Max has a spinner that lands on 1 with a probability of $\frac{1}{2}$, lands on 2 with a probability of $\frac{1}{4}$, lands on 3 with a probability of $\frac{1}{6}$, and lands on 4 with a probability of $\frac{1}{12}$. If Max spins the spinner, and then Zack spins the spinner, then what is the probability that Max gets a larger number than Zack does?
Let $x$ be the probability we are looking for and $y$ be the probability that they both spin the same number. By symmetry, it's clear that the probability of Zack getting a larger number than Max does is also equal to $x$. Furthermore, all possible outcomes can be divided into three categories: Max gets a larger number than Zack does, Max and Zack get the same number, or Zack gets a larger number than Max. The sum of the probabilities of these three events is 1, which gives us the equation $x+y+x=1$.

We can calculate $y$ with a little bit of casework. There are four ways in which they can both get the same number: if they both get 1's, both get 2's, both get 3's or both get 4's. The probability of getting a 1 is $\dfrac{1}{2}$, so the probability that they will both spin a 1 is $\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}$. Similarly, the probability of getting a 2 is $\dfrac{1}{4}$, so the probability that they will both spin a 2 is $\left(\dfrac{1}{4}\right)^2=\dfrac{1}{16}$. The probability of getting a 3 is $\dfrac{1}{6}$, so the probability that they will both get a 3 is $\left(\dfrac{1}{6}\right)^2=\dfrac{1}{36}$ and the probability that they will both get a 4 is $\left(\dfrac{1}{12}\right)^2=\dfrac{1}{144}$. This gives us  $$y=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{144}=\dfrac{25}{72}.$$Substituting this into $2x+y=1$ gives us $2x=\dfrac{47}{72}$, so $x=\boxed{\dfrac{47}{144}}$.