A deck of 52 cards (13 ranks and 4 suits) is randomly shuffled, then split into two piles of 26 cards. Player A and play B each own one of the two piles. They both reveal the top of their piles simultaneously. What is the probability of the two revealed cards have the same rank?
My thought is: what player A reveals doesn't matter, and player B just needs to reveal the same rank. The expected number of cards with that same rank going into player B's pile is $\frac{1}{2}*4=2$ (total 4 cards of the same rank, each chance of going into player B's pile is $\frac{1}{2}$). The chance of that revealed card in player B's pile being the same rank is, therefore, $\frac{2}{26}$. Could I know if my reasoning is correct, or is there exists any overcounting or undercounting? And if mine is wrong, could I know how to correctly approach this problem. Thanks in advance.
Best Answer
The error is that we know one of the cards of that rank is in $A$'s pile. There are now $3$ cards of that rank among the $51$ cards left. The expected number in $B$'s pile is then $\frac {3\cdot 26}{51}$, which is a little greater than $1\frac 12$. The chance a particular card of $A$'s is matched is $\frac 3{51}=\frac 1{17}$ because there are $51$ unknown cards, of which $3$ match.