I'm currently creating a game for class. I know that the probability of one card being higher than another is $8$ out of $17$. What I'm trying to determine is the probability of the following:
A player draws $2$ cards without replacement from a deck. What is the probability that at least one of them will be higher than a third, subsequently drawn, card?
The probability is $\frac23$ if the values are unique, $\frac12$ if two are the same and $0$ if all three are the same.
The probability for the values to be unique is
$$\frac{13\cdot12\cdot11\cdot4^3}{52\cdot51\cdot50}=\frac{352}{425}\;.$$
The probability for exactly two values to be the same is
$$ \frac{13\cdot12\cdot4^2\cdot3\cdot3}{52\cdot51\cdot50}=\frac{72}{425}\;. $$
Just to check that they add up to $1$, the probability for all three values to be the same is
$$ \frac{13\cdot4\cdot3\cdot2}{52\cdot51\cdot50}=\frac1{425}\;. $$
Thus the desired probability is
$$ \frac23\cdot\frac{352}{425}+\frac12\cdot\frac{72}{425}=\frac{812}{1275}\approx64\%\;. $$