11. (12 points) Huanhuan, Yingying, and Nini have a total of 2712 cards in their hands, and there are still some cards on the table. They performed the following operations:
First time: Huanhuan took 1 card from each of Yingying and Nini;
Second time: Nini took 2 cards from the table, and then made the person with fewer cards among Huanhuan and Yingying take 1 card from the person with more cards;
Third time: Yingying took 4 cards from the table, and if the number of cards in her hand was even, she would give half of her cards to the person with fewer cards among Huanhuan and Nini; if it was odd, the game would end.
We call the above three operations "one round of operations". If they successfully completed 50 rounds of operations without the game ending, and at this point the number of cards they have in their hands, in the order of Nini, Huanhuan, and Yingying, forms an arithmetic sequence. How many cards did Huanhuan originally have?
【Solution】Solution: According to the analysis, Yingying is one card short the first time, may have one more or one less card the second time, and has four more cards the third time, but then loses half of them,

So Yingying loses about half of the cards each round. Even if she initially has 2712 cards, after a dozen rounds of operations, she will be down to single digits,

To last 50 rounds, the number of cards Yingying has must eventually be a fixed value, ensuring that she always has an even number of cards for the third operation of each round,

Consider that at this point, Yingying must have fewer cards than Huanhuan. Let's assume Huanhuan has $a$ cards, Yingying has $b$ cards, and $a > b$, then one round of operations is as follows:
\begin{tabular}{|l|l|l|l|}
\hline & Huanhuan & Yingying & NiNi \\
\hline Original & $a$ & $b$ & $c$ \\
\hline First & $a+2$ & $b-1$ & $c-1$ \\
\hline Second & $a+1$ & $b$ & $c+1$ \\
\hline Third & & $b \div 2+2$ & \\
\hline
\end{tabular}

So we have $b=b \div 2+2, b=4$, meaning Yingying ends up with 4 cards;
Each round, the total number of cards increases by $2+4=6$ cards, so the final total number of cards is $2712+6 \times 50=3012$ cards, Huanhuan has $3012 \div 3=1004$ cards, so NiNi has 3012 - $1004=2004$ cards; using the reverse method, we can get the number of cards each person has after one round:
\begin{tabular}{|l|l|l|l|}
\hline & Huanhuan & Yingying & NiNi \\
\hline Third & 1004 & 4 & 2004 \\
\hline Second & 1000 & 4 & 2004 \\
\hline First & 1001 & 3 & 2002 \\
\hline Original & 999 & 4 & 2003 \\
\hline
\end{tabular}

From the data in the table, we can see that Huanhuan gains 5 cards each time, so Huanhuan originally had $1004-5 \times 50=754$ cards. Therefore, the answer is: 754.