While going through the textbook, Discrete Math and its Applications, I found a problem, "what is the probability that a 5-card poker hand contains at least one ace?" It uses a standard deck of 52 cards with 4 aces.
The solution given and that I have found on here uses a complement, which I understand.
But my initial answer, which leads to a wrong value, was
$$\frac{\binom{4}{1}\cdot\binom{51}{4}}{\binom{52}{5}}$$
I still can't figure out why this is wrong, which suggests I have a misunderstanding of combinatorics. I'll write my reasoning and would appreciate if someone can point out how it's flawed.
I thought the size of the event in which our hand has at least one ace would be equal to the number of ways we can choose one ace, which is $\binom{4}{1}$ times the amount of ways we can choose 4 other cards from the remaining cards in the deck, which is $\binom{51}{4}$. And then this goes over the sample space of $\binom{52}{5}$.
I would greatly appreciate an explanation of why this reasoning is flawed or resources that might help me understand this better.
Best Answer
Your answer counts some hands more than once. For example, [A♡ A♠ 3♢ 4♢ 5♢] is counted both as selecting A♡ and then the four cards A♠ 3♢ 4♢ 5♢, but also as selecting A♠ and then the four cards A♡ 3♢ 4♢ 5♢.
To avoid using the complement event, you could instead add together the counts of hands with