I saw this game in The Triangle of Sadness (I wasn't the biggest fan of the movie, but YMVM).
Woody Harrelson picks up a card from a deck ($2N=52$), and asks his opponent red or black. If his opponent guesses correctly, then they are safe. If they guess incorrectly, they must drink. It would seem the winning strategy is to count the number of red and black cards remaining in the deck and guess appropriately. But what percentage of the time can we expect his opponent not to drink? This is easy in simple cases (e.g. $N=1$, guess correctly the first time 100%, guess incorrectly, you know the next card for $E[\text{Correct}] = 75\%$).
Short of enumerating possibilities what is a strategy tractable for humans for large Ns? One could try and enumerate all cases, but the possibility tree seems to blow up. I cheated and used a Monte Carlo:
E[Correct]/Total vs N
It makes sense that the counting advantage approaches 0 the larger the deck, but I wanted a bit more intuition. There are many other card questions, but I did not see this one posted.
Also, in the movie it doesn't matter. Because you can catch Woody Harrelson cheating.
It turns out that there is simple a closed-form expression for $\ E(r,r)\ $. I wrote a Magma script script to calculate the values of $\ E(r,b)\ $ using a version of lulu's recursion. Here are the values of $\ E(r,r)\ $ for $\ r=1\ $ to $\ r=26\ $:
\begin{array}{|r|l|} r&E(r,r)\\ \hline 1 & \frac{3}{2}\\ 2 & \frac{17}{6}\\ 3 & \frac{41}{10}\\ 4 & \frac{373}{70}\\ 5 & \frac{823}{126}\\ 6 & \frac{3565}{462}\\ 7 & \frac{7625}{858}\\ 8 & \frac{129293}{12870}\\ 9 & \frac{272171}{24310}\\ 10 & \frac{1139735}{92378}\\ 11 & \frac{2376047}{176358}\\ 12 & \frac{19743201}{1352078}\\ 13 & \frac{40890483}{2600150}\\ 14 & \frac{168947957}{10029150}\\ 15 & \frac{348259369}{19389690}\\ 16 & \frac{11464229693}{601080390}\\ 17 & \frac{23547218611}{1166803110}\\ 18 & \frac{96587303059}{4537567650}\\ 19 & \frac{197831583443}{8836315950}\\ 20 & \frac{1618881562939}{68923264410}\\ 21 & \frac{3308327420393}{134564468610}\\ 22 & \frac{13508555185547}{526024740930}\\ 23 & \frac{27554570432479}{1029178840950}\\ 24 & \frac{449278087454089}{16123801841550}\\ 25 & \frac{915002452786559}{31602651609438}\\ 26 & \frac{3724430600965475}{123979633237026}\\ \hline \end{array}
The numerators of this sequence are given as entry A322755 in the online encyclopedia of integer sequences, and the denominators as entry A322756. The commentaries on these entries tell us that $$ E(r,r)=r - \frac{1}{2} + \frac{2^{2r-1}}{{2r\choose r}}\ . $$ They also tell us that an equivalent version of the game had been proposed as problem number $\ 630\ $ in the College Mathematics Journal. The solution by John Henry Steelman appears in Vol. 30, No. 3 (May, 1999) of the same journal on pp. 234-235.
An approximate value of $\ E(26,26)\ $ is $\ 30.04066\ $.