掷 5 枚骰子取最大 3 个之和为 18

最大 3 个点数和为 18 当且仅当 5 枚骰子里至少有 3 个 6。

设 6 的个数 $K\sim\mathrm{Binomial}(5,1/6)$：

$$P(K\ge 3)=\sum_{k=3}^5\binom{5}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{5-k}=\frac{23}{648}.$$

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Original Explanation

To obtain a sum of $18$ from three dice, we need to have at least 3 sixes among the 5 dice. Therefore, we really are just asking for the probability of $3,4,$ or $5$ sixes among $5$ die rolls. The probability of $5$ sixes is just $\frac{1}{6^5}$, as it is just $\frac{1}{6}$ probability for each die. The probability of $4$ sixes is $5 \cdot \frac{5}{6} \cdot \frac{1}{6^4}$, as we have $5$ ways to select the die that isn't a $6$, a $\frac{5}{6}$ probability that the selected die is not a $6$, and then $\frac{1}{6}$ probability for each of the other 4 dice to be a $6$. Lastly, the probability that we have exactly $3$ sixes is $\displaystyle \binom{5}{2} \cdot \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6^3}$. This is because there are $\displaystyle \binom{5}{2} = 10$ ways to pick the two dice that aren't a $6$, $\frac{5}{6}$ probability for each of them to not be a $6$, and $\frac{1}{6}$ probability for each of the remaining three dice to be a $6$. Adding all of these up yields

$$\frac{1}{6^5} + \frac{25}{6^5} + \frac{250}{6^5} = \frac{276}{6^5} = \frac{23}{648}$$