Question: What is the probability of getting all of the face cards in one pile when a standard deck of cards is split into two piles?
Additional info: Generally a deck of cards is split into two by putting cards in two different piles one card at a time in an alternating fashion until the entire deck is split into two even piles. A standard deck of cards contains 16 face cards out of 52 total cards. The order of the cards in either pile does not matter, all that matters is that one pile has all the face cards.
My approach and why I think it is wrong: I tried solving this problem by calculating the binomial probability ($_nC_xp^x(1-p)^{n-x}$), where $n=26$, $x=16$, and $p=\frac{16}{52}$, which results in $P(x)=0.00087$. I believe this is incorrect because it does not take into account the decreasing number of face cards and total cards as more are put into the two separate piles (correct me if I am wrong).
Best Answer
The first half of the deck must contain $26$ of the $36$ non-face cards, or it must contain $16$ face cards and $10$ non-face cards. $$\frac{\binom{36}{26}+\binom{36}{10}}{\binom{52}{26}}\approx1.025\times10^{-6}$$