Probability – Showing Probability No Husband Next to Wife Converges to e^-1

combinatoricslimitsprobability

Inspired by these questions:

the more general question of the probability that seating $n$ couples (i.e. $2n$ individuals) in a row at random means that no couples are sat together can be expressed using inclusion-exclusion as

$$\displaystyle\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}$$

which for small $n$ takes the values: 

n   Probability of no couple together           
1   0           0   
2   1/3         0.3333333
3   1/3         0.3333333
4   12/35       0.3481481
5   47/135      0.3428571   
6   3655/10395  0.3516114
7   1772/5005   0.3540460
8   20609/57915 0.3558491

This made me wonder whether it converges to $e^{-1} \approx 0.3678794$ as $n$ increases, like other cases such as the secretary/dating problem and $\left(1-\frac1n\right)^n$ do. So I tried the following R code (using logarithms to avoid overflows)

couples  <- 1000000
together <- 0:couples
sum( (-1)^together * exp( log(2)*together + lchoose(couples,together) + 
     lfactorial(2*couples - together) - lfactorial(2*couples) ) )

which indeed gave a figure of $0.3678794$.

How might one try to prove this limit?

Best Answer

I observe that each term with $i$ fixed approaches a nice limit. We have

$$ 2^i \frac{n(n-1)(n-2)\cdots(n-i+1)}{i!} \frac1{(2n-i+1)(2n-i+2)\cdots(2n)} $$

or

$$ \frac1{i!} \frac{2n}{2n} \frac{2(n-1)}{2n-1} \cdots \frac{2(n-i+1)}{(2n-i+1)} \sim \frac 1{i!} $$

This gives you the series, assuming the limits (defining terms with $i>n$ to be zero) may be safely exchanged,

$$\lim_{n\to\infty} \sum_{i=0}^\infty [\cdots] = \sum_{i=0}^\infty \lim_{n\to\infty} [\cdots] = \sum_{i=0}^\infty (-1)^i \frac1 {i!} \equiv e^{-1}$$

Justifying the limit interchange I haven't thought about but I suspect this can be shown to be fine without too much effort... Edit: You can probably use the Weierstrass M-test.

Related Question