A hand of 13 cards is dealt from a standard deck of 52 cards. What is the probability that it contains more aces than tens? How does this probability change when you have the information that the hand contains at least one ace?
If
$X=[$# aces in hand$]\Rightarrow \mathbb{P}(X=x)=\frac{\binom{4}{x}\binom{52-4}{13-x}}{\binom{52}{13}}$ and
$Y=[$# tens in hand$]\Rightarrow \mathbb{P}(Y=y)=\frac{\binom{4}{y}\binom{52-4}{13-y}}{\binom{52}{13}}$, I should have $\mathbb{P}(X>Y)=\sum_{y=0}^{x}\frac{\binom{4}{y}\binom{48}{13-y}}{\binom{52}{13}}\cdot \sum_{x=y+1}^{4}\frac{\binom{4}{x}\binom{48}{13-x}}{\binom{52}{13}}=\sum_{y=0}^{4}\frac{\binom{4}{y}\binom{48}{13-y}}{\binom{52}{13}}\frac{\binom{4}{y}\binom{48}{13-y}}{\binom{52}{13}}$.
Thus if $X\geq 1$ we obtain $\mathbb{P}(X>Y|X\geq 1)=\frac{\mathbb{P}(X>Y\cap X\geq 1)}{\mathbb{P}(X\geq 1)}=\frac{\mathbb{P}(X>Y)}{\sum_{x=1}^{4}\frac{\binom{4}{x}\binom{48}{13-x}}{\binom{52}{13}}}$,
where $\mathbb{P}(X>Y|X\geq 1)=\mathbb{P}(X>Y)$ because if $X<1\Rightarrow X=0\Rightarrow \mathbb{P}(0>Y)=0$.
Is it correct?