Old Maid Card Game Probability Question I am not a math person but was shocked by something that happened when I played Old Maid for the first time ever with with my two young sons tonight. We were playing with a deck that had 37 cards in total, which means 18 pairs plus the Old Maid. When I dealt the cards into three groups (12+12+13), not one of us had a single pair in our hands. What is the statistical likelihood that this could happen? Can anyone figure out the math for me? I doubt this will ever happen again in our lifetimes, but maybe I'm wrong..
 A: Alternative approach:
I will express the probability as
$$\frac{N}{D} ~: ~D = \binom{37}{13} \times \binom{24}{12} \times \binom{12}{12}. \tag1 $$
That is, in (1) above, the denominator reflects the total number of (equally likely) ways that the $37$ cards can be distributed to the three people.
I will compute $N = [N_1 + (2\times N_2)]$, where $N_1$ represents the number of satisfying distributions where the $(13)$ card hand gets the old Maid, and $N_2$ represents the number of satisfying distributions where one of the $(12)$ card hands gets the old maid.  The factor of $(2)$ represents that two of the three people will have $(12)$ card hands.
Without loss of generality, I will assume that the $(36)$ paired cards are $(18)$ cards numbered from $(1)$ through $(18)$ inclusive, that are Black, and (similarly) $(18)$ such cards that are Red.

To compute $N_1$, first, the $(13)$ card hand must have $(12)$ singletons.  This partial enumeration is
$$\binom{18}{12} \times 2^{(12)}.$$
Here, you can assume, without loss of generality, that the 13 card hand is specifically the old Maid, plus the cards numbered $(1)$ through $(12)$, all Black.
Then, the 2nd player must specifically get exactly one of the cards from each of the ranks numbered $(13)$ through $(18)$.  This can be done in $2^6$ ways.
Further, the 2nd player must then get $6$ more cards, from the remaining Red cards, numbered $(1)$ through $(12)$.  This can be done in $~\displaystyle \binom{12}{6}~$ ways.
So, the overall enumeration of $N_1$ is
$$N_1 = \binom{18}{12} \times 2^{(12)} \times 2^{(6)} \times \binom{12}{6} $$
$$= \binom{18}{12} \times \binom{12}{6} \times 2^{(18)}.  \tag2 $$

To compute $N_2$ first focus on the $(12)$ card hand that does not have the Old Maid.  Then, focus on the $(13)$ card hand, which is assumed to also not have the old Maid.
As in the previous section, the first partial enumeration is
$$\binom{18}{12} \times 2^{(12)},$$
followed (again) by the assumption, without loss of generality, that the pertinent $(12)$ card hand is the $(12)$ Black cards, numbered $(1)$ through $(12)$.
Now, the $(13)$ card hand must (again) specifically get exactly one of the cards from each of the ranks numbered $(13)$ through $(18)$.  Again, this can be done in $2^6$ ways.
Further, there is now the slight modification from the previous section.  The $(13)$ card hand must then get $7$ more cards, from the remaining Red cards, numbered $(1)$ through $(12)$.  This can be done in $~\displaystyle \binom{12}{7}~$ ways.
So, the overall enumeration of $N_2$ is
$$N_2 = \binom{18}{12} \times 2^{(12)} \times 2^{(6)} \times \binom{12}{7} $$
$$= \binom{18}{12} \times \binom{12}{7} \times 2^{(18)}.  \tag3 $$

Putting it all together,
$$N = N_1 + (2N_2) = \binom{18}{12} \times 2^{(18)} \times \left[ ~\binom{12}{6} + \left\langle 2 \times \binom{12}{7}\right\rangle ~\right]$$
and
$$D = \binom{37}{13} \times \binom{24}{12} \times \binom{12}{12}.$$
This yields
$$\frac{N}{D} \approx 1.267 \times 10^{(-3)} \approx \frac{1}{789.302}$$
which agrees with Parcly Taxel's computations.
A: Let players $A,B,C$ have $13,12,12$ cards in their initial hand. Considering hand order to be important and labelling the Old Maid as $0$ and the pairs from $1$ to $18$, but not distinguishing the cards of each pair, the number of initial deals is $T=\frac{37!}{2^{18}}$.
There are two possibilities for a pair-free initial deal. The first is when the Old Maid is held by one of the $12$-handers, say $B$. The numbers of pairs shared between the three players are forced: denote by $N_{PQ}$ the number of pairs shared between $P$ and $Q$, then we have
$$N_{AB}+N_{AC}=13$$
$$N_{AB}+N_{BC}=11$$
$$N_{AC}+N_{BC}=12$$
Solving this gives $N_{AB}=6,N_{AC}=7,N_{BC}=5$. The exact number of pair-free deals with $B$ holding the Old Maid can now be computed as a product of several factors:

*

*Where is the Old Maid in $B$'s hand (remember that we assume ordered hands)? $12$

*Which $6$ ($5$/$7$) positions in $A$'s ($B$'s/$C$'s) hand hold cards whose matches are held by $B$ ($C$/$A$)? $\binom{13}6$, $\binom{11}5$, $\binom{12}7$

*How many ways can the pairs be drawn betwen the ordered hands? $5!6!7!$

*How many ways can the drawn pairs be numbered $1$ to $18$? $18!$
Thus the number of pair-free deals when $B$ has the Old Maid is
$$S_1=12\binom{13}6\binom{11}5\binom{12}75!6!7!18!$$
The other possibility is $A$ having the Old Maid, in which case $N_{AB}=N_{AC}=N_{BC}=6$ and the number of pair-free deals is by a similar computation
$$S_2=13\binom{12}6^36!^318!$$
The final probability, where $S_1$ must be doubled to account for the symmetric case where $C$ has the Old Maid, is
$$\frac{2S_1+S_2}T=\frac{2883584}{2276020775}=0.12669\dots\%=\frac1{789.302747899\dots}$$
Far from being impossibly remote, in $800$ properly randomised deals you should expect your situation to come up at least once.
A: A simple approximation is $2(\frac{2}{3})^{18}\approx0.001353\approx \frac{1}{739}$.
Ignore the presence of the Old Maid (which is reasonable since the number of cards is rather large), and call the number of pairs $3n$.
The probability that no player gets a pair is $P(n)=\dfrac{\binom{3n}{2n}\binom{2n}{n}2^{3n}}{\binom{6n}{2n}\binom{4n}{2n}}$.
Explanation:

*

*The denominator is the total number of ways to divide the $6n$ cards
among the three players. First, among the $6n$ cards, we choose $2n$ cards
to go to the first player, so $\binom{6n}{2n}$. Then among the
remaining $4n$ cards we choose $2n$ cards to go to the second player, so
$\binom{4n}{2n}$. Then the remaining $2n$ cards go to the third
player.

*The numerator is the total number of ways in which each pair is
separated. First, among the $3n$ pairs, we choose $2n$ pairs to
each be represented in the hand of the first player (one card from each pair),
so $\binom{3n}{2n}$. Then we allocate the $n$ remaining pairs to
the second and third players (each player getting one member of each of
these pairs), so then the second player must now get an additional $n$
cards, and these are chosen among the $2n$ cards whose "partner" went
to the first player, so $\binom{2n}{n}$. Then the remaining $n$ cards
go to the third player. Then each pair has two ways of being separated among two players, so $2^{3n}$.

Now it can be shown, either algebraically or by Wolfram, that
$$\lim\limits_{n\to\infty}\frac{P(n)}{2(\frac{2}{3})^{3n}}=1$$
So, for large $n$, $P(n)\approx 2(\frac{2}{3})^{3n}$.
As for an intuitive explanation of why this approximation works, see here.
