I've always wondered about this

An equivalent question is the probability that two randomly chosen integers are relatively prime. (If we choose these integers as coordinates, then if they share a factor $k$ we may divide the line into $k$ equal segments, with a lattice point at the end of each.)

Because of this equivalent question, I will only consider primes. For $p$ , we know that $\frac { 1 }{ { p }^{ 2 } }$ of pairs will have both integers be divisible by p. Thus, the odds that a prime $p$ fails to divide both is $\left( 1-\frac { 1 }{ { p }^{ 2 } } \right)$ . The probability we want is $S= \prod _{ p\in { Z }_{ p } }^{ \quad }{ \left( 1-\frac { 1 }{ { p }^{ 2 } } \right) }$ Then we also know that $\frac { 1 }{ S } =\prod _{ p\in { Z }_{ p } }^{ \quad }{ \frac { { p }^{ 2 } }{ { p }^{ 2 }-1 } }$ Now, it is possible to decompose each $\frac { { p }^{ 2 } }{ { p }^{ 2 }-1 }$ : $\frac { { p }^{ 2 } }{ { p }^{ 2 }-1 } = \sum _{ i=0 }^{ \quad }{ \frac { 1 }{ { p }^{ 2i } } }$ which may be easily shown by multiplying both sides by ${ p }^{ 2 }-1$ . Thus, $\frac { 1 }{ S } =\prod _{ p\in { Z }_{ p } }^{ \quad }{ \sum _{ i=0 }^{ \quad }{ \frac { 1 }{ { p }^{ 2i } } } }$ Expand this out: $(1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 4 } } +\frac { 1 }{ { 2 }^{ 6 } } ...)(1+\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 4 } } +\frac { 1 }{ { 3 }^{ 6 } } ...)...$ This is just the sum of all squares! (Choosing any order results in a square, and we may achieve all prime factorizations)This is known to be $\frac { { \pi }^{ 2 } }{ 6 }$ , therefore the desired probability is the reciprocal, $\frac { 6 }{ { \pi }^{ 2 } }$ .