The numbers $x_1,$ $x_2,$ $x_3,$ $x_4$ are chosen at random in the interval $[0,1].$ Let $I$ be the interval between $x_1$ and $x_2,$ and let $J$ be the interval between $x_3$ and $x_4.$ Find the probability that intervals $I$ and $J$ both contain the number $1/2$.
Let us first find the probability that interval $I$ contains the number $1/2.$
Since $x_1$ and $x_2$ are chosen at random in the interval $[0,1],$ the probability that $x_1$ is less than or equal to $1/2$ is $1/2.$ Similarly, the probability that $x_2$ is greater than or equal to $1/2$ is $1/2.$ Therefore, the probability that interval $I$ contains the number $1/2$ is $(1/2) \cdot (1/2) = 1/4.$
Now, let us find the probability that interval $J$ contains the number $1/2.$
Again, since $x_3$ and $x_4$ are chosen at random in the interval $[0,1],$ the probability that $x_3$ is less than or equal to $1/2$ is $1/2.$ Similarly, the probability that $x_4$ is greater than or equal to $1/2$ is $1/2.$ Therefore, the probability that interval $J$ contains the number $1/2$ is $(1/2) \cdot (1/2) = 1/4.$
Now, to find the probability that both intervals $I$ and $J$ contain the number $1/2,$ we multiply the probabilities together:
$\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}.$
Therefore, the probability that intervals $I$ and $J$ both contain the number $1/2$ is $\boxed{\frac{1}{16}}.$
To find the probability that intervals $I$ and $J$ both contain the number $\frac{1}{2},$ we need to determine the probability that $x_1$ and $x_2$ are on opposite sides of $\frac{1}{2},$ and $x_3$ and $x_4$ are on opposite sides of $\frac{1}{2}.$
First, let's consider the possible positions of $x_1$ and $x_2.$ Since they are chosen at random in the interval $[0,1],$ there are two cases:
1. $x_1 < \frac{1}{2}$ and $x_2 > \frac{1}{2}.$ In this case, the interval $I$ contains $\frac{1}{2}.$
2. $x_1 > \frac{1}{2}$ and $x_2 < \frac{1}{2}.$ In this case, the interval $I$ does not contain $\frac{1}{2}.$
Similarly, for $x_3$ and $x_4,$ there are two cases:
1. $x_3 < \frac{1}{2}$ and $x_4 > \frac{1}{2}.$ In this case, the interval $J$ contains $\frac{1}{2}.$
2. $x_3 > \frac{1}{2}$ and $x_4 < \frac{1}{2}.$ In this case, the interval $J$ does not contain $\frac{1}{2}.$
From the given intervals $[0,1],$ the probability that $x_1 < \frac{1}{2}$ is $\frac{1}{2},$ and the probability that $x_2 > \frac{1}{2}$ is also $\frac{1}{2}.$ Similarly, the probability that $x_3 < \frac{1}{2}$ is $\frac{1}{2},$ and the probability that $x_4 > \frac{1}{2}$ is also $\frac{1}{2}.$
Since $x_1$ and $x_2$ are independent events, the probability that both events occur is the product of their individual probabilities, which is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.$ Similarly, the probability that both $x_3$ and $x_4$ satisfy the conditions is also $\frac{1}{4}.$
Therefore, the probability that intervals $I$ and $J$ both contain the number $\frac{1}{2}$ is the product of the probabilities for each event occurring independently: $\frac{1}{4} \cdot \frac{1}{4} = \boxed{\frac{1}{16}}.$
To make the problem easier, we use geometric probability. Graph the four numbers as points in the plane. Let $x$, $y$, $p$, and $q$ be the coordinates of the four points. Note that $(x,y)$ is uniformly distributed in the unit square, and $x$, $y$, $p$, and $q$ are independent and identically distributed.
