2 player card game involving matching ranks, who is more likely to win? Two players (call them C and D), decide to play a card game using a well shuffled standard 52 card deck.  Player C draws 7 cards from the deck and D draws 4 cards from the deck.  Of the remaining cards in the deck, 1 card at a time is drawn and "shared" by both C and D.  A winner (for the hand) is declared if either C matches 3 of his original 7 cards or D matches 2 of his original 4 cards.  A match is defined as a same rank card that has not already been matched for that same hand and for that same player. Once an original card is matched, it cannot be "rematched" in that same hand.  For example, if C has cards 3 4 5 6 7 8 9 and draws a 3 that is a match but if C draws another 3 in that same hand, it is NOT a match (but the first matching 3 remains a match).  Also, a single card can be a "double" match (for both C and D) such as if C and D both hold an unmatched rank 3 card and a "community" rank 3 card is drawn.  That counts as a match for both C and D.
Some things to note are some original hands can never be winners.  Such as if C gets 333 7777.  There are only 2 unique ranks to match so 3 ranks cannot be matched.  Also if D gets 4444 for example, that is only 1 rank so 2 ranks cannot be matched.  Pairs, triples, and even quads in the original hand make it more difficult (or impossible) to be a winner such as if C draws 3 5 7 3 5 7 3.  In that case, the only way for C to win would be to draw another 3 5 and 7 before D wins the hand but that is "hard" cuz there are at most 2 of each of those ranks remaining in the pile of undrawn cards.  Also, in the rare cases where C and D draw identical "overlapping" ranks, some interesting things can happen.  Suppose C draws 4 5 7 4 5 7 7 and D draws 4 4 5 6.  Here, all the 4s are used up already so they cannot be matched in either hand, but C cannot win cuz there are only 2 more unique ranks to match and C must match 3 to win.  This case will be a win for D as soon as a 5 and 6 appear in the drawn community (shared) cards.
Lastly, ties are possible and end the hand just like a win would except no score is awarded for that hand since it is not a win for either player.
Note I changed this problem slightly from the original post because I realized I had an "error" since there were 41 shared cards and alternating one card each would give one player an "extra" chance to match so I changed it to community (shared) cards instead for the remaining 41.
So the question is who has the winning advantage in the longrun and by how much?
 A: I have a simulation program.  It gives the first 7 cards to C and the next 4 cards to D then it draws community (shared) cards until there is a winner (or tie), or all of the cards in the deck are drawn.  For $250$ million hands I got the following results:
$113,477,236$ : C wins requiring on average $7.3062$ additional cards (counting C wins only).
$116,893,384$ : D wins requiring on average $6.1375$ additional cards (counting D wins only).
$~~19,626,698$ : CD ties requiring on average $9.2320$ additional cards (counting ties only).
$~~~~~~~~~~~~2,682$ : no wins (ran out of cards before either player could win or tie).
So it appears D has a slight advantage in the longrun.  I am a little surprised at how often ties occur, nearly 8% of the time.  That would require each player having a same rank card (such as 9), and then that shared card showing up as the 3rd match for C and the 2nd match for D at the same time.
Also, a real example my computer caught is the following:
C: (sorted) $2,6,6,8,8,8,9$
D: (sorted): $8,9,9,9$
Here, neither player can win since all the 8s and all the 9s are used up.  No wins are rare (about $1$ out of every $100,000$ hands).
One thing that puzzles me somewhat is if D needs 1 less card to win on average, why then does that player appear to only have a slight advantage overall?  Also, f.y.i, when I modded the simulation program to instead use alternating additional cards starting with player C, instead of community (shared cards), I was hoping to make the game more even steven since player C would be about half a card "ahead" of D (on average).  However, when I did this, C is a favorite $52.1$% to D's $47.9$% (excluding ties and no wins).
