等概率和问题

$\boxed{\text{不可能}}$

设一个六面骰各面的概率为 $p_1,p_2,p_3,p_4,p_5,p_6$，另一个六面骰各面的概率为 $q_1,q_2,q_3,q_4,q_5,q_6$。假设这两个骰子的点数和服从均匀分布，即对 $k=2,\ldots,12$，有 $P(\text{sum} = k) = \frac{1}{11}$。那么：

$$p_1q_1 = \frac{1}{11} \quad \text{且} \quad p_6q_6 = \frac{1}{11}$$

同时，

$$\frac{1}{11} = P(\text{sum} = 7) \geq p_1q_6 + p_6q_1$$

因此：

$$p_1\frac{1}{11p_6} + p_6\frac{1}{11p_1} \leq \frac{1}{11}$$

也就是

$$\frac{p_1}{p_6} + \frac{p_6}{p_1} \leq 1$$

令 $x = \frac{p_1}{p_6}$，则有

$$x + \frac{1}{x} \leq 1$$

这不可能成立，因为对任意正实数 $x$，都有 $x + \frac{1}{x} \geq 2$。因此，不存在这样的骰子。

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Original Explanation

$\boxed{No}$

Let $p_1,p_2,p_3,p_4,p_5$ and $p_6$ be the probabilities for one 6-sided die, and $q_1,q_2,q_3,q_4,q_5$ and $q_6$ be the probabilities for another. Suppose that these dice together yield sums with uniform probabilities. That is, suppose $P(\text{sum} = k) = \frac{1}{11}$ for $k=2,...,12$. Then:

$$p_1q_1 = \frac{1}{11} \quad \text{and} \quad p_6q_6 = \frac{1}{11}$$

Also,

$$\frac{1}{11} = P(\text{sum} = 7) \geq p_1q_6 + p_6q_1$$

so:

$$p_1\frac{1}{11p_6} + p_6\frac{1}{11p_1} \leq \frac{1}{11}$$

i.e

$$\frac{p_1}{p_6} + \frac{p_6}{p_1} \leq 1$$

Now, if we let $x=\frac{p_1}{p_6}$, then we have

$$x + \frac{1}{x} \leq$$

which is impossible, since for positive real $x, x + \frac{1}{x} \geq 2$. Thus, no such dice are possible.