Probability à la Tristram Shandy

Problem

Probability a la Tristram Shandy, problem

Solution 1

Hilbert's Hotel! A set with countably infinite number of elements can accommodate finitely many new members! The probability of picking any apple is $1$ and the probability of picking ALL apples is also $1.$ In a way an intuitive connection to Zero-One Laws.

In more detail: the probability of picking one given apple in $N$ trials is

$\displaystyle P(N)=\sum_{cycl}\frac{\displaystyle \left(1-\frac{1}{9i+1}\right)^i}{9i+1}.$

We can show that $\displaystyle \lim_{N\to\infty}P(N)=1.$

Solution 2

The probability of not having selected the first apple after the first selection is $\displaystyle{\frac{9}{10}}$.

The probability of not having selected the first apple after the first two selections is $\displaystyle{\frac{9}{10} \cdot \frac{18}{19}}$.

The probability of not having selected the first apple after the first three selections is $\displaystyle{\frac{9}{10} \cdot \frac{18}{19}\cdot \frac{27}{28}}$.

The probability of not having selected the first apple after the first $n$ selections is

$\displaystyle \prod_{k=1}^n \frac{9k}{9k+1}.$

The limit of this product as $n \to \infty$ is zero.

Then the limit of the probability of selecting the first apple as $n \to \infty$ is one.

Likewise, the limit of the probability of selecting the second apple as $n \to \infty$ is one.

Likewise, the limit of the probability of selecting the $j$-th apple as $n \to \infty$ is one.

Acknowledgment

The problem has been discussed previously two decades ago. It reflects on the Tristram Shandy paradox by Bertrand Russell.

Solution 1 is by N. N. Taleb; Solution 2 is by Jim Henegan.

There's an earlier variant of this topic with all due references.

 

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