# question involving two deck of cards [closed]

Consider two identical card packs: A and B. One card is taken from A and shuffled with pack B. The top card of pack A is QUEEN of HEARTS. What is the probability that the top card of pack B is KING of HEARTS?

• kindly someone help me with this question – mosheni Jun 15 '15 at 7:16
• Welcome to Puzzling! This is a text-book style problem rather than a puzzle, thus a bit off-topic here. Check this: meta.puzzling.stackexchange.com/questions/2783/… – leoll2 Jun 15 '15 at 10:31
• The answer is 0. An unshuffled deck does not have the king of hearts on the top. – Ian MacDonald Jun 15 '15 at 11:39

There was $52$ cards in both pack $A$ and $B$. After the shuffle, there are $51$ in $A$ and $53$ in $B$. Showing the Top card of $A$ leaves, though, with only $51$ possible cards to be taken from $A$ (since we know the Queen hadn't moved).

We have two possible scenarios now. If we did not take the King of Hearts, from $A$ to $B$, and the other were we did.

Be $K$ the event of taking the King of Hearts from $A$ to $B$. Be $B_K$ the event of it being the top card of $B$.

\begin{align} P(B_K)&=P(K)P(B_K\mid K)+P(K^C)P(B_K\mid K^C)\\ &=\frac{1}{51}\frac{2}{53}+\frac{50}{51}\frac{1}{53}\\ &=\frac{52}{2703}\\ &\approx 0.02 \end{align}

In order to respond to @frodoskywalker I'll post a more generic answer.

$r$ will stand for having revealed $r$ cards from $A$ and none being the King. Note that it's important that's done randomly (like simply showing top cards is ok; the problem would be fetching for certain cards).

\begin{align} P(B_K^r)&=P(K_r)P(B^r_K\mid K_r)+P(K_r^C)P(B^r_K\mid K_r^C)\\ &=\frac{1}{52-r}\frac{2}{53}+\frac{51-r}{52-r}\frac{1}{53}\\ &=\frac{53-r}{2756-53r}\\ P(B_K^1)&\approx 0.02 \end{align}

• This answer is subtly wrong. To see why, imagine that the top 50 cards of deck A were revealed (and not king of hearts), and use this method to calculate the odds of KoH on top of deck B... – frodoskywalker Jun 15 '15 at 10:00
• Under your scenario: $P(B_K)=\frac{1}{2}\frac{2}{53}+\frac{1}{2}\frac{1}{53}$. What's the problem with that? – Masclins Jun 15 '15 at 10:21
• Sorry, I'm mistaken. I had a bit of Monty Hall style confusion. – frodoskywalker Jun 15 '15 at 11:10

Different method, same answer.

There are $52+1$ cards in the new deck B. The top card is from deck A with probability $1/53$ and from deck B with probability $52/53$.
If it is from deck A, it is a King of Heart with probability $1/51$. (Any card but the Queen of Heart).
If it is from deck B, it is a King of Heart with probability $1/52$.

The combined probability to be a King of Heart is:
$P(K) = P(A)P(K|A) + P(B)P(K|B)$
$P(K) = (1/53)(1/51) + (52/53)(1/52) = 52/2703$