# Probability of choosing ace of spades before any club

From a deck of $52$ cards, cards are picked one by one, randomly and without replacement. What is the probability that no club is extracted before the ace of spades?

I think using total probability for solve this

$$P(B)=P(A_1)P(B\mid A_1)+\ldots+P(A_n)P(B\mid A_n)$$

But I am not sure how to solve this. Can someone help me?

## 3 Answers

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: $$\sum_{n=1}^{52}\binom{52-n}{13}=\sum_{m=13}^{51}\binom{m}{13}=\binom{52}{14}$$

• Or alternately to your second explanation, $\spadesuit A$ is equally likely to be in each of the fourteen positions and thus has a 1/14 chance of being in the first position. – Mathieu K. Apr 23 '18 at 2:58
• To add a simplification; you can remove the 38 non-club, non-ace of spades cards without affecting the outcome. – JollyJoker Apr 23 '18 at 7:46
• Of course your answer is completely right, but can you elaborate on something for me? Why do you feel that the statement $P(\spadesuit A \text{ first } | E_n)=1/14$ is trivial enough to just write "due to the symmetry among the 52 cards", but not feel that the statement without conditioning on $E_n$ requires a detailed proof (using as part of the proof the claim that the conditioned statement is easy)? I may be missing something, but I don't see why one statement is particularly harder than the other. Thanks! :) – Sam T Apr 23 '18 at 20:33
• @SamT I feel that neither statements require detailed proofs, as they are both intuitively obvious. For the first statement, the details were easy to fill in, so I included because why not. – Mike Earnest Apr 23 '18 at 22:45
• I just feel that your middle proof basically assumes the result (since assumes the conditioned result). I have no problem with your first proof though! – Sam T Apr 24 '18 at 17:15

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: $1/14$.

• How is @MikeEarnest's answer less rigorous than this one? Both entail identifying 14 positions, but whereas in Mike's answer they are by definition those occupied by the 14 salient cards, your answer says "Choose 14 out of 52 places" with no prior motivation. – Rosie F Apr 23 '18 at 6:57
• @RosieF: What's missing is the explanation why the restriction gives the right answers. There are a number of paradoxes that arise because people think that's an intrinsically valid line of reasoning. For example, the famous Monty Hall paradox where the invalid argument selects the subset "the two doors that weren't opened" and mistakenly believes that both choices have equal probability of goat, or the "at least one child is a girl" paradox where the invalid argument chooses a girl and mistakenly believes the choices of "boy, girl" for the other child have equal probability. – Hurkyl Apr 23 '18 at 8:45
• @Hurkyl both of your example mistakes are due to belief that the reveal of information changes the probabilities that existed before the first choice was made. That does not apply to Mike's answer. – OrangeDog Apr 23 '18 at 9:44

Think of the probability that a club has not yet been drawn on draw $d$ as $$\sum _{n=1}^d \left(1-\frac{13}{52-d}\right)$$

And the probability that the ace of spades is drawn on draw $d$ as $$\frac{1}{52-d}$$

Then the probability that the ace of spades is drawn on draw $d$ given that a club has not yet been drawn is $$\left(\sum _{n=1}^d \left(1-\frac{13}{52-d}\right)\right)*\left(\frac{1}{52-d}\right)$$

Does that help?

• And then sum the final expression over all possible values of $d$? I don't think that's going to help. Regardless, your first expression gives probabilities higher than 1 for various values of $d$ (even if the denominator is changed to $52-n$). I think you want a product instead. – Teepeemm Apr 23 '18 at 14:45