Playing cards: Probability pairs

Question

An ordinary deck of $52$ playing cards ($4$ suits, $13$ values) are shuffled. Then player $A$ takes the top two cards. Player $B$ takes the next to top cars. Let $Pr[A]$ denote the probability for player $A$ to have a pair (i.e. two cards with the same value) and $Pr[B]$ the probability for player $B$ to have a pair. Task: Calculate $Pr[A], Pr[B], Pr[A \cap B]$ and $Pr[A|B]$.

My attempt:

$$Pr[A] = 1 \cdot \frac{3}{51} = \frac{3}{51}$$ because $A$ can take any first card but then there are only $3$ out of the remaining $51$ cards left, that match up to a pair.

$$Pr[B] = \frac{3}{49}$$

$$Pr[A \cap B] = Pr[A] \cdot Pr[B]$$ (are they independent?)

$$Pr[A|B] = \frac{Pr[A \cap B]}{Pr[B]}$$

I'm pretty sure that my $Pr[B]$ is wrong (or isn't it??). And because I don't have the exact $Pr[B]$ I can't calculate $Pr[A \cap B]$ and $Pr[A|B]$ just yet.

So, could somebody explain to me how to calculate $Pr[B]$ and if I can then just use the formulas above to calculate the remaining probabilities.

Thank you!

(If anything is unclear, please let me know!)

Answer 1

Here is a perhaps inelegant way to show that $P(B)=P(A)$

Suppose player $A$ gets a pair with probability $\frac{3}{51}$

Then player $B$ gets a pair with probability $$\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}$$

since for his first card, he can either get a different rank than $A$ with probability $\frac{48}{50}$ or get the same rank with probability $\frac{2}{50}$

Suppose player $A$ does not get a pair with probability $\frac{48}{51}$

Then player $B$ gets a pair with probability $$\frac{44}{50}\cdot\frac{3}{49}+\frac{6}{50}\cdot\frac{2}{49}$$

since player $A$ gets one of each rank, and so for his first card, he can either get a different rank than both of $A$'s cards with probability $\frac{44}{50}$ or get the same rank as one of them with probability $\frac{6}{50}$

All together

$$\frac{3}{51}\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)+\frac{48}{51}\left(\frac{44}{50}\cdot\frac{3}{49}+\frac{6}{50}\cdot\frac{2}{49}\right)=\frac{3}{51}$$

From here, the rest is trivial.

We now have

$$P(A\cap B)=\frac{3}{51}\cdot\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)\approx 0.0035$$

and

$$\begin{align*} P(A \mid B) &=\frac{P(A\cap B)}{P(B)}\\\\ &=\frac{\frac{3}{51}\cdot\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)}{\frac{3}{51}}\\\\ &\approx0.0596 \end{align*}$$

Answer 2

Here is one way to look at $P(A)$ and $P(B)$.

There are $52!$ ways to arrange the cards in the deck, all of which we assume are equally likely. The number of those arrangements in which the first two cards have the same rank is $52 \cdot 3 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdots 1$, because there are $52$ ways to choose the first card, then three ways to choose the second card, then $50$ ways to choose the third card, $49$ ways to choose the fourth card, etc.

The number of arrangements in which the third and fourth cards have the same rank is $50 \cdot 49 \cdot 52 \cdot 3 \cdot 48 \cdot 47 \cdots 1$, because there are $52$ ways to pick the third card, then three ways to pick the fourth card, then we can pick the first card in $50$ ways, the second card in $49$ ways, etc.

So the number of arrangements in which the first two cards have the same rank is the same as the number of arrangements in which the third and fourth cards have the same rank, and the probability of either outcome is $$\frac{52 \cdot 3 \cdot 50 \cdot 49 \cdot 48 \cdots 1}{52!} = \frac{3}{51}$$

Answer 3

This is one of the classic questions because B is not independent of A. In fact the third and fourth card do depend on what was selected in the first and second. So, the long hard way to determine this is to use the law of total probability and think about all possibilities for 1st and 2nd pull. To make it easier and still get a handle on how to approach this problem, try to look at a simpler problem where C is the event that the first card is red and D is the event that the second card is red. P(C)=26/52=1/2. P(D)? Well $P(D|C)=25/51.$ $P(D|C^C)=26/51.$ $P(D)=P(C and D) +P(C^C and D)=1/2*25/51+1/2*26/51=1/2$.
What if E is the event that the third card is red. Well, using total probability and thinking of all possible first and second card possibilities it turns out that it again is 1/2. So, without knowing so needing to take in consideration all possible previous results, the probabilities simplify to the same as if it was the first pull. The probability that the 12 card selected was red is also 1/2. So, in your case, P(B)=P(A)=3/51. To determine P(A and B) you need to partition into cases: they were all the same value or they were two distinct pairs. P(A and B)=3/51*([(2*1)+(48*3)]/50*49). For the conditional, Use the formula P(B|A)=P(A and B)/P(A).