掷 5 面骰直到和至少为 5 的期望次数

结果为

$$\left(1+\frac{1}{5}\right)^{4}=\frac{6^{4}}{5^{4}}=\frac{1296}{625}\approx 2.0736.$$

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Original Explanation

We are going to solve the generalized version for $n-$sided die with a sum of at least $n$. Let's denote the expected number of rolls needed to obtain a sum of at least $n$ starting from a sum of $k$ by $E_k$. Clearly we have that $E_n = 0$, as we already have a sum of $n$. Further, we have that $E_{n-1} = 1$, as no matter what is obtained, we have a sum of at least $n$.

Then, we have that $E_{n-2} = 1 + \dfrac{1}{n}E_{n-1} = 1 + \dfrac{1}{n}$, as with probability $\dfrac{1}{n}$, we obtain the value $1$ and we have a total sum of $n-1$. Similarly, $E_{n-3} = 1+ \dfrac{1}{n} E_{n-1} + \dfrac{1}{n}E_{n-2} = 1 + \dfrac{2}{n} + \dfrac{1}{n^2}$.

By continuing with this pattern, one can prove by induction that $E_{n-k} = \displaystyle \sum_{j=0}^{k-1} \dfrac{\binom{k-1}{j}}{n^j}$.

Therefore, by the Binomial Theorem, $E_0 = E_{n-n} = \displaystyle \sum_{j=0}^{n-1} \binom{n-1}{j}\left(\dfrac{1}{n}\right)^j = \left(1 + \dfrac{1}{n}\right)^{n-1}$.

We applied the Binomial Theorem with $x = \dfrac{1}{n}$ and $y = 1$. Therefore, for $n = 5$, our answer is $\dfrac{6^4}{5^4} = \dfrac{1296}{625}$.