掷 5 枚骰子：取最大 3 个之和为 18 的概率

最大 3 个点数之和为 18 当且仅当“最大的 3 个都是 6”，也就是 5 枚骰子里至少有 3 个 6。

设出现 6 的个数为 $K\sim\mathrm{Binomial}(5,1/6)$，则

$$P(K\ge 3)=\sum_{k=3}^5\binom{5}{k}\left(\frac16\right)^k\left(\frac56\right)^{5-k}.$$

也可直接计数：

• 恰好 3 个 6：$\binom{5}{3}5^2=250$；
• 恰好 4 个 6：$\binom{5}{4}5=25$；
• 恰好 5 个 6：$1$。

有利结果共 $250+25+1=276$，总结果 $6^5=7776$，因此

$$P=\frac{276}{7776}=\frac{23}{648}.$$

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Original Explanation

Sum of the top 3 = 18 means the top 3 are all 6s. That implies at least 3 dice show 6, but we also must consider the 4th and 5th dice can be anything $\le6$. Specifically, to get a top-3 sum of 18, we need exactly 3, 4, or 5 of the dice to be 6. Then the leftover dice are $\le6$. Actually, for the sum of the largest 3 to be 18, the largest 3 must be 6,6,6. This can happen if 3,4, or 5 dice are 6: in each scenario, the top three are definitely 6,6,6. Then count the combinations. Probability = $\frac{\#\{\text{ways}\}}{6^5}$. (One can do a small combinatorial sum.)