So if I draw two cards from 52, what is the probability that the second card has a higher face value than the first? The values of the cards are Ace = 1, Two = 2,…, King = 13.
I got as far as $$\frac{13 \choose 1}{52 \choose 2}$$..
i.e. choosing the first card from 13 cards. But I don't how I am supposed to take account of choosing the second card.
Best Answer
OK, I am not going to give you the whole answer as this looks like homework. But let me give you massive hint as to how you can simplify your existence. If you try to compute the probability directly you will find it a very long and tedious calculation.
Call the event you are interested in $E_+$
So think about 2 more events. One being the opposite, so selecting a second card of lower face value, call this $E_-$. The other being, selecting a card of the same face value, $E_0$.
Now, it is easy to see that $\mathbb{P}(E_+) = \mathbb{P}(E_-) = p $. Also, notice that
$\mathbb{P}(E_+) + \mathbb{P}(E_-) + \mathbb{P}(E_0)=1$
since these events are disjoint and cover all posibilities.
So if you can calculate $\mathbb{P}(E_0)$ which is a lot easier than calculating $\mathbb{P}(E_+)$ directly, you are done!