# Interpretation of $\frac{22}{7}-\pi$

Integral and series proofs that $\frac{22}{7}>\pi$

We can prove that $\frac{22}{7}$ exceeds $\pi$ by using Dalzell integral $$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$

or its equivalent series $$\sum_{k=1}^{\infty} \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$

Equivalent expressions

This series may be written in terms of factorials, binomial coefficients or Beta integrals as \begin{align} \frac{22}{7}-\pi &= 3840\sum_{k=1}^\infty \frac{(k+2)!(4k)!}{(4k+8)!k!} \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{{k+2 \choose 2}}{{4k+8\choose 8}} \\ &= \frac{16}{21} \sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)} \end{align} (see A series to prove $\frac{22}{7}-\pi>0$)

Can $\frac{22}{7}-\pi$ be given a combinatorial or probabilistic interpretation?

Some situations where $\pi$ appears are Buffon's needle or the probability that two random integers are relatively prime. See also $\pi$ in random phenomena by Boris Gourévitch and Occorrenze in calcolo delle probabilità e statistica by Mauro Fiorentini.

• A sketch: the numerator in the integral looks like the $p^n(1-p)^m$ in the binomial distribution, while the denominator $1+x^2$and the missing binomial coefficient may be related to some a priori distribution of $p$. So, maybe something like "$\frac{22}{7}- \pi$ is the probability to get 4 heads and 4 tails when flipping a coin 8 times given that $p$ follows a priori distribution... " could make sense. Feb 19, 2016 at 7:57
• $\pi$ appears in the limit of the probability of getting the same number of heads and tails as the number of tosses grows. pi314.net/eng/aleatoire.php Feb 19, 2016 at 8:26
• Feb 22, 2016 at 8:20
• $$\int_0^1 \frac{x^2}{1+x^2}dx = 1-\frac{\pi}{4}$$ is the probability that a point inside a square is outside the inscribed circle. Maybe $\frac{11}{14}-\frac{\pi}{4}$ should be considered. Mar 16, 2016 at 22:22

Let's say the tin has diagonal length 1. Parametrize the possible orientations of the tin with an angle $\theta$ ranging from $0$ to $\pi/4$, where $0$ corresponds to looking at the tin edge-on and $\pi/4$ corresponds to looking at it face-on. Note that $\theta$ is distributed uniformly. Then the apparent width of the tin is $\cos \theta$ and the apparent width of the region you need for a proper hit is $\sin \theta$. So the probability of a proper hit is $\sin \theta \, / \cos\theta = \tan \theta$.
$$\binom{8}{4} \frac1{\pi/4} \int_0^{\pi/4} (\tan \theta)^4(1-\tan\theta)^4\,d\theta \\ = \binom{8}{4} \frac1{\pi/4} \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx \\= \binom{8}{4} \frac1{\pi/4} \left(\frac{22}7 - \pi\right).$$
I can't think of a nice reason to expect a priori that this particular setup would give you a good rational approximation to $\pi$, but there you have it.