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2024 SSMO Team Round Problems/Problem 13

Problem

In a deck of 54 cards (2 identical jokers, 4 identical cards with $1,2,3,\dots,13$), each card is dealt to one of 3 people, each having a $\frac{1}{3}$ chance of receiving each card. If the expected sum of the number of unique cards the three of them have can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n,$ find $m+n.$

Solution

Let $\mathbb{E}(\text{rank})$ denote the expected number of people that have the rank $\text{rank}$ for $\text{rank}\in\{1,2,3\dots,13\}.$ Consider the cards distinguishable for probability calculation purposes. There are 3 ways for one person to have it, $3! \cdot 4 + 3 \cdot \binom42$ ways for two people to have it (3/1 or 2/2 split), and $4 \cdot 3 \cdot 3$ ways for three people to have it. There are 81 total (equally likely) distributions of $\text{rank}$ among the 3 people, so $\mathbb{E}(\text{rank})$ is simply \[\frac{3\cdot1+(24+18)\cdot2+(36)\cdot3}{81} = \frac{65}{27}.\] For $\text{rank} = \text{joker},$ using a similar counting process, we have $\mathbb{E}(\text{rank}) = \frac{3\cdot1+6\cdot2}{9} = \frac{5}{3}.$ So, our answer is \[\frac{5}{3}+13\cdot\frac{65}{27} = \frac{890}{27}\implies 890+27 = \boxed{917}.\]

~SMO_Team

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