number_of_tickets = 0;

foreach($users as $user) {
    for($i = 0; i < $user->tickets; $i++) {
        $tickets[] = $user->id;
        $tickets++;
    }
}

$lucky_draw = rand(0, $number_of_tickets);
$winner = tickets[$lucky_draw] //ID of the user

print("And the winner is...." . $winner);