Imagine the following game: You are throwing two fair dice and you will be awarded with the sum of the points in cents. How much should participants of this game pay so that the game itself is fair.
Let $X$ denote the sum of points then we can easily count the possibilities for each $X=2,\ldots,12$ and the corresponding expected profit for participants is
$$\begin{align}\mathbb{E}[X]=2\cdot\frac{1}{36}+3\cdot\frac{2}{36}+\ldots+7\cdot\frac{6}{36}+\ldots+11\cdot\frac{2}{36}+12\cdot\frac{1}{36}=7\end{align}$$
hence players of the game should pay $7$ cents to make the game fair.
I have seen a different approch to this problem where it is assumed that each of the outcomes has probability $\frac{1}{11}$ interpreting it as one of eleven possible sums. Surprisingly the computation of the expected profit with this probability for each outcome yields
$$\mathbb{E}[X]=\frac{1}{11}\sum_{k=2}^{12}k=7.$$
For this particular scenario this yields the same profit but it feels like it is just coincidence and definitely not correct. The question is now whether the assumption for probabilities $\frac{1}{11}$ is reasonable or is the result really just coincidence - how can either of these be proven?
${1 \over 11}$ solution relies on situation being symmetric (probability of result $3$ equals probability of result $11$ and so on). It breaks easily - consider product of points instead. There are 18 possible values: 1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36. First approach gives (correct) result $12{1 \over 4}$, second - (wrong) result $13{5 \over 9}$.
Actually, in such questions probabilities are assigned to events using some symmetry (usually denoted by word "fair" or "totally random" or the like). Here "fair" are dice, which means that probabilities of different values of a given die are equal. If instead you were told that you get "totally random" number from $2,...,12$, then second approach would be correct.