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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:

P vs N

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.

Dylan Madisetti
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  • Well, there's a simple recursion: Let $E[r,b]$ denote the expected number of drinks assuming you start with $r$ Red and $b$ Black cards in the deck. Then, if $r≥b$ you guess Red and we see that $E[r,b]=\frac r{r+b}\times E[r-1,b]+\frac b{r+b}\times (E[r,b-1]+1)$ with a similar recursive step if $b>r$. And boundary conditions $E[r,0]=0=E[0,b]$. Should be easy to implement...at least then you could solve the problem exactly for many values of $N$. – lulu Jan 23 '23 at 18:56
  • Right, but short of dynamic programming? How do I characterize the curve I posted – Dylan Madisetti Jan 23 '23 at 18:58
  • What do you mean? Looks like a simple decaying exponential. you could fit parameters to your data. As you say, for large $N$ you are mostly guessing blind, so you'd expect to be wrong half the time, so maybe it's just $\frac 1N$ for large $N$. Small $N$ seems a lot harder to get intuition for, but the recursion should help. – lulu Jan 23 '23 at 19:01
  • I mean you're guessing at this point. I can solve the problem (and fit the curve), but I posted this looking more for intuition of the closed form. Also the BC is E[0, b] = b (given b >= r) – Dylan Madisetti Jan 23 '23 at 19:56
  • I don't understand what you are asking. I gave you an explicit recursion so you can solve the problem exactly. Granted, that won't work well as $N$ gets very big, but you can get a lot of exact values easily. And you can fit a curve to your data if you want, but you don't have to. And, of course $E[0,b]=0$...each choice is determined so you never have to drink in that scenario. – lulu Jan 23 '23 at 20:03
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    Should stress: if you want a guess, my guess is that there is no simple closed form for the solution. That could certainly be wrong, but that would be my guess. Best way to find out is to work it out for a lot of $N$ and look for patterns. Then use the recursion to prove that those patterns hold generally. – lulu Jan 23 '23 at 20:04
  • Woops. =0 is fine. I meant to defined it as the percentage of the time that you do not drink (I updated the question, that way it also matches the plot). At what point to we determine there is no simple closed form? This is more a question on how to approach this problem. I can run Monte Carlo or find a explicit recursive definition- but in terms of problem solving approach do I just play around and try to fit the curve and look for patterns? – Dylan Madisetti Jan 23 '23 at 20:16
  • Yeah, that's just about all you can do. If there is a simple closed form, it shouldn't be hard to spot. I suggest: Look at patterns in $E[b,r]$, not just in $E[N]=E[N,N]$. The recursion needs to consider unbalanced decks. Also...might not be a bad idea to post $E[N]$ for $N≤10$ or so. Maybe someone else can spot a pattern even if you can't. – lulu Jan 23 '23 at 20:18
  • Worth saying: the reason I think there is not a closed form solution is that this feels quite similar to the famous red and black card problem which has been studied extensively (and which lacks a closed form). Of course, the problems are not equivalent. – lulu Jan 23 '23 at 20:27

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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\ $.

lonza leggiera
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  • Super cool! Thanks for the reference. Wow OEIS really does have everything- that's a great suggestion to keep everything symbolic and use it as a reference. – Dylan Madisetti Jan 24 '23 at 17:55