Straight up unfair

We want to find the sum of the probabilities of getting tails on a multiple of three:

$Pr(\text{tails on 3rd flip})+Pr(\text{tails on 6th flip})+Pr(\text{tails on 9th flip})+\dots=(\frac{3}{5})^2\frac{2}{5}+(\frac{3}{5})^5\frac{2}{5}+(\frac{3}{5})^8\frac{2}{5}+\dots$ .

This is a geometric series whose initial term is

$a_0=(\frac{3}{5})^2\frac{2}{5}$

and whose ratio is

$r=(\frac{3}{5})^3$ .

So the series sums to $\frac{a_0}{1-r}=\frac{9}{49}$ , and our answer is $9+49=58$ .

This is a geometric distribution with the single flip probability of success $p=\frac{2}{5}$ . A successful flip is a multiple of 3, so the number of failures before the first success is $k = 3 n-1$ , where n = 1, 2 ... $\infty$ . Summing up the probabilities of all successful flips gives the total probability we are looking for:

$Pr(n)=\sum _{n=1}^{\infty } p (1-p)^{3 n-1}$

Plugging in the numbers gives the probability of $\frac{9}{49}$ , so the answer is 9+49= $\boxed{58}$

The probability that you obtain tail during the third attempt is $\frac{18}{125}$ . From now on interesting results occur when your probability is multiplied by a power of $\frac{27}{125}$ , because in this way your $N$ will be a multiple of $3$ . In order to obtain such a probability you notice that this is a geometric series, with ratio $\frac{27}{125}$ . So the solution to the geometric series, obtained by using the geometric series formula, is $\frac{1}{1-\frac{27}{125}} = \frac{125}{87}$ . Now you have to multiply this result with the first you obtained, so $\frac{18}{125} \times \frac{125}{87} = \frac{9}{49}$ . Thus the solution is $\boxed{49}$