Probability of a poker hand contains nothing 
Suppose we are playing 5 card draw. What is the probability that your hand contains nothing not even a pair?

Thoughts: To have a  worthless hand we need no repeated values, no runs of 5, or not all 5 cards have the same suit.
Let
$A$: no repeated values
$B^c$: no runs of 5 
$B$: run of 5
$C^c$: not all 5 cards have the same suit
$C$: all 5 have the same suit
A run of 5 means that we have 5 distinct values. Note a useful property to solve this problem is 
$$P(A\cap B^c\cap C^c) = P(A) - P(A\cap B) - P(A\cap C) + P(A\cap B\cap C)$$
Thus we have $$P(A) = \frac{52\times 48\times 44\times 40\times36}{52\times 51\times 50\times 49\times 48} \approx .51$$
and 
$$P(A\cap B) = P(B) = \frac{10\times 4^5}{{52 \choose 5}} \approx .004$$ 
and 
$$P(A\cap B\cap C) = \frac{10\times 4}{{52\choose 5}} \approx .000015$$
I can't seem to figure out $$P(A\cap C) = ?$$
Any suggestions or how I should think about it are greatly appreciated!
 A: I think it is easier to count possible hands than to try to get into probabilities from the beginning.
Possible sets of ranks. There are $\binom{13}{5}$ ways to select 5 different ranks for a hand, but 10 of these would be straights, so the actual number of desired assignments is $\binom{13}{5}-10$
Possible suit assignments. Once we have chosen the ranks, no matter how we choose them, there are $4^5$ possibilities for which suits they are -- but $4$ of these would be flushes, so we need to subtract them, giving $4^5-4$. (Note that this step is made particularly easy because we already know the ranks are all different, so we never have to worry about using the same card twice).
The number of high-card hands is therefore
$$ ({\textstyle\binom{13}{5}}-10)(4^5-4)  $$
Now divide by $\binom{52}{5}$ to find the probability of such a hand.
A: A different approach, which is much slower and clumsier.  
There are 52 ways to choose the first card. The second card must be of a different rank, so there are only 48 choices and so on.  
So, there are 52*48*44*40*36/5! ways to choose the five cards = 1317888 poker hands with 5 cards all of different ranks. Since  straights, flushes and straights are included in the count , these must all be subtracted from this figure. 
The corrrect figure is 1317888-10240-5108-40 = 1302540 . 
Henning's solution is definitely far superior. 
