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There are many sources online that say that the number of cards you must pull to get your first ace is $10.6$, and the proofs make sense.

Doesn't the number of pulls follow the geometric distribution? There are $4$ aces out of $52$ cards, so there is a success probability of $\frac{1}{13}$. Then, the expected value in this distribution is $\frac{1}{\frac{1}{13}}=13$. Why isn't the expected number of cards to pull to see the first ace $13$? Is it because we aren't replacing the cards, so the sample space cardinality decreases by $1$ after each card pull? If so, what do these calculations look like?

neelsnarayan
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    The draws are not independent. As you draw cards, the make up of the deck changes, so the probability changes. If you drew with replacement then the answer would indeed be $13$. – lulu Jan 01 '25 at 00:30
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    this is clear if you take a smaller deck, so two cards, an Ace and a King. Without replacement, it's a $50-50$ chance whether you get the Ace first or second, so the expected number of draws is $\frac 32$. With replacement, of course it is $2$. – lulu Jan 01 '25 at 00:32
  • Related question https://math.stackexchange.com/questions/4737681/drawing-the-first-ace-using-geometric-distribution The comments there might help you understand why a geometric distribution doesn't quite work (just in case the above comments weren't enough) – JimmyK4542 Jan 01 '25 at 00:45

2 Answers2

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As mentioned in the comments, the Geometric Distribution does not apply here since the trials are not independent. It is true that the first draw is an Ace with probability $\frac 4{52}=\frac 1{13}$ but, if you miss on the first try, the probability that the second is then an Ace is $\frac 4{51}$ which is slightly greater than $\frac 1{13}$.

To do the computation explicitly, let $p_n$ denote the probability that the first Ace is seen on draw $\#n$. Then $p_1=\frac 1{13}$ as you suggest, but after that, the probabilities change.

We can compute $p_n$ by noting that we first need to see $n-1$ non-Aces and then an Ace. Each set of $n-1$ cards is equally likely to be drawn first and there are $\binom {48}{n-1}$ of them that have no Aces. After these $n-1$ non-Aces are drawn, the probability that the next one is an Ace is $\frac 4{52-(n-1)}$ making the answer $$p_n= \frac 4{53-n}\times \binom {48}{n-1}\Big /\binom {52}{n-1}$$

Following that, we can compute the expectation as $$E=\sum_{n=1}^{49}np_n=10.6$$

Where I just used Wolfram Alpha to evaluate the sum.

There are, of course, better ways to compute the expected value, but you asked for the explicit computation.

lulu
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Alternative approach:

Consider the following tableau, where A refers to an Ace.

- - A - - A - - A - - A - -

The distribution of the $~4~$ Aces among the $~52~$ cards, creates $~5~$ islands of non-Aces: before the first Ace, and after each of the $~4~$ Aces. The sum of the size of these $~5~$ islands is $~52 - 4 = 48,~$ and so the average size of these islands is $~9.6.~$

This implies that the expected size of the first island is $~9.6~$ which implies that you can expect to require $~(9.6 + 1)~$ draws, before you hit your first Ace.

$\underline{\text{Addendum}}$

A slightly different wording of the argument:

Assume that the $~5~$ islands are given by $~x_1, ~\cdots, x_5,~$ where $~x_1 + x_2 + x_3 + x_4 + x_5 = 48.$ Now, suppose that you permute these islands in any one of $~5!~$ ways into $~y_1, ~y_2, \cdots, y_5.~$

Each of $~x_1, ~\cdots, ~x_5~$ has a $~1/5~$ probability of being permuted to $~y_1.$

Therefore, the expected number of draws, just before drawing the Ace is

$$\sum_{i=1}^5 \dfrac{x_i}{5} = 9.6.$$

user2661923
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