double decker, 13 card flush vs. 18+ of each color card. Who has the better odds of winning? Two people, call them $A$ and $B$, decide to play a card game. They take 2 standard decks of playing cards and combine them into a "superdeck" of 104 cards, shuffle them well, and then draw 1 card at a time randomly without replacement. $A$ immediately wins if 13 cards of any single suit are drawn.  $B$ immediately wins if at least 18 black and at least 18 red cards are drawn. If they both "win" on the same card draw it is a tie (no decision) and they start over with all 104 cards after reshuffling. They bet even money—dollar for dollar matching. Who, if anyone, has the advantage of winning and by how much?
Note that all cards are shared "community" cards so the players are not drawing their own set of cards.
If someone would like to comment on if and how the probability changes if they draw their cards in tandem, that would be very interesting and informative. For example, two cards would be drawn at a time, giving one to $A$ and one to $B$, and then checked if anyone won or if they tied. This is an optional bonus question and not required to earn the checkmark for best answer.
Simulating 1,000,000,000 (1 billion) decisions (including ties) of the original question (shared cards), I got the following winning percentages:
$$
A\quad 38.7855918\%\\
B\quad 59.7770491\%\\
\textrm{Tie}\quad 1.4373591\%
$$
Minimum # of cards : $14$ (A disappointment as I was expecting it to be $13$)
Maximum # of cards : $42$
Average # of cards : $37.1902$
Note that at "first glance", some people might think that $A$ has the advantage because $A$ can win with as few as 13 cards but B needs 36 cards at a minimum to win. However, look at how much $B$ is actually favored to win. Another thing that somewhat surprised me is the average number of cards for someone to win (or tie) is about 37, only about 1 card more than $B$'s minimum needed to win. That might also imply that $A$ has an advantage. What if I reworded the question to also state that the average number of cards for a decision is slightly over 37, would most people then think at first glance that $A$ has an advantage, possibly a huge one? Yet another thing that might mislead people into thinking that $A$ has a large advantage is that $A$ is guaranteed not to lose if at least 42 cards are drawn. At that point either $A$ has won or tied. Kudos to the person that suspected that $B$ was the favorite from the start because he didn't let these false biases sway him towards $A$ being the favorite.
UPDATE:
I ran the alternate simulation giving each player their own cards and got some interesting results. $B$ remains the favorite at about $57.1\%$ to $A$'s $36.8\%$ which means the ties more than quadrupled to about $6.1\%$. The average number of cards slightly more than doubled from about 37 to about 75 with a low of 28 and a high of 94 (in ten million decisions).
Perhaps one way to simplify getting the answer to this original problem is to start the game by immediately turning over 35 cards since $B$ cannot win with less than 36 cards. Check if $A$ won at 35 cards. If not, then draw cards 36 thru 42 one at a time checking for a winner.  By draw 42 someone will have won or there is a tie. So would the prob of $A$ winning on a 35 card draw be:
$$
\large
\frac{4\times {26 \choose 13} \times {78 \choose 22}}{{104 \choose 35}}
$$
That is about $10\%$. There may be two 13 card flushes in that but we don't care as long as there is at least 1.
Also it is interesting to note that in the original game with shared cards, $42$ cards is the max for a win or tie because of the $25+17$ situation but with separate hands, that doesn't happen so the max # of cards drawn can exceed $84$ (which is $2$ * $42$) and in my simulation is did exceed it as it maxxed out at $94$ cards, not $84$.  For example, B could get $29$ black cards and $17$ red cards but A could have $12, 12, 11,$ and $11$ of each suit (respectively) so at that point, neither is a winner and $2$ more cards would be drawn (almost exhausting the deck of $104$).  The max number of cards ever for the separate card variation of this game seems like $94$ but I am not certain yet.
Another interesting point is that simulation of the separate card variation is considerably slower because the average approximate number of cards needed is about double ($75$ vs. $37$) and can go as high as $94$ which is almost the entire deck so it becomes "harder" (slower).
 A: The Python code below simulates matches with or without replacement.  Based on one million matches with replacement, player A wins $56.8 \pm 0.1 \%$ of the time (and ties occur about $2\%$ of the time), consistent with the two existing answers.  Based on one million matches without replacement, on the other hand -- which is OP's actual question -- player B wins $59.8 \pm 0.1 \%$ of the time (and ties occur about $1\%$ of the time).  So drawing without replacement hurts player A enough to throw the advantage over to player B.
To see why this might be true, note that player A needs half the cards of some suit, while player B needs about a third of the cards of both colors... so if the draws are even across suits, player A needs to see $1/2$ of the deck, while player B only needs to see $1/3$.  Drawing without replacement is "self-correcting", and tends to keep the number of cards drawn from different suits closer together (giving the edge to B).  Drawing with replacement is not, allowing one suit to more easily random-walk into a big lead (giving the edge to A).

