Recently I was playing several rounds of the game of pairs with my children. I was surprised that almost every time, one matching pair was adjacent (either next to each other in a row, or vertically). This led to the following question.

Let $n$ be a positive integer. Consider the set $$C_n = \{1,\ldots, 2n^2\}\times\{0,1\}.$$ We say that $(k,0)$ and $(k,1)$ for $k\in \{1,\ldots,2n^2\}$ is a matching pair in $C_n$.

Let $G_n = \{1,\ldots, 2n\} \times \{1,\ldots,2n\}\subseteq \mathbb{R}^2$ be thought of the "gaming grid" where we place the numbers. Formally: shuffling and distributing the cards corresponds to a bijection $\varphi: C_n \to G_n$. So we can define the minimum distance of a matching pair by $$M_n = \min\big\{||\varphi(k,0), \varphi(k,1)||: k\in \{1,\ldots,2n^2\}\big\}.$$

By $||\cdot||$ we denote the Euclidean distance.

$M_n$ is a random variable, so we can calculate its expected value $E(M_n)$.

Question. In plain English: if we play the game of pairs on an ever growing quadratic play-field, can we expect to find some matching pair in a reasonably close distance? Or more formally: Do we have $\lim_{n\to\infty} E(M_n) < \infty$?

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    $\begingroup$ The probability that the two copies of $k$, formally $(k, 0)$ and $(k, 1)$, are adjacent is about $1/n^2$. Once $(k, 0)$ is placed there are four possible places to put $(k, 1)$, out of $4n^2$. (I'm ignoring edge effects.) There are $2n^2$ different pairs so we should expect about 2 adjacent matching pairs, and these sorts of distributions are "usually" Poisson, so I'd expect $P(M_n = 1) \to 1 - e^{-2}$ as $n \to \infty$. But this sort of reasoning doesn't directly give your desired result since $M_n$ could be very large with non-negligible probability. $\endgroup$ – Michael Lugo Nov 28 '18 at 16:09

The goal here is to show that Michael's "usually Poisson" reasoning can be made rigorous. For $d>0$, let $c_d$ denote the number of other lattice points in $\mathbb{Z}^2$ within distance $d$ of the point $(n,n)$. My claim will be that for fixed $d$ the number of matching pairs of distance at most $d$ is asymptotically Poisson with mean $c_d/2$, and in particular that the probability there is at least one pair within distance $d$ converges to $1-e^{-c_d/2}$.

As pointed out by Michael in his comment, this doesn't directly extend to a guarantee of bounded expectation without some further argument showing there isn't a part of the tail that has probability tending to $0$ but expectation tending to infinity.

Call an unordered pair $\{ (x_1, y_1), (x_2, y_2) \}$ of points in our grid close if the points are within distance $d$ from each other. The number of close pairs is equal to $2n^2 c_d + O(n)$, where here (and throughout) the constant in the $O$ notation may depend on $d$ and $m$. (Equivalently, there's $4n^2 c_d+O(n)$ ordered pairs of close points. The idea is that each of the $4n^2-O(n)$ points $(x_1, y_1)$ not within distance $d$ of the boundary of the grid is the first point in $c_d$ pairs).

For each close pair $p$ of points, let $X_p$ be equal to $1$ if the pair matches, and $0$ otherwise. Then we have $$E(X_p) = \frac{1}{4n^2-1}$$ and the number of matching pairs is $$X= \sum_{p} X_p,$$ Let $\{Z_p\}$ be a family of independent random variables each equal to $1$ with probability $\frac{1}{4n^2-1}$, and let $Z=\sum Z_p$. Then $Z$ converges to the Poisson distribution as $n$ goes to infinity. To show that $X$ also converges to Poisson, it is enough to show that the moments of $X$ are asymptotically equal to the moments of $Z$, i.e. that $E(X^m-Z^m) \rightarrow 0$ for each $m$.

We have \begin{eqnarray*} E(X^m) &=& E\left((\sum_p X_p)^m\right) \\ &=& E \left( \sum_{(p_1, \dots, p_m)} X_{p_1} X_{p_2} \dots X_{p_m}\right) \\ &=& \sum_{(p_1, \dots, p_m)} E(X_{p_1} \dots X_{p_m}) \end{eqnarray*} and similarly for $Z$.

We now divide up our $m-$tuples into two classes:

Class 1: For each $i \neq j$ we either have $p_i=p_j$ or $p_i \cap p_j = \emptyset$ (i.e. no two pairs overlap in exactly one point).

Class 2: There is at least one pair $(i,j)$ for which $p_i$ and $p_j$ intersect exactly in one point.

Notice that in class $2$ it's impossible for both $p_i$ and $p_j$ to be matching pairs (If $(x_1, y_1)$ and $(x_2, y_2)$ match, it's impossible for $(x_1, y_1)$ and $(x_3, y_3)$ to match). So the contribution of Class $2$ to the expectation of $X$ is $0$. The bound we want on the expectation would follow from the following two claims:

Claim 1: For any tuple in class $1$ we have $E(X_{p_1} X_{p_2} \dots X_{p_n}) = E(Z_{p_1} Z_{p_2} \dots Z_{p_n}) (1+o(1))$. (So the total contribution of Class $1$ to $E(X)$ and $E(Z)$ is roughly the same).

