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Min Max Regret Strategy

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| | |

Expected Opportunity Lost

Similar idea:

  • we calculate the expected value on the regret table

Sources