4
$\begingroup$

Suppose $n$ couples are seated in a circle, and let $P_n$ be the probability that no couple is sitting together.

Using Inclusion-Exclusion, I believe it can be shown that

$\hspace{.2 in}\displaystyle P_n=1-\sum_{i=1}^n (-1)^{i+1}\binom{n}{i}2^i\frac{(2n-1-i)!}{(2n-1)!},$

and I would like to find out how to prove that $\displaystyle\lim_{n\to\infty}P_n=\frac{1}{e}\;\;$ (or show that this is not the case).


Here are some numerical values:

$P_3\approx.267,\;P_4\approx.295,\;P_5\approx.310\;, P_6\approx.320,\;P_7\approx.327,\;P_8\approx.332,\;P_9\approx.336,\;\;P_{10}\approx.340$

For a related question, see Showing probability no husband next to wife converges to $e^{-1}$

$\endgroup$
  • 1
    $\begingroup$ This is OEIS A129348 where an asymptotic is given. Combine with Stirling's formula. $\endgroup$ – Marko Riedel Sep 4 '16 at 20:43
1
$\begingroup$

Verification of the Sum

How many ways for $k$ of the $n$ couples to sit together: $$ \overbrace{2^kk!\binom{n}{k}}^1\left[\vphantom{\binom{2n-k}{k}}\right.\overbrace{\binom{2n-k}{k}}^2+\overbrace{\binom{2n-k-1}{k-1}}^3\left.\vphantom{\binom{2n-k}{k}}\right]\overbrace{(2n-2k)!\vphantom{\binom{2n-k}{k}}}^4\tag{1} $$ Explanation:
$1$: the number of ways to select the $k$ couples: $\binom{n}{k}$
$\phantom{1\text{: }}$times the number of ways to arrange them among themselves: $2^kk!$
$2$: the number of ways to arrange $k$ couples and $2n-2k$ singles
$3$: the number of ways to arrange $k-1$ couples and $2n-2k$ singles
$\phantom{1\text{: }}$this is the number of ways to have a couple across the end of the line
$4$: the number of ways to arrange the $2n-2k$ singles among themselves

Noting that $\binom{2n-k}{k}+\binom{2n-k-1}{k-1}=\frac{2n}{2n-k}\binom{2n-k}{k}$, Inclusion-Exclusion gives the probability that no couple sits together to be $$ \begin{align} &\frac1{(2n)!}\sum_{k=0}^n(-1)^k2^kk!\binom{n}{k}\binom{2n-k}{k}\frac{2n}{2n-k}(2n-2k)!\\ &=\sum_{k=0}^n(-2)^k\binom{n}{k}\frac{(2n-k-1)!}{(2n-1)!}\tag{2} \end{align} $$


Verification of the Limit $$ \begin{align} \sum_{k=0}^n(-2)^k\binom{n}{k}\frac{(2n-k-1)!}{(2n-1)!} &=\sum_{k=0}^n\frac{(-1)^k}{k!}\frac{2^k\frac{n!}{(n-k)!}}{\frac{(2n-1)!}{(2n-k-1)!}}\\ &=\sum_{k=0}^n\frac{(-1)^k}{k!}\frac{2n(2n-2)\cdots(2n-2k+2)}{(2n-1)(2n-2)\cdots(2n-k)}\\ &\to\sum_{k=0}^\infty\frac{(-1)^k}{k!}\\[3pt] &=\frac1e\tag{3} \end{align} $$

$\endgroup$
  • $\begingroup$ Thank you for your answer. If you don't mind, would you please explain why the sum in the 2nd line of the Verification of the Limit converges to the series in the 3rd line. (Does $\lim_{n\to\infty}a(n,k)=b_k\implies\lim_{n\to\infty}\sum_{k=0}^{n}a(n,k)=\sum_{k=0}^{\infty}b_k$?) $\endgroup$ – Monty Hall Sep 5 '16 at 20:27
  • $\begingroup$ $\sum\limits_{k=0}^\infty\frac{(-1)^k}{k!}$ converges absolutely. For each $k$, $\frac{2^k\frac{n!}{(n-k)!}}{\frac{(2n-1)!}{(2n-k-1)!}}[k\le n]\le2$ converges to $1$. Apply Dominated Convergence. $\endgroup$ – robjohn Sep 5 '16 at 21:01
  • $\begingroup$ Thank you for your reply -- you made it look easy! $\endgroup$ – Monty Hall Sep 5 '16 at 22:35
1
$\begingroup$

Start with the inclusion-exclusion argument. We compute the cardinality of a set of arrangements where a given set of $q$ couples or more sit together. There are $2^q$ possibilities of ordering the $q$ couples. These are now fused and we have $2n-2q$ people left over which are joined by $q$ couples in a circular permutation, so there are $(2n-2q+q)!/(2n-2q+q) = (2n-q-1)!$ such arangements. This yields by inclusion-exclusion the closed formula

$$\sum_{q=0}^n {n\choose q} (-1)^q 2^q (2n-q-1)!$$

which we divide by $(2n-1)!$ to get the probability. For the asymptotics we use

$$(2n-q-1)! = \Gamma(2n-q) = \int_0^\infty x^{2n-q-1} e^{-x} dx.$$

This yields for the sum

$$\int_0^\infty x^{2n-1} e^{-x} \sum_{q=0}^n {n\choose q} (-1)^q 2^q x^{-q} dx \\ = \int_0^\infty x^{2n-1} e^{-x} \left(1-\frac{2}{x}\right)^n dx = \int_0^\infty x^{n-1} (x-2)^n e^{-x} dx.$$

