Very Heavy Coin

First, we need to find the probability we flip the first tail. The chance that we flip a head is $\frac{n^2-1}{n^2}$ , meaning that the probability of getting only heads through $p$ turns is $\prod \limits_{n=k}^{p}\frac{n^2-1}{n^2}$ . After infinite turns, the probability is $\prod \limits_{n=k}^{\infty}\frac{n^2-1}{n^2}$ . First, we will use the property $\prod \limits_{n=p}^qa_n=\exp \left(\ln \left(\prod \limits_{n=p}^qa_n\right)\right)=\exp \left(\sum \limits_{n=p}^q\ln \left(a_n\right)\right)$ to make the product into the form $\sum \limits_{n=2}^{\infty}\ln \left(\frac{n^2-1}{n^2}\right)$ . Using properties of logarithms, this simplifies to $\exp\left(\sum \limits_{n=2}^{\infty}\left(\ln \left(n+1\right)+\ln \left(n-1\right)-2\ln \left(n\right)\right)\right)$ . This is a telescoping sum which converges to $\exp\left(\ln\left(1\right)-\ln\left(2\right)+\lim \limits_{n\to\infty} -\ln\left(n\right)+\ln\left(n+1\right)\right)=\exp\left(\ln\left(\frac{1}{2}\right)\right)=\frac{1}{2}$ .

This means that there is a $1-\frac{1}{2}=\frac{1}{2}$ chance we flip at least one tail.

The probability of flipping a second tail once we have flipped the first is $\prod \limits_{n=3}^{\infty}\frac{n^2-1}{n^2}$ . Using the same strategy as the first, we see this is equal to $\frac{2}{3}$ , and more generally, $\prod \limits_{n=k}^{\infty}\frac{n^2-1}{n^2}=\frac{k-1}{k}$ .

This means the probability of getting exactly $a$ tails is $\frac{1}{\left(a+1\right)!}\cdot\frac{a+1}{a+2}=\frac{a+1}{\left(a+2\right)!}$ .

By multiplying the previous expression by $a$ and summing it from $0$ to $\infty$ , we get the average number of tails flipped. This is equivalent to:

$\sum \limits_{i=2}^{\infty}\left(\frac{\left(i-1\right)\left(i-2\right)}{k!}\right)$ by substituting $i=a+2$ to make the denominator $i!$

By expanding and splitting the sum, we get $\sum \limits_{i=2}^{\infty}\frac{i^2}{i!}-3\sum \limits_{i=2}^{\infty}\frac{1}{\left(i-1\right)!}+2\sum \limits_{k=2}^{\infty}\frac{1}{i!}$

The second and third sums can be quickly evaluated as $e-1$ and $e-2$ , respectively, leaving only the first sum. Substituting $j=i-1$ , the sum becomes $\sum _{j=1}^{\infty}\frac{j+1}{j!}$ . Splitting this apart and evaluating, we get $\sum \limits_{i=2}^{\infty}\frac{i^2}{i!}-3\sum \limits_{i=2}^{\infty}\frac{1}{\left(i-1\right)!}+2\sum \limits_{k=2}^{\infty}\frac{1}{i!}=\left(e\right)+\left(e-1\right)-3\left(e-1\right)+2\left(e-2\right)=e-2=\boxed{0.7182}$