Bayes Theorem Question about drawing two cards from a choice of four

Question

Let's take two black cards and two red cards. I draw two of these cards out at random and look at them. You ask me "Is one of your cards red?" I answer "Yes." What is the probability that I am holding a pair of red cards?

Answer 1

There are three possible, unordered, hands: $RR,RB,BB$. Let's get the probability of each:

$P(RR)=P(BB)$ (by symmetry). The first is red with probability $\frac 12$, and the second matches with probability $\frac 13$. Thus $$P(RR)=P(BB)=\frac 16$$

We deduce that $$P(RB)=1-\frac 13=\frac 23$$ We can see that directly since the first is whatever color and the second has the opposite color with probability $\frac 23$.

You get a YES response if the hand is $RR$ or $RB$, a combined probability of $$\frac 23+\frac 16=\frac 56$$ It follows that the desired answer is $$ \frac 16\Bigg /\frac 56\quad =\quad \frac 15$$

Answer 2

The trap is that the second pick of card is not independent from the first pick because it changes the card pool, the pick is without replacement. It is as simple as that. This kind of riddle always build a context that tricks you into thinking that two successive experiments are independent when they are not.

An answer based on formulas rather than explanation. I believe it could help for more complex and similar riddle:

$R_1$ the event "the first card picked is red", $B_1$ the event "the first card picked is black"

Let A be the event where I answer "Yes" to the question "Is one of your cards red?"

Note that $A = R_1 \cup R_2 $

$P(R_1 \cap R_2| A) = P(R_1 \cap R_2 \cap A)/P(A) $

$P((R_1 \cap R_2) \cap A) = P(R_1 \cap R_2) = P(R_2|R_1)P(R_1) = \frac13 \frac12 = \frac16$

$P(R_1 \cup R_2) = 1- P(B_1 \cap B_2) = 1 - P(R_1 \cap R_2) = \frac56 $

$P(R_1 \cap R_2| A) = \frac{\frac 16}{\frac 56} = \frac 15$