As they say, "the only winning move is not to play." I assume there are two dice. In that case, the expected value of all three bets is the same, and it's negative: on average, you lose $\frac{1}{6}$ of what you bet.
Some details: when rolling $2$ dice the probabilities of getting a sum of $2, 3, 4, ... 12$ are $\frac{1}{36}, \frac{2}{36}, \frac{3}{36}$, and so on, increasing until $7$ and then decreasing. This is a nice exercise if you've never done it; you just need to count the number of ways a pair of numbers between $1$ and $6$ can add up to a given number. It follows that the probability of getting a sum of $7$ is $\frac{1}{6}$, and by symmetry the probability of getting above $7$ or below $7$ are each $\frac{5}{12}$.
It follows that if you bet $7$ then on average you multiply your stake by $\frac{5}{6}$, whereas if you bet either below or above then on average you also multiply your stake by $\frac{5}{6}$. So all three options are equally bad, in the long run.
Some meta-lessons to learn from this computation:
- Generally the expected value of a game like this is negative, or the host would never bother to offer it. For example, the expected value of every game at a casino and the lottery has to be negative, or else the casino (resp. the institution running the lottery) would not make money. This is as much a lesson about incentives as it is about probability.
- If one of the options was better than the others, at some point people would be aware of this and play only that option.