Special 64 card 2 player game. Who has the advantage? Two people (call them E and F), decide to play a special card game.  They use a special deck of playing cards which is a standard $52$ card deck, but with $12$ extra cards added in.  You can think of them as $3$ extra ranks, each having $4$ suits.  So the new ranks (in ascending order) are $123456789TJQKABC$ (T = ten).  Ranks $2$ thru $A$ are the normal ranks found in a standard deck of cards.  $A$ is considered the rank just above king and never the same as $1$ for this game.  We added a new low $1$ rank which comes before $2$, and we added $2$ new high ranks ($B$ and $C$) after the $A$ (Ace).  So for each hand, we shuffle well and then deal out only half ($32$ out of $64$) of the cards.  They are community cards shared between E and F.  That is, both E and F use the same $32$ dealt cards to determine winners.    Both E and F can win in the same hand multiple times.
E gets a win for each occurrence of a quad dealt in order, including multiple wins per hand such as $77772439999$ which would count as $2$ wins for E.  Best case for E would be something like $11112222333344445555666677778888$ which would be $8$ wins.
F gets a win if any $5$ card straight is dealt in order (such as $23456$).    Wraparound straights such as $ABC12$ are not a win for F.  Only ascending straights are wins for F.  For something like $234567$, that counts as a double win for F since there are $2$ overlapping straights ($23456$ and $34567$).  Best case for F then would be $24$ wins in a single hand ($123456789TJQKABC123456789TJQKABC$).  To make it even easier for F to win, we will assign a point multiplier.  Each win (as described above) will have points awarded.  E gets $8$ points for each win but F gets $15$ points for each win.
So best case for E would be $64$ points in a single hand but F can get up to $360$ points in a single hand.
If there is no winner in a hand, the cards are simply returned to the deck, reshuffled well, and a new hand is dealt.
So who has the point advantage long term and by how much?
 A: Let's start with E.  We number the cards in the hand from 1 to 32.  Let
$$X_i = \begin{cases} 1  &\text{if there are four of a kind starting at card i}\\
0  &\text{otherwise}
\end{cases}$$ for $i = 1,2, \dots ,29$.  Then
$$\Pr(X_i = 1) = \frac{3}{63} \cdot \frac{2}{62} \cdot \frac{1}{61}$$ and the expected number of "wins" per hand for E is
$$E(\sum_{i=1}^{29} X_i) = \sum_{i=1}^{29} E(X_i) = 29 \cdot \frac{3}{63} \cdot \frac{2}{62} \cdot \frac{1}{61} \approx 0.0007303$$
So E's expected score is about $8 \times 0.0007303 \approx 0.005842$.
Now for F.  Let $$Y_i = \begin{cases} 
1 &\text{if there is a 5-card straight starting at card i} \\
0 &\text{otherwise} 
\end{cases}$$ for $i = 1, 2, \dots ,28$.  First, the card at i must be in the range from 1 to 12 in order for a straight to be possible; then the next 4 cards must line up appropriately. So
$$\Pr(Y_i = 1) = \frac{12}{16} \cdot \frac{4}{63} \cdot \frac{4}{62} \cdot \frac{4}{61} \cdot \frac{4}{60}$$ and the expected number of "wins" per hand for F is
$$E(\sum_{i=1}^{28} Y_i) = \sum_{i=1}^{28} E(Y_i) = 28 \cdot \frac{12}{16} \cdot \frac{4}{63} \cdot \frac{4}{62} \cdot \frac{4}{61} \cdot \frac{4}{60} \approx 0.0003760$$ So F's expected score is about $15 \times 0.0003760 \approx 0.005641$.
Therefore E's expected score is a little higher than F's (0.005842 vs. 0.005641); but the game must be boring to play, because most of the time no-one scores.