[asy]
unitsize(2 cm);
draw((0,0)--(2,0));
draw((0,0)--(0,2));
draw((1,0.1)--(1,-0.1));
draw((1,0.1)--(1,0.9));
draw((1,0.1)--(0.8,0.9)--(1.2,0.9)--cycle);
draw((0.1,1)--(-0.1,1));
draw((0.1,1)--(0.9,1));
draw(rotate(45)*(0.8,0.9)--rotate(45)*(1.2,0.9)--rotate(45)*(1,1.1)--cycle);
label("$x$", (2,0), S);
label("$y$", (0,2), W);
label("$x_1$", (1,-0.2), S);
label("$x_2$", (1,0.9), E);
label("$x_3$", (-0.2,1), S);
label("$x_4$", (0.9,1), W);
label("$x_1 < x_2$", (1.2,1.4), E);
label("$x_3 < x_4$", (1.7,0.9), N);
dot("$(x,y)$", (1,1), NE, UnFill);
dot("$(p,q)$", (0.8,0.5), NW, UnFill);
[/asy]
Then the probability that interval $I$ contains $\frac{1}{2}$ is the probability that the point $(x,y)$ lies on one side of the line $y = x - \frac{1}{2}$ (inclusive), and similarly for the probability that interval $J$ contains $\frac{1}{2}$, substituting $(p,q)$ for $(x,y)$. Therefore, we want to find the probability that $(x,y)$ lies on one side of the line $y = x - \frac{1}{2},$ and simultaneously, that $(p,q)$ lies on one side of the line $q = p - \frac{1}{2}.$
[asy]
unitsize(2 cm);
draw((0,0)--(2,0));
draw((0,0)--(0,2));
draw((1,0.1)--(1,-0.1));
draw((1,0.1)--(1,0.9));
draw((1,0.1)--(0.8,0.9)--(1.2,0.9)--cycle);
draw((0.1,1)--(-0.1,1));
draw((0.1,1)--(0.9,1));
draw(1.5*dir(135)--(0,0.5),red);
draw(1.5*dir(135)--(2,1.5),red);
label("$x$", (2,0), S);
label("$y$", (0,2), W);
label("$x_1$", (1,-0.2), S);
label("$x_2$", (1,0.9), E);
label("$x_3$", (-0.2,1), S);
label("$x_4$", (0.9,1), W);
label("$x_1 < x_2$", (1.2,1.4), E);
label("$x_3 < x_4$", (1.7,0.9), N);
dot("$(x,y)$", (1,1), NE, UnFill);
dot("$(p,q)$", (0.8,0.5), NW, UnFill);
[/asy]
Take a point in this region, and let $x_1 < x_2$ and $x_3 < x_4.$ Call this point $(r,s).$
On the line $s = r - \frac{1}{2}$ and within the unit square, it is clear that the point $(r,s)$ divides the bottom half of the unit square symmetrically (about the line $s = \frac{1}{2}$). In addition, the probability that $r > s$ (i.e. $y > x - \frac{1}{2}$) is $1/4,$ and the probability that $r < s$ (i.e. $y < x - \frac{1}{2}$) is $3/4.$ Furthermore, the line $s = r - \frac{1}{2}$ divides the region $r > s$ into two parts of probability $1/2$ each, and divides the region $r < s$ into two parts of probability $3/2$ each.
Therefore, if we let $P(r,s)$ represent the probability that the intervals $I$ and $J$ satisfy $x_1 < x_2$ and $x_3 < x_4,$ we find that
\[P(r,s) = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{4} + \left( \frac{3}{4} \cdot \frac{3}{2} \cdot \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{3}{4} \right) = \frac{11}{32}.\]Thus, $P(r,s) = \frac{11}{32},$ which is the probability that the intervals $I$ and $J$ contain $\frac{1}{2}.$
Finally, we have to multiply by $\frac{1}{2}$ for the probability that $x < y.$ Therefore, the probability we seek is $\frac{11}{32} \cdot \frac{1}{2} = \boxed{\frac{11}{64}}.$