Combining random cards from two decks and calculating probability

Question

Suppose you have 2 fair decks, let's call them deck A and deck B. Now take two cards from deck A and add them to the deck B, and shuffle; thus you now have a deck with 54 cards. Now draw 2 cards from deck B, what's the probability that you draw an ace?

EDIT: Let's say you draw exactly 1 ace.

This is kind of an arbitrary question I made up, I don't care so much about the specific answer but about the process. I'm really trying to understand conditional probability and I feel like if I can answer a question like this I'll be on my way.

My guess is that first we calculate the odds that one or both of the cards from deck A were aces. Then you would weigh the possibilities of two aces having been transferred, one ace having been transferred, or no aces having been transfered across drawing from deck B?

Can someone point me in the right direction?

Also as a complete aside, how do you calculate the odds of drawing exactly one ace when drawing two cards from a deck of 52? I've answered the question using complelemntary probability of drawing 2 aces, and no aces and then subtracting that from 1 but is there a "direct" way of doing it? i feel that I understand better when it's not through complements

Answer 1

We transferred $0$, $1$, or $2$ Aces.

The probability $p_0$ that we transferred $0$ is $\frac{\binom{4}{0}\binom{48}{2}}{\binom{52}{2}}$.

The probability $p_1$ that we transferred $1$ is $\frac{\binom{4}{1}\binom{48}{1}}{\binom{52}{2}}$.

The probability $p_2$ that we transferred $2$ is $\frac{\binom{4}{2}\binom{48}{0}}{\binom{52}{2}}$.

Given that we transferred $0$, the probability $a$ we draw exactly one Ace from the modified Deck B is given by $a=\frac{\binom{4}{1}\binom{50}{1}}{\binom{54}{2}}$.

Given that we transferred $1$, the probability $b$ we draw exactly one Ace from the modified Deck B is given by $b=\frac{\binom{5}{1}\binom{49}{1}}{\binom{54}{2}}$.

Given that we transferred $2$, the probability $c$ we draw exactly one Ace from the modified Deck B is given by $c=\frac{\binom{6}{1}\binom{48}{1}}{\binom{54}{2}}$.

Our required probability is $p_0 a+p_1b+p_2c$.

Answer 2

Compute as follows, letting $Y$ be number of aces drawn from deck B, and letting $X$ be the number of aces drawn from deck A: \begin{align} P[Y = 1] &= \sum_{i=0}^2P[Y=1 \mid X = i]P[X=i] \\ &= \sum_{i=0}^2 \frac{\binom{4+i}{1}\binom{50-i}{1}}{\binom{54}{2}} \frac{\binom{4}{i}\binom{48}{2-i}}{\binom{52}{2}} . \end{align} You can compute $P[Y=2]$ in a similar fashion.