最后一掷

设停止阈值为 $T=1000$。当累计和接近阈值时，最后一掷要“跨过”阈值。

用递推可得：从和为 0 开始，最后一次掷出 $k\in\{1,2,3,4,5,6\}$ 的概率分别为

$$\frac{1}{21},\frac{2}{21},\frac{3}{21},\frac{4}{21},\frac{5}{21},\frac{6}{21}.$$

因此最可能的是 $6$。

直观上：越大的点数越容易作为“跨过阈值”的最后一步出现。

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英文解析

Set the stop threshold to $T=1000$. When cumulative and close to the threshold, the last throw is to “cross” the threshold.

Recursively: starting with and 0, the probability of the last throw of $k\in\{1,2,3,4,5,6\}$ is

$$\frac{1}{21},\frac{2}{21},\frac{3}{21},\frac{4}{21},\frac{5}{21},\frac{6}{21}.$$

So the most likely outcome is $6$.

Intuitively: the larger the number of points, the easier it is to appear as the last step in “crossing the threshold”.