Two players draw $m=14$ cards each without replacement from two decks of cards shuffled together. Two cards are considered a pair if and only if they are of the same suit and the same number. The initial $104$ cards are therefore $p_0 = 52$ pairs of cards. The first player draws $m=14$ cards. Then the second player draws another $m=14$ cards from the remaining $2p_0-m=104-14=90$ cards. Who has a higher probability of drawing only unique cards (no pairs among the $m=14$ cards)?
The problem came from the game of Rummikub, which, excluding jokers, uses the equivalent of two decks of cards (52 Rummikub tile pairs numbered 1 to 13 of four colors). Having pairs is usually considered a bad thing in Rummikub. The game starts with 14 tiles. I wonder whether the first player who draws 14 tiles before the second player does has any advantage.
The answer appears to be "no." Empirically, they have the same probability but I don't know how to prove it mathematically. I probably missed something very basic.
I can show by simulation that the probability is $36\%$ for both players. I also have the probabilities in closed form for both players. They are numerically the same for as many decimal places as my computer can show ($36.44119\dots\%$).
Can someone prove that the two closed-form formulas below are equivalent?
The probability of getting exactly $n$ unique cards by drawing $m$ cards out of a pool of $d$ unique cards and $p$ card pairs is
\begin{equation} f(m, n, d, p) = {{\sum_{k=0}^{d}{2^{n-k} {p \choose {(m+n)/2-k}}{(m+n)/2-k\choose {n-k}}{d\choose k}}} \over{{d+2p}\choose m}}. \end{equation}
The probability of the first player drawing only unique cards is \begin{equation} f(m, m, 0, p_0) = {{2^m {p_0 \choose m}} \over{{2p_0}\choose m}}. \end{equation}
Suppose that the first player draws $n$ unique cards, leaving (the other set of) $n$ unique cards and $p_0 - (m+n)/2$ card pairs for the second player. ($n$ and $m$ must share the same parity.) The probability of the second player drawing only unique cards after the first player has removed $m$ cards, exactly $n$ of which are unique, is \begin{equation} \begin{aligned} &\sum_{n\in S(m)}{ f(m, m, n, p_0 - (m+n)/2) f(m, n, 0, p_0) } \\ &= {2^m \over { {2p_0 \choose m} {{2p_0-m}\choose m}}}\sum_{n\in S(m)} { { {2^n {p_0 \choose {(m+n)/2}}} {{(m+n)/2} \choose {n}} } {\sum_{k=0}^{n} { {2^{-k}} {{p_0 - (m+n)/2}\choose {m-k}} {n\choose k} } } }, \end{aligned} \end{equation} where $S(m) = \begin{cases} \{0, 2, 4, \dots, m\} \text{ if } m \text{ is even,} \\ \{1, 3, 5, \dots, m\} \text{ if } m \text{ is odd.} \end{cases}$
The above two formulas are numerically equivalent given $m=14$ and $p_0=52$. Wonder if anyone can prove that they are the same in general.
Hint: Symmetry. For every possible arrangement of the 104 tiles so that the first 28 are drawn, there is a bijection to an arrangement where we swap the first 14 and second 14 tiles.
It follows that they have the same probability. The probability that player 1 (or 2) has more unique tiles is $ \frac{ 1 - p } { 2 } $, where $p$ is the probability that they have the same number of unique tiles.