We will compute the asymptotic with Laplace's method. To do this put $x=nz$ to obtain

$$n^{n-1} \times n^n \times n \times \int_0^\infty z^{n-1} (z-2/n)^n e^{-nz} dz \\ = n^{2n} \times \int_0^\infty z^{n-1} (z-2/n)^n e^{-nz} dz .$$

The integrand is $$\exp(n(\log(z)+\log(z-2/n)-\log(z)/n-z)) = \exp(n f(z)).$$

To find the saddle point we solve the saddle point equation $f'(z)=0$ which produces $$z_0 \approx 2 + \frac{1}{n^2}.$$

The asymptotic is then given by

$$\sqrt{\frac{2\pi}{n|f''(z_0)|}} e^{nf(z_0)}.$$

We have for $f(z_0)$

$$\log 2 + \log(1+1/n^2/2) + \log 2 + \log(1-1/n+1/n^2/2) \\ - 1/n \log 2 - 1/n \log (1 + 1/n^2/2) - 2 - 1/n^2.$$

The first two terns are

$$f(z_0) \approx 2\log 2 -2 - 1/n \log 2 - 1/n.$$

Furthermore we have

$$f''(z_0) \approx -\frac{1}{2}$$

so $f(z)$ does indeed have a maximum there.

Putting it all together we obtain

$$\sqrt{\frac{4\pi}{n}} e^{2n\log 2 - 2n - \log 2 - 1} = 2\sqrt{\frac{\pi}{n}} \frac{2^{2n}}{2e^{2n+1}} = \sqrt{\frac{\pi}{n}} \frac{2^{2n}}{e^{2n+1}}.$$

This yields for our sum

$$n^{2n-1/2} \sqrt{\pi} \frac{2^{2n}}{e^{2n+1}}.$$

Now Stirling's formula yields

$$(2n-1)! \sim \sqrt{2\pi(2n-1)} (2n-1)^{2n-1} e^{-(2n-1)}.$$

This gives for the probability

$$\frac{1}{e^2} \frac{2\sqrt{n}}{\sqrt{2(2n-1)}} \frac{2^{2n-1} n^{2n-1}}{(2n-1)^{2n-1}} \sim \frac{1}{e^2} \left(1+\frac{1}{2n-1}\right)^{2n-1} \sim \frac{1}{e}$$

and the conjecture is proved.

Remark. These are the details of the inclusion-exclusion argument. The nodes of the underlying poset represent arrangements where some set of $q$ couples sit together, possibly more. The Mobius function of the poset is $(-1)^q.$ A specific arrangement with exactly $p$ couples sitting together is included in the ${p\choose q}$ nodes where $q$ couples from the $p$ couples sit together, or more. It is not included in any nodes representing more than $p$ couples sitting together because this is impossible when the count is precisely $p$. Therefore the total weight with the given Mobius function is

$$\sum_{q=0}^p {p\choose q} (-1)^q.$$

This is zero when $p\ge 1$ and one when $p = 0.$ These weights correctly represent the requirement of arrangements with no couples sitting together being counted once and all others being counted zero times.

Addendum. An approach to this problem using the methods of experimental mathematics might begin by computing the values for small $n$ using total enumeration and hence verify the formula from inclusion-exclusion, and then use that to compute the probabilities for larger $n$, leading to a conjecture that we have $1/e.$ The enumeration task is practicable to about $n=5$ and confirms the correctness of the formula. This is the code, which consumes very little memory. We get starting with one couple

$$0, 2, 32, 1488, 112512,\ldots$$

Observe that we got the correct value for $n=1$ unlike from the formula, which does not work here because when there is only one couple the two orderings of its constituents are the same, but the factor $2^q$ produces a value of two instead of a value of one.

with(combinat);

X :=
proc(n)
option remember;
local perm, circ, count, pos;

    count := 0;

    perm := firstperm(2*n-1);

    while type(perm, `list`) do
        circ := [op(perm), 2*n, perm[1]];

        for pos to 2*n do
            if abs(circ[pos]-circ[pos+1]) = 1
            and type(min(circ[pos], circ[pos+1]), `odd`)
            then
                break;
            fi;
        od;

        if pos = 2*n + 1 then
            count := count + 1;
        fi;

        perm := nextperm(perm);
    od;

    count;
end;


Q := n -> add(binomial(n,q)*(-1)^q*2^q*(2*n-q-1)!, q=0..n);
$\endgroup$
  • $\begingroup$ Thanks for your very detailed answer. Could you please explain how you got $z_0\approx 2+\frac{1}{n^2}$? $\endgroup$ – Monty Hall Sep 5 '16 at 20:49
  • $\begingroup$ You get the two by solving the equation $g'(z) = 0$ with $g(z) = 2\log z - z$ which is an asymptotic representation of $f(z).$ The second term was obtained by solving $f'(z)=0$ exactly and computing the first two terms of the asymptotic expansion of the solution. $\endgroup$ – Marko Riedel Sep 5 '16 at 20:53
  • $\begingroup$ Thank you for explaining this. $\endgroup$ – Monty Hall Sep 5 '16 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.