I have spent some time playing a 3 person card game called "12" with my family. There are 36 cards in the deck and each player is dealt 12 cards. If you get the queen and the king in the same suit (clubs, hearts, diamonds or spade) you get extra points (I do believe it is called trump?). Let's denote the event "being dealt a king and a queen in the same suit" as $T$. I tried to calculate $P(T>0)$, thinking that it would be an easy thing to do. However, it wasn't. I resolved it by doing $10^7$ Monte-Carlo simulations, resulting in $P(T>0)\approx 0.37$. There must be a more satisfying way of solving this (satisfying in the meaning that I won't be needing Monte-Carlo to solve it). Can you help me?

Do I really have to use the multivariate hypergeometric distribution with five classes (four classes for the suited pairs and one class for the rest of the cards) and go through all combinations that will yield $P(T>0)$ or is there an easier solution?

  • $\begingroup$ Inb4 president joke. This does hint at hypergeometric or multinomial, but I'm not entirely sure. $\endgroup$ – Sean Roberson Aug 16 '17 at 17:05

You are correct that to use multi-hypergeometric distribution to count. Let $S, H, D, C$ be the event that having KQ in the suits respectively. Then by inclusion-exclusion principle,

$$ \begin{align} &~ \Pr\{S \cup H \cup D \cup C\} \\ = &~ \binom {4} {1} \frac {\displaystyle \binom {2} {2} \binom {34} {10}} {\displaystyle \binom {36} {12} } - \binom {4} {2} \frac {\displaystyle \binom {2} {2} \binom {2} {2} \binom {32} {8}} {\displaystyle \binom {36} {12} } +\binom {4} {3} \frac {\displaystyle \binom {2} {2} \binom {2} {2}\binom {2} {2}\binom {30} {6}} {\displaystyle \binom {36} {12} } \\ &~- \binom {4} {4} \frac {\displaystyle \binom {2} {2} \binom {2} {2}\binom {2} {2}\binom {2} {2}\binom {28} {4}} {\displaystyle \binom {36} {12} } \\ = &~ \frac {339749} {916980} \\ \approx &~ 0.3705 \end{align}$$

Each term corresponding to the generic case with $1, 2, 3, 4$ pairs of KQ in particular suit, where the leading combinatoric coefficient is counting the number of particular suit combinations for each generic case.

  • $\begingroup$ Great! Thank you very much! Now I will be unbeatable in "12" :-) $\endgroup$ – Ture Aug 16 '17 at 17:35

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