# Waiting for an Ace

### Problem

### Solution 1

Four ace divide the deck into five parts, each of which may contain from $0$ to $48$ cards. According the Symmetry Principle, all five have on average the same length which is, therefore, $48/5=9.6.$ The next card after an aceless run should be an ace - a card number $10.6.$

### Solution 2

Note the succesive waiting times, excluding events. We have the following probabilities:

$\frac{4}{52}$, waiting time $0$

$\frac{4}{51} \left(1-\frac{4}{52}\right)$,waiting time $1$

$\frac{4}{50} \left(1-\frac{4}{52}\right) \left(1-\frac{4}{51}\right),$

The expectation becomes:

$\displaystyle \sum _{i=1}^{48} i \frac{4 }{52-i}\prod _{j=0}^{i-1} \left(1-\frac{4}{52-j}\right)=\frac{48}{5}.$

(The expression could also be written as $\displaystyle \sum _{i=0}^{47} \frac{4 (i+1) \prod _{j=0}^i \left(1-\frac{4}{52-j}\right)}{52-1-i}=\frac{48}{5}$. The former seems self-explanatory.)

### Acknowledgment

The problem and Solution 1 are from Fifty Challenging Problems in Probability with Solutions, (Dover, 1987, problem 40) by F. Mosteller. Solution 2 is by N. N. Taleb.

Mosteller defines the Principle of Symmetry exactly in accordance with the above usage:

When $n$ (uniformly) random points are picked on a segment, the $n+1$ parts so created have the same length distribution.

- What Is Probability?
- Intuitive Probability
- Probability Problems
- Sample Spaces and Random Variables
- Probabilities
- Conditional Probability
- Dependent and Independent Events
- Algebra of Random Variables
- Expectation
- Probability Generating Functions
- Probability of Two Integers Being Coprime
- Random Walks
- Probabilistic Method
- Probability Paradoxes
- Symmetry Principle in Probability
- d'Alembert's Misstep
- Random Points on a Segment
- Waiting for an Ace

- Non-transitive Dice

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