# The First Ace

## Problem

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?

## Solution

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:

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