I know to calculate the probability in a 5 card hand of at least one ace, one could calculate the probability of no aces, and take 1 - that. but if i wanted to do it directly without the complement to test my understanding, how would I do this?
The way i see it is this is a union problem, so its one ace, or 2 aces, or 3 aces etc..
so what I came up with is: $4 \choose 1$$48 \choose 4$ + $4 \choose 2$$48 \choose 3$ + $4 \choose 3$$48 \choose 2$ + $4 \choose 4$$48 \choose 1$
in the numerator (the total number of cards with at least one ace? and then the denominator is $52 \choose 5$
But This answer is different than what i get with the 1- pr(no ace). so I am wondering where I am going wrong?