我唔知大家有無玩過以下新年抽利是遊戲，不過都在此推介一下。

English version:

Suppose `n` red packets with distinct monetary values are placed in a sealed lucky draw box and drawn out one by one at random to find the packets with the largest value in at most `k` [#990000]replacements[/#990000]. At each stage the packet drawn is compared to the packet with the largest value found so far. If the packet value drawn is smaller, we discard it; if it is larger we replace the previous largest packet value by it.

A. If `n = 10 and k = 4`, what is the probability that we pick the red packet with the largest value in the box?
B. What is the general form of the probability for any `n and k` that we cannot pick the red packet with the largest value in the box within `k` replacements?
C. If we know that `k = 6` and assure that we can successfully pick the best red packet in the box with at least `60%`, how many `n` red packets at most should we have in the box?

中文版:

假設有 `n` 封有不同面值的利是放係一個未開封的抽獎箱裡面，而每次都係隨意抽出利是。每次抽出的新利是，都會同對上一封最大面值那封比較。抽出的新一封面值較細就放棄，並不會放回入箱；抽出面值較大就會保留，放棄舊有那封。重定要取走的利是最多只有`k`次，但並不會限制幾多次由抽獎箱抽出新利是。即是，若果第一次重抽那次利是面值比原先那封面值細，就當然不會重定，仍舊視作 0 次重定。

A. 若果 `n = 10 and k = 4`, 成功抽到整個箱最大面值那封利是的概率係幾多呢?
B. 可否有一條通用公式對任何 `n and k`計出抽唔到整個箱最大面值那封利是的概率?
C. 若果知道左最多只可以重定 `6`次，而成功抽到整個箱最大面值那封利是的概率最少係 `60%`, 整個箱裡面的利是最多有幾多?