Min Max Regret Strategy
How to choose an alternative?
This principle is also called the Savage's Opportunity Loss principle.
Idea:
- instead of calculating the maximal gain (like in Max Max Strategy) or maximal loss (Max Min Strategy) we calculate the regret
- use this measure to decide which option to choose
- we don't want to experience a lot of regret, so we will minimize the maximal regret we have
- so it's similar to Max Min Strategy, but instead of utility we use regret
Regret
- regret is a measure that shows how we regret choosing some alternative $a$ to another alternative $a^*$ after $e \in E$ happens
- imagine that $e$ happens and the best alternative in this case is $a^*$
- but we chose $a$
- so our regret is $c(e, a^*) - c(e, a)$
- it we chose $a^*$ then our regret is 0
So define regret as
- $R(a, c) = max_{b \in A} \big[ c(b, e) - c(a, e) \big]$
We choose such $a \in A$ that:
- $\min_{a \in A} \max_{e \in E} R(a, e)$
Remarks
- it must be meaningful to make differences for calculating regret
- so the scale should be numerical, not ordinal
Example
Suppose we have the following matrix:
| $c$ |
$e_1$ |
$e_2$ |
$e_3$
|
| $a_1$
|
40 |
70 |
-20
|
| $a_2$
|
-10 |
40 |
100
|
| $a_3$
|
20 |
40 |
-5
|
- if $e_1$ happens, the regret of choosing $a_1$ is 0
- $a_1$ is the best for $e_1$
- if $e_1$ happens, the regret of choosing $a_2$ is 50
- 40-(-10) = 50
- the best value for $e_1$ is the value for $a_1$, which is 40
So this way we compute a regret matrix:
| $R$ |
$e_1$ |
$e_2$ |
$e_3$ |
max
|
| $a_1$
|
0 |
0 |
120 |
120
|
| $a_2$
|
50 |
30 |
0 |
50
|
| $a_3$
|
20 |
30 |
105 |
105
|
In this case the alternative $a_2$ minimizes the maximal possible regret, so we choose it
Manipulation
This method does not satisfy the principle of Independence to Third Alternatives from the Voting Theory
- adding or removing alternatives may alter the choice in unpredicted ways
Example
| $c$ |
$e_1$ |
$e_2$
|
| $a_1$
|
8 |
0
|
| $a_2$
|
2 |
4
|
|
| $R$ |
$e_1$ |
$e_2$ |
max
|
| $a_1$
|
0 |
4 |
4
|
| $a_2$
|
6 |
0 |
6
|
|
Now $a_1$ wins - so we choose it
But what if we add a third alternative?
| $c$ |
$e_1$ |
$e_2$
|
| $a_1$
|
8 |
0
|
| $a_2$
|
2 |
4
|
| $a_3$
|
1 |
7
|
|
| $R$ |
$e_1$ |
$e_2$ |
max
|
| $a_1$
|
0 |
7 |
7
|
| $a_2$
|
6 |
3 |
6
|
| $a_3$
|
7 |
0 |
7
|
|
But now $a_2$ wins!
Similar idea:
- we calculate the expected value on the regret table
Sources