You and your friend are each dealt two cards: hers face up and yours face down. Which of the following scenarios are you more likely to have a pair? I am self-teaching probability, could someone please help me determine if I've solved this problem correctly, as I don't have the answers to these exercises.
The following question is taken from Dennis Sun's Probability course at Cal Poly SLO: "You and your friend Amy are each dealt two cards: hers face up and yours face down. In which of the following scenarios are you more likely to have a pair: 1) When she has a pair of Queens, 2) Or when she has a Queen and a 5?" https://dlsun.github.io/probability/conditional.html
I modeled this the same way as if I were drawing 4 cards. Let A be the statement: I have a pair. B: Amy has Queens. C: Amy has Q5
I calculated the prior probability that Amy has Queens as $$P(B) = {{4 \choose 2}\over{52 \choose 2}} = 0.00452$$ And the prior that she has Q5 as: $$P(C) = {{4 \choose 1}{4 \choose 1}\over{52 \choose 2}} = 0.0121$$Then the probability that Amy has Queens and I have a pair is
$$P(AB) = {{4 \choose 4} + {4 \choose 2}{12 \choose 1}{4 \choose 2}\over{52 \choose 4}} = 0.00160$$
where the 1st term in the numerator is the ways that she can have Queens and I have the other 2 queens, and the 2nd term is the ways that she can have Queens and I have any other pair. Then the conditional probability would be: $$ P(A|B) = {{P(AB)}\over{P(B)}} = .353$$ And the probability that Amy has Q5 and I have a pair is
$$P(AB) = {{4 \choose 1}{4 \choose 3}\cdot2 + {4 \choose 1}{4 \choose 1}{11 \choose 1}{4 \choose 2}\over{52 \choose 4}} = 0.00402$$ where the 1st term in the numerator is the probability that I have 5's or Q's. And the 2nd term is I have any other pair.
And the conditional probability would be: $$ P(A|C) = {{P(AC)}\over{P(C)}} = .333$$
Did I make a mistake? It seems strange that the probability of me making a pair would go down if she has Q5.
 A: Your probability of having a pair does in fact decrease if Amy has Q5, rather than QQ.
The easiest way to stretch your intuition around this is to assume that you are playing with an $8$ card deck, $4$ Q's and $4$ 5's.
Then, if Amy has a Q5, the number of ways that you can have a pair are $~\displaystyle \binom{3}{1} + \binom{3}{1} = 6. ~$  That is,
there are $3$ ways of forming a pair with the three remaining 5's, and so on.
If Amy has QQ, then the number of ways that you can have a pair are $~\displaystyle \binom{2}{2} + \binom{4}{2} = 7.$
So, Amy's having paired up has increased your chances.

With the specific problem around a 52 card deck, you are confused about how to use Bayes Theorem.
Let $E_1$ denote the event that Amy has Q5.
Let $E_2$ denote the event that Amy has QQ.
Let $E_3$ denote the event that you have a pair.
In effect, you are being asked to compute and compare 
$p(E_3|E_1)$ versus $p(E_3|E_2)$.
From the intuition stretching in the first part of the answer, you should expect that $p(E_3|E_2)$ will be slightly larger than $p(E_3|E_1).$
You know that in general, $~\displaystyle p(R|S) = \frac{p(R,S)}{p(S)}.$  Here, I use the syntax $(R,S)$ to denote that events $R$ and $S$ both occur.
So, you will need to compute:
$$p(E_3|E_1) = \frac{p(E_1,E_3)}{p(E_1)} \tag1 $$
and
$$p(E_3|E_2) = \frac{p(E_2,E_3)}{p(E_2)}. \tag2 $$

Problems like these are best attacked combinatorically, where the probability is expressed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
$\displaystyle p(E_1) = \frac{\binom{4}{1}\times \binom{4}{1}}{\binom{52}{2}}.$
The easiest way to determine $p(E_1,E_3)$ is to reason it out.  With a Q5 given to Amy, there are
$\displaystyle \left[11 \times \binom{4}{2}\right] + \left[2 \times \binom{3}{2}\right] = 72$ different ways that your cards may be paired up.  The $(11)$ refers to the $(11)$ ranks other than Q or 5, and the $(2)$ represents the Q and 5 ranks.
So, $\displaystyle p(E_1,E_3) = \frac{\binom{4}{1}\times \binom{4}{1} \times 72}{\binom{52}{2} \times \binom{50}{2}}.$
In the above expression, the denominator represents that first there are $\binom{52}{2}$ ways that Amy is dealt two cards, and then $\binom{50}{2}$ ways that you are dealt two cards from the remaining cards in the deck.
Putting this all together, you have that
$$p(E_3|E_1) = \frac{p(E_1,E_3)}{p(E_1)}$$
$$= \frac{\frac{\binom{4}{1}\times \binom{4}{1} \times 72}{\binom{52}{2} \times \binom{50}{2}}}{\frac{\binom{4}{1}\times \binom{4}{1}}{\binom{52}{2}}}
$$
$$= \frac{72}{\binom{50}{2}}. \tag3 $$

The Mathematics around $~\displaystyle \frac{p(E_2,E_3)}{p(E_2)}~$ will be very similar.
$\displaystyle p(E_2) = \frac{\binom{4}{2}}{\binom{52}{2}}.$
The easiest way to determine $p(E_2,E_3)$ is to reason it out.  With a QQ given to Amy, there are
$\displaystyle \left[12 \times \binom{4}{2}\right] + \left[1 \times \binom{2}{2}\right] = 73$ different ways that your cards may be paired up.  The $(12)$ refers to the $(12)$ ranks other than Q, and the $(1)$ represents the Q rank.
So, $\displaystyle p(E_2,E_3) = \frac{\binom{4}{2} \times 73}{\binom{52}{2} \times \binom{50}{2}}.$
Putting this all together, you have that
$$p(E_3|E_2) = \frac{p(E_2,E_3)}{p(E_2)}$$
$$= \frac{\frac{\binom{4}{2}\times 73}{\binom{52}{2} \times \binom{50}{2}}}{\frac{\binom{4}{2}}{\binom{52}{2}}}
$$
$$= \frac{73}{\binom{50}{2}}. \tag4 $$
So, as expected,
$$p(E_3|E_1) = \frac{72}{\binom{50}{2}} ~~\text{and}~~ p(E_3|E_2) = \frac{73}{\binom{50}{2}}.$$
A: In either case, there are $50$ cards left in the deck and $\binom{50}{2}$ sets of $2$ cards you can draw, all of which we assume are equally likely.  It only remains to count the number of two-card sets which are pairs in the two cases.
Case 1: Amy has drawn a pair of Queens.  In this case, the number of pairs left in the deck is
$$12 \binom{4}{2} + 1 \binom{2}{2} = 73$$
Case 2: Amy has drawn a Queen and a Five.  The number of pairs left in the deck is
$$11 \binom{4}{2} + 2 \binom{3}{2} = 72$$
So your chance of drawing a pair is higher in the first case, in which Amy has drawn a pair of Queens.  If you want to know the probabilities involved, divide the number of favorable cases by $\binom{50}{2}$.
