Probability of choosing ace of spades before any club
The event that you find $\spadesuit A$ before any $\clubsuit$ is entirely determined by the order in which the $14$ cards $\spadesuit A,\clubsuit A,\clubsuit 2,...,\clubsuit K$ appear in the deck. There are $14!$ possible orderings of these $14$ cards, and each of these orderings are equally likely.
How many of these orderings have $\spadesuit A$ appearing first? The first card must be $\spadesuit A$, there are $13$ choices for the second card, $12$ for the third, and so on, so there are $13!$ such orderings. Therefore, the probability is $13!/14!=\boxed{1/14}$.
Put even more simply: of the fourteen cards $\spadesuit A,\clubsuit A,\clubsuit 2,...,\clubsuit K$, each is equally likely to appear earliest in the deck, so the probability that you find $\spadesuit A$ first is $1/14.$
Added Later: There is also a way to solve this using the law of total probability. We may as well stop dealing cards once the $\spadesuit A$ or any $\clubsuit$ shows up. Let $E_n$ be the event that exactly $n$ cards are dealt. Then $$ P(\spadesuit A\text{ first})=\sum_{n=1}^{39}P(\spadesuit A\text{ first }|E_n)P(E_n) $$ Now, given that the experiment ends on the $n^{th}$ card, we know that the $n^{th}$ card is one of $\spadesuit A,\clubsuit A,\clubsuit 2,...,\clubsuit K$, and none of the previous cards are. Each of these is equally likely (due to the symmetry among the 52 cards), so $P(\spadesuit A\text{ first }|E_n)=1/14$. Therefore, $$ P(\spadesuit A\text{ first})=\sum_{n=1}^{39}\frac1{14}P(E_n)=\frac1{14}\sum_{n=1}^{39}P(E_n)=\frac1{14}\cdot 1, $$ using the fact that the events $E_n$ are mutually exclusive and exhaustive.
I offer one final method which is more direct, but leads to a summation which is difficult to simplify. Let $F_n$ be the event that the $n^{th}$ card is the $\spadesuit A$. Then $$ P(\spadesuit A\text{ first})=\sum_{n=1}^{52}P(\spadesuit A\text{ first }|F_n)P(F_n)=\sum_{n=1}^{52}\frac{\binom{52-n}{13}}{\binom{51}{13}}\cdot\frac1{52} $$ You can simplify this to $1/14$ using the hockey-stick identity: $$\sum_{n=1}^{52}\binom{52-n}{13}=\sum_{m=13}^{51}\binom{m}{13}=\binom{52}{14}$$
These sorts of questions are often solved via "shortcut", such as in the other answer. The reason why the shortcut works is rarely discussed.
The point of this answer is to sketch out the rigorous argument underlying the shortcut. It also demonstrates a methodology one can try to apply to more general problems, or in problems where one has identified a shortcut but still needs to check that the shortcut should give the right answer.
We can describe a choice of how to order a deck of cards in the following way:
- Choose 14 out of 52 places
- Choose an arbitrary ordering of the 14 cards consisting of the 13 clubs and the ace of spades
- Choose an arbitrary ordering of the remaining 38 cards
The ordering this describes is given by placing the 13 clubs and the ace of spades into the chosen places, in the chosen order, and the remaining 38 cards in the remaining places, in the chosen order.
The important thing for this to be a good description is that it has the following properties:
- Every ordering of a deck of cards can be described in this fashion
- Every such description determines a unique ordering of the deck
So, we can determine probabilities simply by counting.
The reason for choosing this description is that:
- the three choices are completely independent from one another
- the problem depends only on the second choice: how to order the 13 clubs and the ace of spades
So, we can (rigorously!) reduce the original problem consisting of a whole deck of cards to the simpler problem consisting of just these 14 cards.
By a similar analysis, we can describe choices of how to order these 14 cards by:
- Choose 1 place
- Choose an arbitrary ordering of the 13 clubs
The ordering so described puts the ace of spades in the chosen place and the clubs in the remaining places, in the chosen order.
Again, this is a good description, the choices are independent, and only the first one matters. So we've reduced the original problem to:
What are the odds of a chosen place among 14 cards is the first?
which is easy to answer: $1/14$.