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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.

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Imagine taking two cards. One card has a higher value than the other. For every pair, there is one ordering where the first card is higher than the second, one ordering where the second is higher than the first. Note that you can assign any ordering after you pick a pair. –  Herp Derpington Nov 30 '12 at 13:25
    
So $13 \choose 2$ gives me the number ways of picking a pair where the numbers on the cards aren't the same and order isn't important. Then I multiply by 16 to give me the ways of choosing the suits. So the answer is $$\frac{16 {13 \choose 2}}{52 \choose 2}$$ Is that it? –  csss Nov 30 '12 at 13:33

2 Answers 2

up vote 10 down vote accepted

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!

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It's not homework it's exam preparation. –  csss Nov 30 '12 at 13:34
    
ok, but you should be able to get the idea. hope it helps! :) –  chango Nov 30 '12 at 13:35
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Definitly best way to calculate this. +1 –  user127.0.0.1 Nov 30 '12 at 13:40

The probability that the second card has the same face value as the first (i.e. that they both have the same) is $\frac{3}{51}$, because no matter what value the first card has, there will be exactly three cards left in your pile of 51 cards, that have that same value.

So the probablilty that the two cards have different values is $1 - \frac{3}{51} = \frac{48}{51} = \frac{16}{17}$.

Clearly, the chance that the first card has a higher value than the second card is the same as the chance that the second card has a higher value than the first. So each of them must be $\boxed{ \frac{8}{17}}$.

edit: by the way this is exactly the same as Chango's approach, only without the fancy probability-theory notation.

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