An illustrative example for why minimzing regret is interesting is the Traveler's Dilemma:
An airline lost the luggage of two of their passengers. Both passengers happen to have identical luggage and the airline know it is worth some value between $2$ and $100$ USD which they now have to reimburse. To find out the value of the luggage they ask both passengers $1, 2$ for their number $a, b$, respectively. Then they give a reimbursement of the lower number (because the airline will now claim that this is the correct one) but give the passenger who reported the lower number a $2$ USD honesty reward, while the passenger who reported the higher number gets a $2$ USD penalty (mind: this penalty is also from the lower reported number). If both passengers report the same number, then both receive exactly that amount.
Thus, we have the following utility functions
$$u_1(a, b) = \begin{cases}a+2 & \text{if } a < b,\\a & \text{if } a = b,\\b-2 & \text{if } a > b,\end{cases}$$
and $u_2(a, b) = u_1(b, a)$.
Now, if both player play cooperatively, they could just both report the number $100$ and receive that amount. This is clearly a welfare-maximizing strategy.
However, maximizing utility is not something one can just do in game theory, as you have multiple agents whose decision are not yours but influence what is possible for you. That is, the state $(100, 100)$ is not stable (i.e. not a Nash equilibrium) as both players have an incentive to deviate: If $1$ knows that $2$ is reporting $100$, the $1$'s best response is to report $99$ to receive $99+2=101$ instead of just $100$.
Iterating this reasoning yields that the (unique) Nash equilibrium is $(2, 2)$, which is indeed the welfare-minimizing state.
Certainly, this is not really a rational outcome to this game. In some sense, a regret minimizing player is rather looking for a more robust solution against irrational play.
Given a strategic game $(N, (S_i)_{i \in N}, (u_i)_{i \in N})$, we denote the maximum utility that a player $i \in N$ can achieve against a profile $s_{-i}$ by
$$u_i^\ast(s_{-i}) := \max_{s_i \in S_i} u_i(s_i, s_{-i}).$$
We define the regret that player $i$ experiences for playing $s_i$ against $s_{-i}$ by
$$\mathsf{regret}_{u_i}(s_i, s_{-i}) := u_i^\ast(s_{-i}) - u_i(s_i, s_{-i}).$$
The maximum regret that a player can experience by playing a strategy $s_i \in S_i$ is
$$\mathsf{maxreg}_{u_i}(s_i) := \max_{s_{-i} \in S_{-i}} \mathsf{regret}_{u_i}(s_i, s_{-i}).$$
A regret minimizing player is a player who aims to minimize the maximum regret, i.e. she wants to find a strategy $s_i \in S_i$ such that $\mathsf{maxreg}_{u_i}(s_i)$ is minimized.
Applying these definitions, will yield that the strategies $\{96, \ldots, 100\}$ will minimze the maximum regret. To most people, these number should seem to be much more reasonable strategies for the Traveler's Dilemma than the Nash equilibrium.
Moreover, if we restrict the strategy space to these numbers, then $97$ will be the only strategy that survives a procedure one may call iterated regret minimization, presented in [Halpern, Pass (2012)] in which you can also find the calculations for the example above (it is also instructive and not too difficult to do it yourself, however).