Unclear step in derivation of optimality equation for state-value function By substituting the optimal value function into the Bellman equation, we get the Bellman equation for $v_{\star}$
$$ v_{\star}(s) = \sum\limits_a \pi_{\star}(a|s) \sum\limits_{s'} \sum_r p(s', r | s, a)[r + \gamma v_{\star}(s')]$$
From the above equation, how can we obtain the this one?:
$$ v_{\star}(s) = \max\limits_a \sum\limits_{s'} \sum\limits_r p(s', r|s,a) [r + \gamma v_{\star}(s')])$$
 A: Maybe you should provide some context. The second equation can only be derived from the first one, if the policy is deterministic, so for a poker game this would not hold.
The second equation is then obtained from the first by the policy
\begin{equation}
\pi_{\star}(a | s) = \begin{cases}
1  & \text{if a = }\text{a}_{\star},\\
0  & \text{otherwise}\\
\end{cases}\end{equation}

Where $a_{\star}$ is the best action that can be taken in $s$, hence $a_{\star} = \text{argmax}_a q(s,a)$.
Then
\begin{equation}
v_{\star}(s) = \sum_{a} \pi_{\star}(a | s) \sum_{s',r}p(s', r|s,a)[r + \gamma v_{\star}(s')] = \sum_{a} \pi_{\star}(a | s)q_{\star}(s, a) = q_{\star}(s, a_{\star}) = \max\limits_a q_{\star}(s, a) = \max\limits_a \sum_{s',r}p(s', r|s,a_{\star})[r + \gamma v_{\ast}(s')]
\end{equation}
I used the equality $\sum_{s',r}p(s', r|s,a)[r + \gamma v_{\star}(s')] = q_{\star}(s, a)$ which is given by Sutton's "Reinforcement Learning: An Introduction" (second edition) at equation (3.20) combined with (3.19) in straight-forward fashion.
