A probability problem by Trung Le

Let there be $a$ face cards in the small pile of $b$ cards, and $c$ face cards in the bigger pile of $d$ cards. Then since there are $12$ face cards in a standard deck of $52$ cards, we have that

$a + c = 12$ and $b + d = 52 \Longrightarrow c = 12 - a$ and $d = 52 - b,$ (A).

Now we are given that $\dfrac{a}{b} = \dfrac{4}{11} \Longrightarrow 11a = 4b$ and that $\dfrac{c}{d} = \dfrac{2}{15} \Longrightarrow 15c = 2d.$

Substituting the results from (A) into this last equation gives us that

$15*(12 - a) = 2*(52 - b) \Longrightarrow 180 - 15a = 104 - 2b \Longrightarrow 2b = 15a - 76 \Longrightarrow 4b = 30a - 152.$

But we also have that $4b = 11a,$ and so $30a - 152 = 11a \Longrightarrow 19a = 152 \Longrightarrow a = \boxed{8}.$