Computation of maxmin strategy in a two-players game

Question

I am taking the Stanford Coursera course on Game Theory. Teacher explained the concept of maxmin strategy and tries to give an example with the following game.

Consider a goalkeeper and kicker game where actions for both players are Left and Right:

kicker\goalie |       L      |      R
-----------------------------|---------------
            L |    0.6, 0.4  |    0.8, 0.2
-----------------------------|---------------
            R |    0.9, 0.1  |    0.7, 0.3

The game is explained in this lecture at 9:08.

By substituting $s_i\left(R\right)=1-s_i\left(L\right)$, he reduce the max-min strategy to

$$ \max_{s_1}\min_{s_2}\left( 0.2-s_1\left(L\right)\cdot 0.4 \right) \cdot s_2\left(L\right)+0.7 +s_1\left(L\right)\cdot 0.1$$

Basically he is taking the first derivative with respect to $s_2(L)$ and setting it to $0$, to find the minimum:

$$\left( 0.2-s_1\left(L\right)\cdot 0.4 \right) =0$$

and then it solves for $s_1\left(L\right)$:

$$s_1\left(L\right) = 0.5, s_1\left(R\right)=0.5$$

First of all, since it is a first-degree equation, isn't it wrong to use the derivative to find the minimum?

Secondly, he wants to find minimize over $s_2\left(L\right)$: why does he end up with a minimizing value for $s_1(L)$?

Furthermore, should not the notation for the max-min strategy be $$\arg\max_{s_1}\min_{s_2}$$?

Maybe I am completely wrong, but can someone explain his reasoning? The video is totally unclear to me.

Answer 1

This slides explain it better. When the taking the derivative there are three possible cases:

1. the derivative is always $<0$, whatever the strategy of player 1. Then player 2 should always play the first strategy (with probability one) to minimize player 1's payoff. The derivative is always $<0$; it means the line is decreasing to the minimum value of the line will be at the end of the range $[0,1]$, i.e. $s_2(L)=1$. Player 1 will take the action that maximises her reward when Player 2 takes his first action.
2. the derivative is always $>0$, whatever the strategy of player 1. Then player 2 should always play the second strategy (with probability one) to minimize player 1's payoff. The derivative is always $>0$; it means the line is increasing to the maximum value of the line will be at the beginning of the range $[0,1]$, , i.e. $s_2(L)=0$. Player 1 will take the action that maximises her reward when Player 2 takes his second action.
3. There exist a value of $s_1(L)$ for which the derivative is $0$. When Player 1 chooses this strategy, his payoff does not depend on Player 2's action: this is his max-min strategy. If he deviates from this strategy, the derivative is not $0$ anymore and we fall back in one the two previous cases where Player 2's action can minimize Player 1's payoff (and hence reducing his utility).