Question

我需要一些我真不知道如何制作它的东西。

以下是一个例子：

用户1 = 3
用户2 = 1
用户3 = 6

我想要的是，用户1，有30％的几率，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，但也有机会的百分比。我真的不知道如何解释它。然后，它为用户3提供60％的机会，用户2的机会为10，用户1的机率为30％。但是，这最后可以尽可能长。这该怎么做？对不起，我的解释真的很糟糕。

谢谢！

Answer 1

您想要的可以被描述为“加权概率分布”或技术性“discrete distribution”和“Categorical distribution”：

分类分布是一种离散概率分布，它描述了随机变量的可能结果，该随机变量可以承担K个可能的基本事件之一，并且每个基本事件的概率都是单独指定的。

- 维基百科

如果您有random variable with uniform distribution in the range 0 to 1，则可以使用Inverse Method构建所需的任何发布。

您的第一步是规范化分发。这意味着要确保曲线下方的面积等于一（即，你的体重总和不超过100％）。对于离散分布，它意味着确保权重之和等于1。 [这与将值作为向量并在同一方向上计算单位向量相同]，只需取每个值并将其除以值的总和。

因此，你可以这样做：

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

对此：

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，您不需要它（标准化）的权重，而是它的权重加上之前的所有权重。

因此，你可以从这个

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

对此：

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

最后，您获得random variable with uniform distribution in the range 0 to 1并查看下面的值：

$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：范围是包容性的，因为PHP is weird。

好的，一起去，（丑陋的）PHP：

$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

让我们重新实现MySQL，因为原因。

首先我们需要一个包含输入的表，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`

然后我们总结值

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

然后我们正常化

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＆＃34; PHP ....＆＃34;不是标签＆＃34; mysql＆＃34;

每个用户都有n 门票。在你的情况下，你有10张票，因此需要一个0到9之间的随机数。一个不优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);

Php随机偶然发现

2 个答案: