Edit for "off topic" "correction".
I don't know why this question was considered "off topic". It seemed to draw a lot of attention. The "puzzle" was to figure out which person has the advantage which is not obvious. The pieces that need to be put together are the checks for each winner.
Also, if this type of question is "off topic", then why do they have both probability and card tags available on this site? I used both of those tags to classify my question here.
Four people, (call them C, D, E and F), decide to play a card game for fun. They use an ordinary fair deck of $52$ cards, shuffled well immediately before each hand is drawn, and randomly draw cards from it one a time without replacement, all 4 using (sharing) the same drawn cards to determine who wins. A win is defined as follows:
C wins if he gets at least one of all $13$ ranks of the cards (regardless of suit as they can be mixed suits or even all the same suit).
D wins if he gets either $6$ red cards or $6$ black cards in a row (consecutive).
E wins if she gets $4$ of a kind of ANY ONE odd rank ($A,3,5,7,9,J,K$).
F wins if she gets at least $12$ black cards and at least $12$ red cards.
Each win is based on the current hand only as there is no "carryover" from a previous hand. Each new hand starts "fresh".
It is possible for ties to occur but the rule is any and all ties are awarded as a half win for D and a half win for E. That is, if each player bet 1 dollar for a hand so that there were 4 dollars in the pot for that hand, D and E would split any ties so that they would get 2 dollars each. Note that D and E split ANY ties so even if only C and F tie, they both lose and D and E split the pot with a half win each.
So the question is who has the highest probability of winning and by how much over the competition?
Note that the minimum # of cards needed to win differs for each player:
(C: $13$), (D: $6$), (E: $4$), (F: $24$).
Another game rule is if nobody wins by the $28$th card drawn in the hand, then that hand has no winner and a new hand will be drawn. That is, $28$ cards max will be drawn per hand.
F.Y.I., my simulation has lots of "buckets" to count up interesting things and the most rare thing I see (that shows up as nonzero) out of $10,000,000$ iterations is D and E tying in only $8$ cards drawn. This is not a "bonus" tie because neither C nor F can tie with only $8$ cards drawn. It is so rare that out of $10$ million decisions, only $29$ were ties between D and E with exactly $8$ cards drawn. That one special case can probably be calculated mathematically since there are only $52 \choose 8$ card combos which is about $3/4$ billion. Even simulation of all of those hands is possible on a computer to get the exact count of winners.
$UPDATE:$ I am curious about how many actual ways there are for D and E to tie in exactly $8$ cards since it seems to be the most rare event that is happening in my simulation. I was able to quickly generate the 0.75 billion card combinations which is $52 \choose 8$ so I just have to insert the check for the ties and count them up. I have another simulation running now overnight so I don't want to stress my CPU at $100$% for many hours so I will wait until the first one finishes and then code the 2nd one and report back. Based on the simulation getting only $29$ out of $10,000,000$, I could think out of $750$ million possible $8$ card hands there should be a little over $2000$ ways to tie.
$UPDATE - 2$ Out of 1 billion simulated hands, I am seeing only $2547$ D/E ties on the $8$th drawn card. That is the most rare event I see of all the buckets I am viewing. It is almost as rare as a royal straight flush in a $5$ card poker hand ($5$ cards only).
$UPDATE - 3$ I made a mistake in that to simulate all the possible $8$ card hands to check for a D/E tie on that exact card, it is not just $52 \choose 8$ combinations because order is important, so it is $8$! more than that which is a huge number and too large to easily simulate on a computer. An example where D and E tie on the 8th card (using C=Club, H=Heart, S=Spade, D=Diamond) is AH,AD,AC,2C,3C,4C,5C,AS. That completes both the $6$ blacks in a row and quad aces on the $8$th card. So since $52!$ / $44!$ is such a huge number of hands to simulate/generate (it is over $30$ trillion), probably the best I can do (easily) is just simulate a $7$ card hand, check to make sure neither D nor E won at that point, then draw the 8th card and check if they tied on that card. Again that is only an approximation but will be faster than drawing up to $28$ cards and checking for $4$ possible winners. Note that there is no way there can be a tie on the $7$th card because a quad has $2$ of each color card so we would need those plus $6$ in a row of the same color card so that is $8$ cards minimum.