Waiting for an Ace

Problem

Waiting for an Ace

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:

$\displaystyle \frac{4}{52}$, waiting time $0$
$\displaystyle \frac{4}{51} \left(1-\frac{4}{52}\right)$, waiting time $1$
$\displaystyle \frac{4}{50} \left(1-\frac{4}{52}\right) \left(1-\frac{4}{51}\right),$ waiting time $2$
and so on.

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.)

Generalization

The kind of experiment we deal with on this page is usually referred to as drawing without replacement. More specifically, we are concerned with an expected number of trials to the first success. This is quite different from a much simpler experiment with replacement.

N. N. Taleb and Michael Wiener derived a general formula:

For a total sample size of $n$ with the initial probability of $p$ the expected time to the first success is

$\displaystyle E(\tau)=\frac{n+1}{pn+1}.$

Note that as $n\to\infty,$ this converges to the solution $\displaystyle \frac{1}{p}$ of the aforementioned problem with replacement.

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.

 

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