For $0\le k\le n$, let $p(n,k)$ be the probability to win the game if it starts with $n+1$ cards $0,1,\ldots, n$ and the first card shown is card $k$. By symmetry, $p(n,k)=p(n,n-k)$. If $k\ge \frac n2$, then the first guess will be "lower" (where both possible guesses a just as good in case $k=\frac n2$) and succeed with $\frac{k}{n}$, namely if the next card $l$ is one of $0,\ldots,k-1$. This gives us the recursion
$$ p(n,k)=\frac1n\sum_{l=0}^{\max(k,n-k)-1}p(n-1,l)$$
(with the bases case $p(0,0)=1$).
If we start with $n$ cards and have not seen the first card yet, the winning probability is then $$p(n)=\frac1n\sum_{k=0}^{n-1} p(n-1,k)$$
I doubt that there is a simple closed formula for $p(n)$.
One can combute by hand that with a five card deck the probability is $p(5)=\frac{31}{60}$, which is nicely close to $50:50$.
Just for fun, I let my computer evaulate this for a deck of $32$, where the overall winning probability is
$$p(32)=\frac{593119019981747761359995677819}{4111419327088961408862781440000000}\approx 0.00014 \approx0.7585^{32}$$
as well as for for a deck of $52$, where it is
$$p(52)=\\\frac{2502415780357808863930924231118727082667638137165232707577843}{10082271896367984821457579607050470871911188180110409728000000000000}\\\approx 2.48\cdot10^{-7}\approx 0.7454^{52} $$
This may be surprising because the first guess will be correct with at least $\frac 34$. So how come that the $0.7585$ per round decrease to $0.7454$ per round when the deck is enlarged from $32$ to $52$ cards? The problem is that after a correct guess of "lower", say, all cards above the previuos card are excluded. Thus the next card is a bit more likely to be of medium size - but that's the situation where guessing right is harder!
In this answer I am assuming that the player is obliged to make his guess of the 5 card sequence before any cards are revealed.
Lets assume that the cards are labeled 1,2,...,52 and that the player's ordered guess is 1,2,3,4,5 ( since this guess is equally likely to be correct as any other guess).
What we want to count is the number of injective functions from {1,2,3,4,5} into {1,2,...,52} that have exactly k fixed points. Here k will correspond to the exact number of cards guessed correctly by the player. We must be familiar with inclusion/exclusion.
Here is the Mathematica code that gives the probability that the player guesses exactly k cards correctly for k=0,1,2,3,4,5.
Table[Binomial[5, j] Sum[Binomial[5 - j, k] FactorialPower[52 - k - j, 5 - j - k] (-1)^k/FactorialPower[52, 5], {k, 0, 5 - j}], {j, 0, 5}]
The exact probabilities are: $\frac{16649407}{18345600}, \frac{5541121}{62375040}, \frac{6511}{1834560}, \frac{2257}{31187520}, \frac{47}{62375040}, \frac{1}{311875200}$.
Best Answer
The best way to see this is to ask yourself what can happen in each scenario. Suppose you asked the first question, about if the card was red or not.
If the card holder says yes, you now only have $26$ cards to guess from (a $\frac{1}{26}$ chance of guessing correctly). If the holder says no, you also have $26$ cards to guess from (because you would know it's a black card, also with $\frac{1}{26}$ chance of success).
Combining these probabilities, we have that $$P({\text{correct guess}})=\frac{1}{2}\left(\frac{1}{26}\right)+\frac{1}{2}\left(\frac{1}{26}\right)=\frac{1}{26}.$$In other words, either the card holder reveals that it's a red card AND (multiplication) you guess correctly, OR (addition) it's a black card AND you guess correctly.
Apply this same thinking to the second question: what happens if the holder says yes, it's a $10$ of spades? What are the chances of them saying that? What are the chances you guess the correct card if they say that? What if they said no? You should see that you also arrive at $P({\text{correct guess}})=\frac{1}{26}.$