Simple Casino War Probability

Question

This question is from QuantGuide(Casino War):
You are playing a game with the casino. You and the dealer are each dealt a card from a fair, shuffled deck of 52 cards. If you have a strictly larger number than the dealer's, you win- else, you lose. What is the probability you win?
My Approach:
Let my probability of getting the greater value be S. It will be the same for my dealer. Now let the probability that I get the same value as the dealer be X. The equation will be: \begin{equation} 2S+X=1 \end{equation} For X we get select 1 set of 4 cards from 13 sets of 4 cards. From that, we will select 2 cards. For the total number of cases, it will be 52*51.
Hence $X=\frac{13*6}{51*52}=\frac{1}{34}$ and $S=\frac{33}{68}$.This answer doesn't pass.

Answer 1

For $X$, the first case has no restriction, after the first card is chosen, we then have $3$ options out of $51$ cards.

Hence $X=\frac{3}{51}=\frac{1}{17}$.

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Your mistake is your denominator should be $\binom{52}{2}$.

$X=\frac{13\cdot 6}{\binom{52}{2}}=\frac{13\cdot 12}{52\cdot 51}=\frac1{17}$