三骰子点数和为 10 的概率

不考虑顺序的组合为：

$(1,6,3)$、$(1,5,4)$、$(2,6,2)$、$(2,5,3)$、$(2,4,4)$、$(3,4,3)$。

其中 3 个是三值全不同：$(1,6,3)$、$(1,5,4)$、$(2,5,3)$，各有 $3!=6$ 种排列，共 $3\cdot 6=18$。

另外 3 个包含重复（两值）：各有 3 种排列，共 $3\cdot 3=9$。

总有利结果 $18+9=27$，样本空间大小 $6^3=216$，因此

$$P=\frac{27}{216}=\frac{1}{8}.$$

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Original Explanation

The most efficient approach here would be generating functions. However, we are going to solve via casework instead. The possibilities of values (not ordered in any way currently) that result in a sum of $10$ are

$$(1,6,3), (1,5,4), (2,6,2), (2,5,3), (2,4,4), (3,4,3)$$

Three of these combinations have all distinct values. Namely, these are $(1,6,3), (1,5,4)$, and $(2,5,3)$. For these sets of outcomes, there are $3! = 6$ ways they can be assigned to the three dice. These yield $3 \cdot 6 = 18$ possibilities.

The other three outcomes have two distinct values. For these, there are only $3$ ways they can each be assigned to the dice, as you only select the die that shows the value that differs from the other two. Then, the other two dice are fixed. These yield $3 \cdot 3 = 9$ possibilities.

Combining these two cases yields $18 + 9 = 27$ possibilities that result in a sum of $10$. We divide by $6^3 = 216$ to get the probability, yielding a final answer of $\frac{1}{8}$.