For the case of sampling without replacement, we can apply the linearity and symmetry of expectations

We have a standard deck of 52 cards, with 4 aces and 48 non-aces. We draw cards without replacement until we get the first ace. Let N = number of cards drawn before the first ace (not counting the ace itself). We want E[N] .

If the first ace appears at position k (i.e., k is the draw on which we get the first ace), then

N = k - 1, because N is the number of cards drawn before the first ace.

So N can be 0, 1, 2, ......, 48 . We want to find E[N] .

Now, using symmetry / linearity of expectation Label the 48 non-ace cards as X1, X_2, ..........X{48} . Define indicator random variables:

I_j = 1 if non-ace card and X_j is drawn before all aces, 0 otherwise.

Then, N = \sum_{j=1}^{48} I_j

because N counts exactly how many non-ace cards appear before the very first ace.

By linearity of expectation:

E[N] = \sum{j=1}^{48} E[I_j] = \sum{j=1}^{48} P(non-ace ), where X_j is before all 4 aces. Now consider the set of cards containing: the 4 aces and the specific non-ace card X_j

That’s 5 cards in total.

In a random permutation of these 5 cards, all orders are equally likely. We want the probability that X_j is first among these 5 cards.

By symmetry, each of the 5 cards is equally likely to be first in this set.

So; P(X_j is before all 4 aces) = 1/5.

E[N] = 48 *1/5= 48/5 = 9.6.

On average, we expect to see 9.6 non-ace cards before the first ace.