John and Jane are taking the cards from the well mixed pack of $16$ cards. What is the probability that John in the hands has card of boy?

Question

John and Jane are taking the cards from the well mixed pack of $16$ cards. In pack of cards are $4$ aces $(A)$, $4$ kings $(K)$, $4$ queens $(Q)$ and $4$ boys $(J)$. First John take one card from the top of the pack. But if he take a boy then he quickly, before Jane see, take another card and return card with boy on the top of the pack. Otherwise he keep the card. What is the probability that John in the hands has card of boy?

So my solution is $$P(\text{John in the hands has card of boy}) = 1/4$$

Is that correct?

Answer 1

No, since John takes a look at his first card, we should condition on the event $J_1$ i.e. the event that the first card is $J$. If his first card is boy (event $J_1$ with probability $4/16$) then he changes to a second card different from the first. So, there is now a $3/15$ chance to draw a boy again. Contrary, if his first card is not a boy (event $J_1^c$ with probability $12/16$) he keeps it and therefore there is $0$ probability that he will land with a boy in his hands in this case. Lets denote with $J_2$ that the final card is a boy. Putting these together, we then have \begin{align}P(J_2)&=P(J_2\mid J_1)P(J_1)+P(J_2\mid J_1^c)P(J_1^c)=\frac{3}{15}\cdot\frac{4}{16}+0\cdot\frac{12}{16}=\frac{12}{240}=\frac1{20}\end{align}

Answer 2

No, that would be the probability if John couldn't swap the first card in case it was a $J$.

Note that John has a $J$ if and only if both top cards were $J$. The probability for this is

$$ \frac{\binom42}{\binom{16}2}=\frac{4\cdot3}{16\cdot15}=\frac1{20}\;. $$