Two cornucopias and infinitely many gold coins

Question

There're two cornucopias of infinite volume, each with one gold coin in them. Right now $t=0$. A new coin will be born at $t=\frac{2^n-1}{2^n}$ for every positive integer $n$. Each new born coin will appear in the first cornucopia with probability $\frac{x}{x+y}$ and in the second with probability $\frac{y}{x+y}$, where $x$ and $y$ are the current number of coins in the two cornucopias respectively.

At $t=1$, infinitely many gold coins have been born, so we know at least one cornucopia contains infinite coins.

Question: What is the probability that both of them contain infinitely many gold coins?

Answer 1

The probability that both of them contain infinitely many coins is

1

Let us label the two cornucopias A and B and let us look at an example of a finite sequence:

n    2    3    4    5    6    7    8    9   10   11   12   13   14   15
A    1              2    3    4         5    6         7         8    9
B    1    2    3                   4              5         6
P        1/2  2/3  1/4  2/5  3/6  3/7  4/8  5/9  4/10 6/11 5/12 7/13 8/14

Now,

We can get the probability of this particular sequence by multiplying the probabilities of each step. However, we should notice here that the product of the probabilities only depends on the final numbers of coins in the cornucopias and doesn't actually depend on the order in which the coins appeared! If we denote the final numbers of coins by a and b, the denominators will always take every value between 2 and a+b-1 and the numerators will take the values [1,a-1] and [1,b-1] so the final product will be

$\frac{(a-1)!(b-1)!}{(a+b-1)!}=\frac{1}{(a+b-1)\binom{a+b-2}{a-1}}$

To get the total probability that the final number of coins ends up as a and b, we should multiply this probability with the number of possible orderings $\binom{a+b-2}{a-1}$ so that the probability becomes

$P(n_A=a, n_B=b)=\frac{1}{a+b-1}=\frac{1}{n_{tot}-1}$,

which doesn't depend on the actual values of a and b, only the number of total coins! This means that every end result is actually equiprobable!

Now, the probability that either one of the cornucopias will have no more than m<n/2 coins is clearly

$P(n_A\leq m \text{ OR } n_B\leq m)=\frac{2m}{n-1}$.

If we now look at the infinite case, then the probability that one of the cornucopias will have at most some finite m number of coins is

$P(n_A\leq m \text{ OR } n_B\leq m)=\lim_{n\rightarrow\infty}\frac{2m}{n-1}=0$

Answer 2

I will use the term "urn" instead of "cornucopia," in line with Pólya's urns.

In order to show there are infinitely many coins in both urns, we just need to show that if there are currently $x$ coins in the left urn, then with probability one, there will eventually be $x+1$ coins in the left urn. So, suppose there are currently $x$ and $y$ coins in the left and right urns, respectively. Let us find the complimentary probability that every subsequent coin will be added to the right urn. This is given by the following infinite product: $$ \frac{y}{x+y}\cdot\frac{y+1}{x+y+1}\cdot\frac{y+2}{x+y+2}\cdot\frac{y+3}{x+y+3}\cdots $$ An infinite product is an infinite limit, so at this point, we need a little calculus. It can be shown that an infinite product of the form $\prod_{i=1}^\infty (1-r_i)$, where $r_i$ are real numbers between $0$ and $1$, converges to a nonzero number if and only if $\sum_{i=1}^\infty r_i<\infty$. So, we should add up one minus all of the fractions being multiplied above, which is $$ \frac{x}{x+y}+\frac{x}{x+y+1}+\frac{x}{x+y+2}+\dots $$ This sum indeed diverges, because it is $x$ times a tail sum of the Harmonic series. Therefore, the original product is zero, so the left urn will get infinitely many coins.