Evaluating $\sum_{r=0}^{1010} \binom{1010}r \sum_{k=2r+1}^{2021}\binom{2021}k$ I need to find the summation
$$S=\sum_{r=0}^{1010} \binom{1010}r \sum_{k=2r+1}^{2021}\binom{2021}k$$
I tried various things like replacing $k$ by $2021-k$ and trying to add the 2 summations to a pattern but was unable to find a solution. For more insights into the question, this was essential to solve a probability question wherein there were 2 players, A and B. A rolls a dice $2021$ times, and B rolls it $1010$ times. We had to find the probability of A having number of odd numbers more than twice of B. So if B had $r$ odd numbers, A could have odd numbers from $2r+1$ to $2021$, hence the summation. I can get the required probability by dividing this by $2^{2021+1010}$.

For $N=1010$ or others
\begin{equation}
S_N\equiv \sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
=
\end{equation}
\begin{equation}
\sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
+
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2N+1}\binom{2N+1}{k}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
\sum_{r=0}^N\binom{N}{r}2^{2N+1}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
2^N 2^{2N+1}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
2^{3N+1}
-
% \sum_{k=0}^{2N}\sum_{r=k}^{N}
\sum_{r=0}^{N}\sum_{k=0}^{2r}
\binom{N}{r}\binom{2N+1}{k}
\end{equation}
substituing $r'=N-r$
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k=0}^{2N-2r'}
\binom{N}{N-r'}\binom{2N+1}{k}
\end{equation}
and substituting $k'=2N+1-k$
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k'=2N+1}^{2N+1-(2N-2r')}
\binom{N}{N-r'}\binom{2N+1}{2N+1-k'}
\end{equation}
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k'=2N+1}^{1+2r'}
\binom{N}{r'}\binom{2N+1}{k'}
\end{equation}
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=0}^{N}\sum_{k'=2N+1}^{1+2r'}
\binom{N}{r'}\binom{2N+1}{k'}
=
2^{3N+1}-S.
\end{equation}
Adding $S$ to both sides gives $2S=2^{3N+1}$, therefore
\begin{equation}
S=2^{3N}.
\end{equation}