Claim 2: The total contribution of class $2$ to $E(Z)$ is $o(1)$.

The intuition here is that because $m$ is fixed and the size of the grid tends to infinity, most collections of points won't overlap at all, so class $2$ isn't that significant.

For claim $1$: If $r$ is the number of distinct pairs in $(p_1, p_2, \dots, p_m)$, then we have \begin{eqnarray*} E(Z_{p_1} Z_{p_2} \dots Z_{p_m}) &=& \left(\frac{1}{4n^2-1}\right)^r \\ E(X_{p_1} X_{p_2} \dots X_{p_m}) &=& \left(\frac{1}{4n^2-1}\right)\left(\frac{1}{4n^2-3}\right) \dots \left(\frac{1}{4n^2-(2r-1)}\right)\end{eqnarray*} (Knowing certain pairs match makes it slightly more likely that other pairs match too).
In particular, we have $$E(Z_{p_1} Z_{p_2} \dots Z_{p_m}) \leq E(X_{p_1} X_{p_2} \dots X_{p_m}) \leq E(Z_{p_1} Z_{p_2} \dots Z_{p_m})\left(\frac{4n^2-1}{4n^2-2m}\right)^m$$

A sketchier argument for Claim $2$: Let $r$ be as in claim $1$. If we consider all $m$-tuples containing $r$ distinct pairs, then only an $O(\frac{1}{2n^2c_d})$ of them have any overlap between the $r$ pairs. (the denominator here coming from how there's $2n^2c_d$ close pairs to choose from. Again the constant in the $O$ notation can depend on $m$ or $d$). All tuples with $r$ pairs have the same contribution to $E(Z)$, so the contribution from pairs with overlap is only $O(\frac{1}{n^2})$ the contribution from all pairs. The claim follows since $E(Z)=O(1)$.

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I wrote a program that repeatedly generates a random grid and counts the number of adjacent pairs, and finally outputs the distribution of this number. Here are some results:

10*10 grid, 50 pairs, 1000000 games

     0      1      2      3      4      5      6      7     8    9   10  11  12
158437 295726 272342 164827  73472  25585   7400   1784   350   67    8   2   0

100*100 grid, 5000 pairs, 10000 games

   0      1      2      3      4      5      6      7      8      9     10
1361   2794   2729   1725    891    335    117     37      9      2      0

1000*1000 grid, 500000 pairs, 100 games

 0      1      2      3      4      5      6      7      8      9     10
13     25     23     25     10      3      1      0      0      0      0

So these results confirm Kevin's post. The probability that there is no adjacent pair converges to $e^{-2}$ = 13.53%.

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  • $\begingroup$ Hi Mark - thanks for the program and the results -- and welcome to MathOverflow! $\endgroup$ – Dominic van der Zypen Dec 4 '18 at 10:10

I think the exact expected number of adjacent pairs is $\dfrac{4n}{2n+1}$, which is $\frac43$ when $n=1$, while it is $\frac85$ when $n=2$ and converges to $2$ as $n$ increases, consistent with Michael Lugo's almost Poisson comment. When $n=1$ the probability of at least one (in this case exactly two) adjacent pairs is $\frac23$, and it seems likely to increase from there

I suspect that the $1-e^{-2}$ asymptotic probability of a minimum distance of $1$ can also be extended to $e^{-2}-e^{-4}$ of minimum distance $\sqrt{2}$, then $e^{-4}-e^{-6}$ of minimum distance $2$, and $e^{-6}-e^{-10}$ of minimum distance $\sqrt{5}$ [note the jump in the exponent as a chess knight can move to up to $8$ squares], etc.

Adding these up, and a couple more, suggest a limit for $E(M_n)$ of approximately $1.06727$. The worst case is again when $n=1$ with $E(M_1)=\frac{2+\sqrt{2}}{3} \approx 1.138$ while simulation suggests $E(M_2) \approx 1.088$

Like Mark Dettinger, I attempted a simulation, mine of $100000$ games with a $200\times200$ grid, i.e. with $n=100$. I got the simulated proportion of games with a minimum distance of $1$ to be $0.86322$ which is indeed close to $1-e^{-2}\approx 0.86466$. I also got a sample average minimum distance of about $1.0683$, and the full list of frequencies of minimum distances compared with the predicted asymptotic probabilities was

mindist   sample frequency   prediction 
   1          86322           0.8646647
 sqrt(2)      11798           0.1170196
   2           1626           0.0158369
 sqrt(5)        247           0.0024334
 sqrt(8)          6           0.0000393
   3              1           0.0000053
>=sqrt(10)        0           0.0000008

which also looks reasonably close to me

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