非常重要的编辑： 所有 _A i_都是 唯一的 。

问题

我有一个列表 一个 的 ñ 独特的 对象。每个对象 A i_具有可变的百分比 _P i。

我想，以形成产生一个新的列表的算法 乙 的 ķ 对象（ ķ < ñ / 2，并且在大多数情况下 ķ 是显著小于 ñ / 2。
例如，N = 231，K = 21 ）。列表 B 应该没有重复项，并且将填充具有以下限制的源自列表 A的 对象：

物体 A i_出现在 _B中 的概率为 P i。

我尝试过的

（这些代码段仅出于测试目的而在PHP中）我首先列出了列表 A

$list = [
    "A" => 2.5, 
    "B" => 2.5, 
    "C" => 2.5, 
    "D" => 2.5, 
    "E" => 2.5, 
    "F" => 2.5, 
    "G" => 2.5, 
    "H" => 2.5, 
    "I" => 5,   
    "J" => 5,   
    "K" => 2.5, 
    "L" => 2.5, 
    "M" => 2.5, 
    "N" => 2.5, 
    "O" => 2.5, 
    "P" => 2.5, 
    "Q" => 2.5, 
    "R" => 2.5, 
    "S" => 2.5, 
    "T" => 2.5, 
    "U" => 5,   
    "V" => 5,   
    "W" => 5,   
    "X" => 5,   
    "Y" => 5,   
    "Z" => 20   
];

最初，我尝试了以下两个算法（这些仅在PHP中用于测试）：

$result = [];

while (count($result) < 10) {
    $rnd = rand(0,10000000) / 100000;

    $sum = 0;
    foreach ($list as $key => $value) {
        $sum += $value;
        if ($rnd <= $sum) {
            if (in_array($key,$result)) {
                break;
            } else {
                $result[] = $key;
                break;
            }
        }
    }
}

和

$result = [];

while (count($result) < 10) {
    $sum = 0;
    foreach ($list as $key => $value) {
        $sum += $value;
    }

    $rnd = rand(0,$sum * 100000) / 100000;

    $sum = 0;
    foreach ($list as $key => $value) {
        $sum += $value;
        if ($rnd <= $sum) {
            $result[] = $key;
            unset($list[$key]);
            break;
        }
    }
}

两种算法之间的唯一区别是，一种算法在遇到重复项时会重试，而另一种则在选择对象列表 A 时将其删除。事实证明，这两种算法具有相同的概率输出。

我运行了第二个算法100,000次，并跟踪了每个字母被选择了多少次。以下数组确定了从100,000个测试中选出任何列表 B 中的字母的机会百分比。

[A] => 30.213
[B] => 29.865
[C] => 30.357
[D] => 30.198
[E] => 30.152
[F] => 30.472
[G] => 30.343
[H] => 30.011
[I] => 51.367
[J] => 51.683
[K] => 30.271
[L] => 30.197
[M] => 30.341
[N] => 30.15
[O] => 30.225
[P] => 30.135
[Q] => 30.406
[R] => 30.083
[S] => 30.251
[T] => 30.369
[U] => 51.671
[V] => 52.098
[W] => 51.772
[X] => 51.739
[Y] => 51.891
[Z] => 93.74

当回顾算法时，这很有意义。该算法错误地将原始百分比解释为针对任何给定位置（而不是任何列表 B ）选择对象的机会百分比。因此，例如，实际上，在列表
B中 选择Z的机会是93％，但是在索引 B n上_选择Z的机会是20％。这不是我想要的。我希望从列表 _B中 选择Z的机会是20％。

这有可能吗？如何做呢？

编辑1

我尝试简单地让所有 P i = k的和，如果所有 _P i_都相等，则可以工作，但是修改它们的值后，它开始变得越来越错误。

初始概率

$list= [
    "A" => 8.4615,
    "B" => 68.4615,
    "C" => 13.4615,
    "D" => 63.4615,
    "E" => 18.4615,
    "F" => 58.4615,
    "G" => 23.4615,
    "H" => 53.4615,
    "I" => 28.4615,
    "J" => 48.4615,
    "K" => 33.4615,
    "L" => 43.4615,
    "M" => 38.4615,
    "N" => 38.4615,
    "O" => 38.4615,
    "P" => 38.4615,
    "Q" => 38.4615,
    "R" => 38.4615,
    "S" => 38.4615,
    "T" => 38.4615,
    "U" => 38.4615,
    "V" => 38.4615,
    "W" => 38.4615,
    "X" => 38.4615,
    "Y" =>38.4615,
    "Z" => 38.4615
];

10,000次运行后的结果

Array
(
    [A] => 10.324
    [B] => 59.298
    [C] => 15.902
    [D] => 56.299
    [E] => 21.16
    [F] => 53.621
    [G] => 25.907
    [H] => 50.163
    [I] => 30.932
    [J] => 47.114
    [K] => 35.344
    [L] => 43.175
    [M] => 39.141
    [N] => 39.127
    [O] => 39.346
    [P] => 39.364
    [Q] => 39.501
    [R] => 39.05
    [S] => 39.555
    [T] => 39.239
    [U] => 39.283
    [V] => 39.408
    [W] => 39.317
    [X] => 39.339
    [Y] => 39.569
    [Z] => 39.522
)