Suppose one suit is a trump suit. What is the probability that the second card would beat the first card in a trick? 
Question: Consider a deck of $52$ cards, ordered such that $A>K>Q>⋯>2$. I pick one first, then you pick one,
  Suppose one suit is a trump suit. What is the probability that the second card would beat the first card in a trick? (i.e. it is a trump and the first card is not, or it is higher than the first card and in the same suit.)

My attempt: 
Suppose we want to pick a trump suit at our second card and first card is not a trump. It has probability $\frac{13}{52}\times \frac{39}{51}.$
Now, if we do not want to pick a trump suit but it is higher than the first card and in the same suit, then the probability is $\frac{39}{52}\times \frac{?}{51}.$
I do not know what to input for $?$.
 A: Given that the two cards are the same suit, the chance the second wins is $\frac 12$ by symmetry.  Compute the chance that the two cards are the same suit (including trumps) and divide by $2$.
A: I suggest another approach which gives you also the $\color{green}{\mbox{missing number}}$ in your expression "$\frac{39}{52}\times \frac{\color{green}{?}}{51}$".


*

*number of possible ordered pairs from the deck: $\color{blue}{52\cdot 51}$

*number of suits: $\color{blue}{4}$

*number of winning pairs within a suit: $\color{blue}{\frac{13\cdot 12}{2}}$
Note that there are $13\cdot 12$ ordered pairs within a suit. So, half of them are winning. So, the number you are looking for in your expression is $\color{green}{6}$:
$$\frac{39}{52}\times \frac{\color{green}{?}}{51} = 3\cdot \frac{13}{52}\times \frac{\color{green}{6}}{51}$$


*

*number of winning pairs consisting of 1st card non-trump and second card trump: $\color{blue}{39\cdot 13}$
All together
$$P\left(\mbox{trick}\right) = \frac{\color{blue}{4\cdot \frac{13\cdot 12}{2} +  39\cdot 13}}{\color{blue}{52\cdot 51}}$$
