Probability of drawing an ace from a reduced set of cards Let's say you have a deck of $52$ cards, $4$ of which are aces.  The odds of drawing an ace from this deck are $4/52 \approx 8\%$.
But you're not drawing from the full deck.  Instead, you have a deck of only $10$ cards, randomly pulled from the full $52$-card deck.  What's the chance of drawing an ace from this reduced deck?
All $4$ aces might have made it into this deck, in which case the odds would be $4/10 = 40\%$.  But maybe none of the aces made it, leaving the odds of drawing one at $0\%$.
What is the distribution of the odds of drawing an ace from the reduced deck?
 A: For the distribution of the odds of drawing an ace from the reduced deck, the odds is 0 if the reduced deck contains no ace, i.e.
$$\Pr \left( odds = 0 \right) = \frac{\binom{48}{10}}{\binom{52}{10}}$$
for the $\binom{48}{10}$ combinations out of the $\binom{52}{10}$ combinations which gives no aces in the reduced set.
Similarly,
$$\Pr \left( odds = 10\% \right) = \frac{\binom{4}{1}\binom{48}{9}}{\binom{52}{10}}$$
Hope you now know the rest, and check the probability of getting an ace at last is still $\frac{1}{13}$.
A: Let $p_S$ be the probability of randomly picking an ace from a set $S$ of $10$ cards.  We note that $p_S$ varies only with the number of aces in $S$.  More specifically, $$p_S=\frac{\text{nr aces in } S}{10}.$$
The number of aces in $S$ must be in $\{0,1,2,3,4\}$, since there are only $4$ aces in a deck of cards.  Hence $p_S \in \{0,1/10,2/10,3/10,4/10\}$.
If we choose $S$ uniformly at random from the set of all $10$-subsets of a deck of $52$ playing cards, then, for $i \in \{0,1,2,3,4\}$,
\begin{align*}
\mathrm{Pr}(p_S=i/10) &= \mathrm{Pr}(\text{nr aces in } S \text{ is } i) \\
&=\frac{\text{nr 10-subsets with } i \text{ aces}}{\text{total nr 10-subsets}} \\
&= \frac{\binom{4}{i}\binom{48}{10-i}}{\binom{52}{10}}
\end{align*}
since $S$ comprises of $i$ aces and $10-i$ non-aces.
A: The probability of drawing an ace out of ten randomly selected cards is 1/13, as peterwhy explains in his comment:  The card drawn is equally likely to be each of the fifty-two cards.
Please be careful in your use of the terms "probability" and "odds."
A: Another way of thinking that may convince you the correct value is $\left(\frac{1}{13}\right)$:
Before you start with the standard deck, take the $1-10$ of some suit, mix them up and pick one at random.  Say it's the $7$. Remember that!
Put the complete deck back together and shuffle it well.  You're going to take the top ten as your random subset of $10$, and then take the $7-th$ card as your pick from the subset (That's why you did the first pick, remember?).  The top ten cards are random, 'cause you shuffled well.  The choice of a particular one from the 10 is random.  All the specs have been met.
So, what's the probability that the $7-th$ card down in the shuffled deck is an ace?
