Max of algebraic equations

Question

I am going through Dynamic Programming and Optimal Control text by Dimitri Bertsekas, on p. 18 - he derives these equations ...

$$\begin{equation}\begin{aligned} \text{open-loop probability of win} &= \max(p_w^2(3-2p_w), p_wp_d+p_w^2(1-p_d)) \\ &= p_w^2+p_w(1-p_w)\max(2p_w,p_d) \end{aligned}\end{equation}\tag{1}\label{eq1}$$

and then...

$$\begin{equation}\begin{aligned} \text{value of information} &= p_w^2(2-p_w)+p_w(1-p_w)p_d \\ &- p_w^2-p_w(1-p_w)\max(2p_w,p_d) \\ &= p_w(1-p_w)\min(p_w,p_d-p_w) \end{aligned}\end{equation}\tag{2}\label{eq2}$$

Please can someone help explain:

1. In equation 1 - how one would go about deriving line# 2 from line# 1, and
2. In equation 2 - how and why max turns into min?

Where: $p_w$ and $p_d$ are probabilities of winning and drawing the game respectively.

Answer 1

1. $\max(a+b,a+c) = a+\max(b,c)$, and $\max_i \{k f_i\} = k \max_i \{f_i\}$ for $k\ge 0$
2. $\max_i\{f_i\} = -\min_i \{-f_i\}$