# 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..

• Maybe , there is a direct way to solve this, but my approach would be to make a simulation. Commented Oct 2, 2022 at 9:15
• @JoséCarlosSantos Really? This is actually a good question because there's a real-life experience behind it (not a homework problem). Commented Oct 2, 2022 at 9:32
• Seems that the odds are about $1:800$ according to my simulation with PARI/GP Commented Oct 2, 2022 at 10:07
• This time I agree that it is an interesting question and should not be closed , let alone deleted , despite of the policy contents. Commented Oct 2, 2022 at 10:56

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.

• Incredible. Thanks so much for solving this. I am in awe. Commented Oct 2, 2022 at 11:50

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 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.