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` replacements. The largest value is known beforehand. 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 `50%`, 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，但最多只有`k`次重抽機會。事先標明最大面值係幾多。當然，係`k + 1` 次中抽得最大面值那封利是就當作正式利是可以取走，但只可以取走一封利是。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可以放棄，並不會放回入箱；抽出面值較大就會保留，放棄舊有那封。

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抽 `6`次，而成功抽到整個箱最大面值那封利是的概率最少係 `50%`, 整個箱裡面的利是最多有幾多?[/quote]