Two coins, each with P(Head)$\to 0$ as $n\to\infty,$ where $n=n-$th coin toss. Under what conditions is P(both coins Head) infinite times $<1?$

Question

Suppose we have two coins. The probability that the first coin lands on heads is $p_1(n)$ and depends on $n,$ the $n-$th coin toss of this first coin. Similarly, the probability that the second coin lands on heads is $p_2(n)$ and depends on $n,$ the $n-$th coin toss of this second coin. All coin tosses are mutually independent.

If $\displaystyle\lim_{n\to\infty} p_1(n) > 0 $ and $\displaystyle\lim_{n\to\infty} p_2(n) > 0,$ then it is somewhat clear that if we toss the coins an infinite number of times, then the probability that the number of coin tosses whose result is that both coins land on heads (e.g. both coins land on heads on the third toss, both coins land on heads on the third toss $17$th toss, etc) is infinite, is $1.$

Suppose $\displaystyle\lim_{n\to\infty} p_1(n) = 0 = \lim_{n\to\infty} p_2(n).$

It is (somewhat?) clear that if both functions converge to $0$ very slowly, then the result in the previous paragraph remains.

How quickly do $p_1$ and $p_2$ have to tend to $0$ (/what are some conditions on $p_1,p_2$) so that the probability that the number of coin tosses whose result is that both coins land on heads is infinite, is less than $1?$

Answer 1

I show the following satisfying result.

CLAIM: Let $S=\sum_{n=0}^\infty p_1(n)p_2(n)$.

• If $S<\infty$, then the probability of having infinitely many tosses where both coins show heads is zero.
• If $S=\infty$, then the probability of having infinitely many tosses where both coins show heads is one.

Let $E_n$ be the event that both the first and the second coin show heads in the $n$-th toss (of both coins simultaneously). Then we have $\mathbb P(E_n)=p_1(n)p_2(n)$. So, in the first case we have $\sum_{n=0}^\infty \mathbb P(E_n)<\infty$. By Borel-Cantelli the probability that infinitely many of the $E_n$ occur is zero, meaning the probability that infinitely many tosses show two heads is zero.

For the second case we have $\sum_{n=0}^\infty \mathbb P(E_n)=\infty$. Now, we use that the events $E_n$ are mutually independent and the second (or converse) Borel-Cantelli to obtain that the probability that infinitely many of the $E_n$ occur is one, which means that the probability that infinitely many tosses show two heads is one.

This completes the proof. This phenomenom that the probability is either zero or one is quite common. If you want to dive deeper into the subject, take a look at Kolmogorov and tail events.