The First Ace


Shuffle an ordinary deck of \(52\) playing cards containing \(4\) aces. Then turn up cards from the top until the first ace appears. On the average, how many cards are required to produce the first ace?


Let \(X\) represent the number of cards that are turned up to produce the \(1\)st ace. For this problem, we cannot apply the Geometric Distribution because cards are sampled without replacement.

Instead, we begin by considering the probabilities of drawing the \(1\)st ace on the \(1\)st card, \(2\)nd card, and so on:

\[\begin{split} P(1st\;Card) &= \frac{4}{52} \\ P(2nd\;Card) &= \left(\frac{48}{52}\right)\left(\frac{4}{51}\right) \\ P(3rd\;Card) &= \left(\frac{48}{52}\right)\left(\frac{47}{51}\right) \left(\frac{4}{50}\right) \\ P(n^{th}\; card) &= 4\left[\frac{48!}{(49-x)!}\right]\left[{\frac{(52-x)!}{52!}} \right] \end{split}\]

Then, we can calculate the average number of cards by applying the definition of expected value:

\[E[X] = \sum_{x=1}^{52} 4x\left[\frac{48!}{(49-x)!}\right]\left[{\frac{(52-x)!}{52!}} \right] = \frac{53}{5} = 10.6\]

Thus, on average it will take \(10.6\) cards to get the \(1\)st ace.

Alternative Solution

The solution above is complex due to the unwieldy summation. Another approach is to apply the Principle of Symmetry, which states that \(n\) randomly placed points will divide a segment into \(n+1\) pieces, each of which has the same distribution.

This problem is an application of the principle with \(n=4\), since each ace in the deck represents a division point. Then, the average length of the \(5\) segments (stretches of cards without an ace) is \(\frac{52-4}{5} = \frac{48}{5}\). Each of these segments is immediately followed by an ace, so the expected number of cards until the \(1\)st ace is the following:

\[E[X] = \frac{48}{5} + 1 = \frac{53}{5} = 10.6\]