Tim plays a game with 3 outcomes having the score of 2, 5 and 6. The corresponding probabilities are $\dfrac{2}{7}$, $k$ and $\dfrac{4}{7}$ respectively, where $k$ is a constant. Tim pays $m in total to play the game twice. The outcome of each game is independent. He receives an amount corresponding to twice the absolute difference of the two scores from each game.

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 $m > 0$.
$0 < m < \frac{24}{7}$