蒂姆（Tim）玩一个有 3 种结果的游戏，得分分别为 2、5 和 6。相应的概率分别为 $\dfrac{2}{7}$、$k$ 和 $\dfrac{4}{7}$，其中 $k$ 是常数。 蒂姆总共支付 $m 玩两次该游戏。每次游戏的结果是独立的。他收到的金额相当于两次游戏得分之差的绝对值的两倍。

1. Set the sum of the probabilities of all possible outcomes to 1 and solve for $k$.

$\frac{2}{7} + k + \frac{4}{7} = 1$
$k = 1 - \frac{6}{7}$
$k = \frac{1}{7}$

1. Determine the value of $k$ and the probability distribution of the absolute difference between the two scores.

From the sum of probabilities, $k = 1 - \frac{2}{7} - \frac{4}{7} = \frac{1}{7}$.

Let $D$ be the absolute difference between the two scores. The possible differences are $0, 1, 3$, and $4$.
$\mathrm{P}(D = 1) = \mathrm{P}(5,6) + \mathrm{P}(6,5) = 2 \left(\frac{1}{7}\right)\left(\frac{4}{7}\right) = \frac{8}{49}$
$\mathrm{P}(D = 3) = \mathrm{P}(2,5) + \mathrm{P}(5,2) = 2 \left(\frac{2}{7}\right)\left(\frac{1}{7}\right) = \frac{4}{49}$
$\mathrm{P}(D = 4) = \mathrm{P}(2,6) + \mathrm{P}(6,2) = 2 \left(\frac{2}{7}\right)\left(\frac{4}{7}\right) = \frac{16}{49}$

2. Calculate the expected amount Tim receives.

The expected absolute difference is:
$\mathrm{E}(D) = 1\left(\frac{8}{49}\right) + 3\left(\frac{4}{49}\right) + 4\left(\frac{16}{49}\right) = \frac{8+12+64}{49} = \frac{84}{49} = \frac{12}{7}$

Tim receives twice the absolute difference.
$\text{Expected Amount Received} = 2 \times \mathrm{E}(D) = 2 \times \frac{12}{7} = \frac{24}{7}$

3. Determine the range of values of $m$ for an expected profit.

For Tim to be expected to make a profit, his expected amount received must be strictly greater than the cost $m$.
$m < \frac{24}{7}$
Since he pays a positive amount to play, we must also have $m > 0$.
$0 < m < \frac{24}{7}$