From Jason Zimba comes a surprisingly sandy puzzle:
In the Great Riddlerian Desert, there is a single oasis that is straight and narrow. There are N travelers who are trapped at the oasis, and one day, they agree that they will all leave. They independently pick a random location in the oasis from which to start and a random direction in which to travel. Once their supplies are packed, they all head out.
What is the probability that none of their paths will intersect, in terms of N? (For the purposes of this puzzle, assume the oasis is a line segment, while the desert is an infinite Cartesian plane.)
For simplicity, define the oasis as running North <-> South. Consider $k$ travelers who will travel East.
Given $k$ travelers with paths on the same side of the oasis, the probability that they do not intersect is $\dfrac{1}{k!}$.
In other words, the order of the travelers along the oasis is quite simply the same as the order of the slopes of their paths.
Consider $N$ travelers at the oasis. Travelers' paths being East or West are bionomially distributed.
A traveler going West can ignore travelers going East, and vice versa.
The probability that none of their paths will intersect is:
$$\sum {(\text{Probability of East-West combination})\cdot(\text{Probability of East travelers not intersecting})\cdot(\text{Probability of West travelers not intersecting})}$$
$$ = \sum\limits_{k=0}^{N} \left(\dfrac{N \choose k}{2^N}\right)\cdot\left(\dfrac{1}{k!}\right)\cdot\left(\dfrac{1}{(N-k)!}\right)$$
$$ = \sum\limits_{k=0}^{N} \dfrac{N!}{2^Nk!(N-k)!} \cdot \dfrac{1}{k!} \cdot \dfrac{1}{(N-k)!}$$
$$ = \dfrac{1}{2^N}\sum\limits_{k=0}^{N} \dfrac{N!}{\big(k!(N-k)!\big)^2}$$
$$ = \dfrac{1}{2^NN!}\sum\limits_{k=0}^{N} \left(\dfrac{N!}{k!(N-k)!}\right)^2$$
$$ = \dfrac{1}{2^NN!}\sum\limits_{k=0}^{N} {N \choose k}^2$$
Using the Sum of Squares of Binomial Coefficients,
$$ = \dfrac{1}{2^NN!}{2N \choose N}$$
$$ = \dfrac{(2N)!}{2^N(N!)^3}$$
This value approaches 0 quite quickly.