import random

rng = random.Random()

def getShuffledDeck(m, n, x):
  deck = []
  for i in xrange(m):
    for j in xrange(n * x):
      deck.append((i, j%n))
  rng.shuffle(deck)
  return deck

def playGame(repl=False, m=13, n=4, x=2, aa=13, bb=18):
  deck = getShuffledDeck(m, n, x)
  suitCts = [0 for suit in xrange(n)]
  colorCts = [0, 0]
  pos = -1
  while True:
    if repl: pos = rng.randrange(0, len(deck))
    else: pos += 1
    (rank, suit) = deck[pos]
    suitCts[suit] += 1
    colorCts[suit % 2] += 1
    awin = (max(suitCts) >= aa)
    bwin = (min(colorCts) >= bb)
    if (awin or bwin): return (awin, bwin)
  return (False, False)

def playGames(num, repl=False):
  ret = {}
  for i in xrange(num):
    win = playGame(repl)
    ret[win] = ret.get(win, 0) + 1
  return ret


A: I ran a Python simulation for an infinite deck, which is the same as if the draws are with replacement.  Out of $10,000$ games I got $5728$ victories for A, $4069$ victories for B, and $203$ ties, close to David's calculation.  I verified one game counted the individual cards correctly and twenty games counted the winner based on the card counts.  A slight win for A.  Here is the code if you care.  plev is just the diagnostic print level.
import random  
def game(plev=0):  
    counta=[0,0,0,0] #count of cards in each suit#  
    countb=[0,0] #count of black and red#  
    awin=False  
    bwin=False  
    while True:  
        i=random.randint(0,3) #pick a suit#  
        counta[i]+=1 #add 1 to the proper suit#  
        countb[i/2]+=1 #add 1 to the proper color#  
        if max(counta)==13: awin=True  
        if min(countb)==18: bwin=True  
        if plev > 19:  
            print i, counta, countb, awin, bwin  
        if awin and not bwin: #a has wone and b has not#  
            if plev > 9: print counta, countb, awin, bwin  
            return 'a'  
        if bwin and not awin: #b has won and a has not#  
            if plev > 9: print counta, countb, awin, bwin  
            return 'b'  
        if awin and bwin: #it is a tie#  
            if plev > 9: print counta, countb, awin, bwin  
            return 'tie'  
def match(games,plev=0):  
    awins=0  
    bwins=0  
    ties=0  
    for i in range (games):  
        result=game(plev)  
        if result=='a': awins+=1  
        if result=='b': bwins+=1  
        if result=='tie': ties+=1  
    print awins, bwins, ties  

A: While this is only heuristic, we can get an interesting first take by considering how many cards of the 'larger' color B will have when they first satisfy the win condition.  We treat the red vs. black cards as heads vs. tails coin flips (i.e., simplifying to draw with replacement); then after $N$ 'flips' the expected number of red cards is $\frac12N$ and the variance is $\sigma^2=\frac14N$ (i.e. $\sigma=\frac12\sqrt{N}$).  The next approximation is to treat this as approximately normal, and to use results on the Half-normal distribution to figure out what the expected value of min(red, black) is (since min(red,black) and max(red,black) are symmetric).  Since the expectation of a half-normal variate is $E=\frac{\sigma\sqrt{2}}{\sqrt{\pi}}$, we should figure on min(red,black) being approximately $\frac{N}{2}-\frac{\sqrt{N}}{\sqrt{2\pi}}\approx .5N-.4\sqrt{N}$.  Setting this equal to 18 and solving suggests $N\approx 41$; in other words, when B gets 18 of their 'minimum' color they'll have (on average) approximately 23 of their 'maximum' color.  Then there need to be no more than 13 cards in any individual pile, or else A has already won; we can use a normal approximation to the binomial again to figure out these probabilities.  For $N=18$, $\sigma=\frac32\sqrt{2}\approx2.12$; this means that $A$ wins on the 'small' side if the result is more than approximately 1.9 standard deviations from the norm; the probability of this is (very approximately) 5%.  Similarly, on the $N=23$ 'side', $\sigma \approx 2.4$ and so the extremal result is approximately .625 SDs from the mean; the probability of A winning on this 'side' is then about 53%.
Combined, these suggest that A possibly has a small edge, but of course the lack of replacement throws a wrench into the work here; overall, the odds look remarkably even.
