$$求:\frac{1}{m^n} \cdot \sum_{i=1}^{n} i^k \cdot f(i) $$
$$i为抽出鬼牌的次数， f(i)为此情况的数量$$
$$其中f(i)=\binom{i}{n} \cdot (m-1)^{n-i}$$
$$原式=\frac{1}{m^n} \cdot \sum_{i=1}^{n} i^k \cdot \binom{i}{n} \cdot (m-1)^{n-i}$$
$$=\frac{1}{m^n} \cdot \sum_{i=1}^{n} \binom{i}{n} \cdot (m-1)^{n-i} \sum_{j=0}^{k} {j \brace k} \cdot i^{\underline j}$$
$$= \frac{1}{m^n} \cdot \sum_{j=0}^{k} {j \brace k} \sum_{i=1}^{n} \cdot \binom{i}{n} \cdot i^{\underline j} \cdot (m-1)^{n-i} $$
$$= \frac{1}{m^n} \cdot \sum_{j=0}^{k} {j \brace k} \sum_{i=1}^{n} \binom{i}{n} \cdot i^{\underline j} \cdot (m-1)^{n-i} $$
$$= \frac{1}{m^n} \cdot \sum_{j=0}^{k} {j \brace k} \cdot n^{\underline j} \sum_{i=1}^{n} \binom{i-j}{n-j} \cdot (m-1)^{n-i} $$
$$= \frac{1}{m^n} \cdot \sum_{j=0}^{k} {j \brace k} \cdot n^{\underline j} \sum_{i=1}^{n} \binom{n-i}{n-j} \cdot (m-1)^{n-i} \cdot 1^{n-i} $$
$$由二项式定理得：$$
$$= \frac{1}{m^n} \cdot \sum_{j=0}^{k} {j \brace k} \cdot n^{\underline j} \cdot m^{n-j}$$