Probability of drawing ace of spade given that 2 aces are drawn from two hands

Question

So my initial thoughts are:

\begin{align} \binom{4}{2} = 6 \end{align}

There are 6 pairs with 2 aces. And by inspection( since the numbers are small): I know there's 3 pairs:

(spade, club), (spade, heart), (spade, club)

So \begin{align}P(\text{Ace of spade | both are aces}) = 3 / 6 = 1/2 \end{align}

But I couldn't figure out how to express them as combinatorics terms or a more systematic way to express it.

Say I also want to find: \begin{align}P(\text{Ace of spade, Ace of heart| both are aces}) = 5/6 \end{align}

I know it by inspecting all the outcomes.But I doubt I'll recognize the pattern in more complicated cases.

I suppose I can try find it using complement:

\begin{align} P(\text{Ace of spade, Ace of heart| both are aces}) &= 1 - \frac{\text{# aces pair without spade or heart}}{\text{# ace pairs}} \\ &= 1 - \frac{\binom{2}{2}}{\binom{4}{2}} \\ &= 1 - \frac{1}{6} \\ &= \frac{5}{6} \end{align}

But if I want to build the solution from the other direction in a formalized way. How should I do that?

Edit #1.

It was one of my first guesses. But I don't know if I was falling to the trap of "wishing" the answer is true.

My first interpretation was like

\begin{align} \dfrac{\text{(only 1 way to choose a Ace spade) (3 ways to choose the remaining Aces, pick1)}}{(\text{6 pairs with 2 aces)}}=\dfrac{\binom{1}{1}\binom{3}{1}}{\binom{4}{2}} \end{align}

This question came up while I was studying the classic question.

\begin{align}P(\text{both are aces | Ace of spade })\end{align}

I was trying to figure out all the angles I can look at the question.

From here

As for the problem of $Pr(\text{Two aces}\mid \text{Ace of spades})$, we approach similarly as above, noting that the ace of spades might be in the first or the second position, yielding the final calculation of:

$$\dfrac{2\times \frac{1\times 3}{52\times 51}}{\frac{1}{52}+\frac{1}{52}-0}=\frac{1}{17}$$

In above, the numerator has the probability multiply by 2, given the ace of spade could be 1st draw or 2nd draw. I've been going through many problem sets, but I still have trouble determining whether or should multiply by factors or not. It feels more like guessing and memorizing the question pattern. I'm only sure after looking at the solution.

Like for

\begin{align} \dfrac{\text{(only 1 way to choose a Ace spade) (3 ways to choose the remaining Aces, pick1)}}{(\text{6 pairs with 2 aces)}}=\dfrac{\binom{1}{1}\binom{3}{1}}{\binom{4}{2}} \end{align}

I would also be second guessing whether I should multiply by 2 or not? What intuition could I be missing to see through it clearly?

Can't embed images yet... My sketch

Answer 1

"but I couldn't figure out how to express them as combinatorics terms" If you insist you could write this as $\dfrac{\binom{1}{1}\binom{3}{1}}{\binom{4}{2}}$, which follows the pattern one should recognize for the Hypergeometric Distribution.

With $N$ objects, $K$ of which have a desired feature, having drawn $n$ of the objects without replacement and without bias, the probability of $k$ of those draws having the desired feature will be $\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$, here for your example $K=k=1, N=4, n=2$.

That said, there is not a requirement that you rephrase things in such a way if it is perfectly understandable where the numbers are coming from using a more plebian approach.

As for your second half, you are not being clear whether you are wanting to have both a spade and a heart or not in which part of the calculation and presentation of your hypothetical problem. Having both a spade and a heart simultaneously should be the probability $\dfrac{1}{6}$ and you can formalize this by simply stating "there is only one possibility having both a spade and a heart simultaneously out of six equally likely possibilities for the what the combinations are of two suits"

If you wanted to find the value of $\frac{5}{6}$ in a direct way... you could have seen this by inclusion-exclusion as $\frac{3}{6}+\frac{3}{6}-\frac{1}{6}$ or you could have seen this by breaking into cases as "spade without heart, heart without spade, both" as $\frac{2}{6}+\frac{2}{6}+\frac{1}{6}$. This is not a terribly useful thing to have bothered with however, the indirect approach is almost always going to be the more convenient to use.