Best strategy to optimize probability of winning in card drawing game with opponent

Question

I am curious about the solution to a probability question that was asked in a trading interview:

You and your opponent choose a suit (of a standard 52 cards deck). Then one card after another is drawn (without replacement) until either your suit or your opponents suit have appeared 5 times (not necessarily in a row). You win if your suit is the one which was drawn 5 times first. Now, your opponent lets you decide if you want to choose your suit before the game starts, after 1, or after 2 cards are revealed and he will choose his suit afterwards. Which strategy gives you the best chances to win?

So as an example, I decide to choose after the first card, it comes hearts, I choose hearts, he chooses any of the other suits and from that point I will only need 4 more hearts whereas he still needs 5 of his suit.

This makes it obvious that choosing before the game starts is not optimal, as you get an advantage by choosing the suit of the first card.

But I am not sure how to determine whether choosing after the first or after the second card is better. In the scenario where I choose after the second there are two possibilities: Either the first 2 cards have the same suit, then I will only need 3 more or they have a different suit. In that case I choose one of the two suits and my opponent chooses the other and our chances will be equal again.

To summarize, my chances of winning if I choose after the second card are $$ P(\textrm{first 2 cards have same suit}) P(\textrm{3 out of 11 before 5 out of 13}) \\ + \frac{1}{2}P(\textrm{first 2 cards have different suit}) \\ = \frac{12}{51}P(\textrm{3 out of 11 before 5 out of 13}) + \frac{1}{2}\frac{39}{51} $$ And if I choose after the first card my chances are simply $$ P(\textrm{4 out of 12 before 5 out of 13}) $$

But in both cases I have no idea how to come up with solutions for the missing probabilities, especially because the game is played without replacement and I can not use a binomial distribution approach as it were possible if the game was for example played with coin tosses and you can choose head or tails.

Answer 1

Update: Notice that if we had only two suits to begin with, the two strategies would yield the same winning probability. So in effect, having two more suits diluted the weight of $p$ in $P_2$.

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There's little calculation needed.

Denote $P_i$: probability you win if you choose suit after seeing $i$ card(s).

Denote $p$: probability you win if you have already seen two cards of your suit, and no cards from opponent's suit. Then $p>\frac 12$.

And

$$P_1 = \frac{12}{25}p+\frac{13}{25}\cdot \frac 12\\ P_2 = \frac {12}{51} p + \frac{39}{51} \cdot \frac 12\\ \implies P_1 - P_2 = \left(\frac{12}{25}- \frac{12}{51} \right) p + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right)\\ > \left(\frac{12}{25}- \frac{12}{51} \right) \frac 12 + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right) = 0. \blacksquare $$

Answer 2

Let $f(i, j)$ be probability of you winning if there already are $i$ of your cards and $j$ of your opponents ($0 \leq i, j \leq 5$, at most one of $i$ and $j$ is equal to $5$). $f(5, j) = 1$ and $f(i, 5) = 0$ by definition, also $f(i, i) = \frac{1}{2}$ by symmetry.

Probability of winning if you choose before any card draw is $f(0, 0)$, if you choose after the first - $f(1, 0)$ and after the second - $p \cdot f(1, 1) + (1 - p) \cdot f(2, 0)$, where $p = \frac{39}{51}$ is probability of first two cards having different suits.

Your question is if $f(1, 0) > \frac{39}{102} + \frac{24}{102} \cdot f(2, 0)$.

It's not too hard to calculate using $f(i, j) = \frac{13 - i}{26 - i - j} \cdot f(i + 1, j) + \frac{13 - j}{26 - i - j} \cdot f(i, j + 1)$. In theory, it can be calculated by hands if needed, but I'll cheat a bit and just print the table:

$$\begin{array}{|c|c|c|c|c|c|} \hline j\backslash i & 4 & 3 & 2 & 1 & 0 \\ \hline 4 & \frac{1}{2} & \frac{5}{19} & \frac{11}{76} & \frac{11}{133} & \frac{13}{266} \\ \hline 3 & \frac{14}{19} & \frac{1}{2} & \frac{44}{133} & \frac{29}{133} & \frac{442}{3059} \\ \hline 2 & \frac{65}{76} & \frac{89}{133} & \frac{1}{2} & \frac{1117}{3059} & \frac{923}{3496} \\ \hline 1 & \frac{122}{133} & \frac{104}{133} & \frac{1942}{3059} & \frac{1}{2} & \frac{169}{437} \\ \hline 0 & \frac{253}{266} & \frac{2617}{3059} & \frac{2573}{3496} & \frac{268}{437} & \frac{1}{2} \\ \hline \end{array}$$

(we actually need just left-bottom half of the table)

So $f(1, 0) = \frac{268}{437} \approx 0.61$, $f(2, 0) = \frac{2573}{3496}$ and $\frac{39}{102} + \frac{24}{102} \cdot f(2, 0) \approx 0.56$.