If I draw two cards, what is the probability that either is an ace?

Question

I thought to solve this problem as follows:

$P_{atLeast} = P_{AceFirstOnly} + P_{AceSecondOnly} + P_{BothAces}$

The actual solution:

$P_{atLeast} = P_{AceFirst} + P_{AceSecond} - P_{BothAces}$

Why is my approach wrong and the solution correct.

My equation considers the appropriate cases without, more importantly, letting one case partially coincide with another.

My thoughts on the solution? When you enumerate all hands with an ace first, those hands also include aces occurring second. We shouldn't add $P_{AceSecond}$, as a result.

EDIT: I have included below what I think the resolution is as to why the solution doesn't overcount.

The following arrangements and their permutations represent the hands involving two aces from $P_{AceFirst}$ alone. The trick here is to realize that the 12 arrangements already already take double aces into account. If we then add $P_{AceSecond}$ to $P_{AceFirst}$, $P_{AceSecond}$, we will again take double aces into account. As a result, we must subtract this "double consideration" by subtracting $P_{BothAces}$ from the foregoing expression.

\begin{matrix} A_{1}A_{2} & A_{2}A_{3} \\ A_{1}A_{3} & A_{2}A_{4} \\ A_{1}A_{4} & A_{3}A_{4} \end{matrix}

and their permutations.

Answer 1

Your answer is correct, because $P_{AceFirstOnly}=P_{AceFirst}-P_{BothAces}$.

Answer 2

Both should work. In your case $$P_{FirstOnly}=P_{SecondOnly}=\frac{4\cdot 48}{52\cdot 51}\\P_{Both} = \frac{4\cdot 3}{52\cdot 51}$$

So your approach gives $\frac{396}{52\cdot 51}$

On the other hand, the other answer has: $P_{First}=P_{Second}=\frac{1}{13}=\frac{4\cdot 51}{52\cdot 51}$. So that approach gives:

$$\frac{2\cdot 4\cdot 51 - 4\cdot 3}{52\cdot 51} = \frac{396}{52\cdot 51}$$

So both answers are right.

The second approach is slightly easier to compute - those $\frac{1}{13}$ are fairly trivial - but it does require the more complicated inclusion-exclusion logic.

There is a third approach: $$P_{First}+P_{SecondOnly} = \frac{1}{13}+\frac{48\cdot 4}{52\cdot 51} = \frac{396}{52\cdot 51}$$