Draw two cards what is the probability the second is higher than the first? Is my approach correct? I've seen similar questions posted here before but I was wondering if my method/answer was correct
My reasoning was let's say you draw a 2 as your first. Card there are 12 cards with higher values, across 4 suits. Similarly if you draw a 3 there are 11 cards with higher values across 4 suits....son and so forth. So what I did was:
$ \dfrac{4*(12+11+10+......2)}{{52}\choose{2}}$ and I got $\dfrac{4}{17}$ as my answer. I saw a different post on this site where the answer was $\dfrac{8}{17}$ and I was confused what may or may not have missed.
 A: I assume drawing is done without replacement. The probability that the values are equal is $\frac{3}{51}$. This is because whatever card we draw first, the probability of matching it is $\frac{3}{51}$.  
So the probability they are not equal is $\frac{48}{51}$.
Thus by symmetry the probability the second is higher than the first is $\frac{24}{51}$. 
Remark: Your counting procedure, with slight modification, will work. Note that there are $(52)(51)$ ordered pairs of cards.
If the first is $2$, there are $48$ good choices for the second. If the first is $3$, there are $44$, and so on for a total of $4(48+44+\cdots+4)$. Now divide by $(52)(51)$. 
A: A different approach:
Number the cards as so: $1,2,3,...,13$.
If the first card is a $1$, with a $\frac{1}{13}$ chance, then there are 12 numbers that can be higher the second draw, each with four suits:
$$P(card\ 2>1) = \frac{12*4}{51} = \frac{48}{51}$$
If the first card is a $2$, with a $\frac{1}{13}$ chance, then using the same logic it's:
$$P(card\ 2>2) = \frac{11*4}{51} = \frac{44}{51}$$
So we simply want the sum of all these scenarios:
$$P(card\ 2>\ card\ 1) = \frac{1}{13} (P(card\ 2>1) + P(card\ 2>2) + ... + P(card\ 2>13)$$
This sums up to:
$$P(card\ 2>\ card\ 1) =  \frac{1}{13} \sum_{i=1}^{13} \frac{(13-i)*4}{51}$$
It's easy to see that summing up $13-i$ from 1-->13  is the same as summing up $i$ from 12-->0. (plug in for i). We then take out the multiplication. So we are left with:
$$P(card\ 2>\ card\ 1) =  \frac{1}{13} \frac{4}{51} \sum_{i=0}^{12} i $$
$$P(card\ 2>\ card\ 1) =  \frac{1}{13} \frac{4}{51} \sum_{i=1}^{12} i $$
You should also know this neat rule:
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$
So we have:
$$P(card\ 2>\ card\ 1) =  \frac{1}{13} \frac{4}{51} \frac{12*13}{2} $$
$$P(card\ 2>\ card\ 1) =  \frac{4}{51} \frac{12}{2} $$
$$P(card\ 2>\ card\ 1) =  \frac{2}{51} \frac{12}{1} $$
$$P(card\ 2>\ card\ 1) =  \frac{24}{51} $$
