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 $2^{2021+1010}$.
 A: Finding that summation...., too lazy for that. But solving that puzzle you gave in your description is tempting enough to give it a try.

First a modulation.

Let it be that $B$ throws $n$ dice and $A$ throws $2n$ dice.

If $O_B$ and $O_A$ denote the number of odds thrown by $B$ and $A$ respectively then it can be shown that: $$P(O_A>2O_B)=P(O_A<2O_B)\tag1$$
I will prove this assertion below but let us first look at the consequences of $(1)$.
Suppose that $A$ and $B$ want to avoid a draw and agree that after the $3n$ throws player $A$ always throws a die that is decisive in the case that there is indeed a draw. It has no impact if there is no draw but if there is one then $A$ wins iff this final throw results in an odd number.
If $(1)$ is true then evidently this agreement provides equal chances for $A$ and $B$ to win.
But this agreement also boils down to:

$B$ throws $n$ dice and $A$ throws $2n+1$ dice and $A$ wins if he throws a number of odds more than twice as $B$.

which is exactly the game you describe in your question.
So if $(1)$ is correct then we can conclude that $A$ will win the game with probability $\frac12$ or equivalently that:$$\sum_{r=0}^n\binom{n}r\sum_{k=2r+1}^{2n+1}\binom{2n+1}k=2^{3n}$$

Now a proof of $(1)$ which is surprisingly simple.
Let $E_B$ and $E_A$ denote the number of evens thrown by $B$ and $A$ respectively.
Then by symmetry we find:$$P(O_A>2O_B)=P(E_A>2E_B)=P\left(2n-O_A>2(n-O_B)\right)=P(O_A<2O_B)$$and we are ready.
A: 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}
A: A variation. We obtain
\begin{align*}
\color{blue}{S_n}&=\sum_{r=0}^{n}\binom{n}{r}\sum_{k=2r+1}^{2n+1}\binom{2n+1}{k}\\
&\,\,\color{blue}{=\sum_{r=0}^n\binom{n}{r}\sum_{k=2n-2r+1}^{2n+1}\binom{2n+1}{k}}\tag{$r\to\ n-r$, (1)}\\
\\
\color{blue}{S_n}&=\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}^{2n-2r}\binom{2n+1}{2r+1+k}\tag{index $k$ starts with $0$}\\
&=\sum_{r=0}^n\binom{n}{r}\sum_{k=0}^{2n-2r}\binom{2n+1}{2n+1-k}\tag{$k\to 2n-2r-k$}\\
&\,\,\color{blue}{=\sum_{r=0}^n\binom{n}{r}\sum_{k=0}^{2n-2r}\binom{2n+1}{k}}\tag{2}\\
\end{align*}

Adding (1) and (2) and division by two gives
\begin{align*}
\color{blue}{S_n}&=\frac{1}{2}\sum_{r=0}^n\binom{n}{r}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\\
&=\frac{1}{2}\cdot 2^{n}\cdot 2^{2n+1}\\
&\,\,\color{blue}{=2^{3n}}
\end{align*}

