Flip the coin yourself

As each person is independent of the others we can consider each set of tosses separately. We refer to tossing a $\text{head}$ as $H$ and $\text{tail}$ as $T$

There are two ways of ending up with a $\text{H}$ . You can either go $\text{H}$ or $\text{TH}$ :

$P(H)+P(TH)=\dfrac{1}{2}+\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{3}{4}$

We now use this to find the probability of $6$ people ending up on $\text{heads}$ :

$\left ( \dfrac{3}{4} \right)^6=\dfrac{729}{4096} \implies x=729,y=4096 \implies x+y= \boxed{\boxed{4825}}$

Consider each coin individually. The probability for a given coin to end as tails requires the first flip to be tails, as well as the second flip to be tails, which occurs with probability $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ . Therefore the probability of ending with heads is $1 - \frac{1}{4} = \frac{3}{4}$ . Since there are six coins, the probability for all coins to be heads is $\left(\frac{3}{4}\right)^6 = \frac{729}{4096}$ , so the answer is $729 + 4096 = \boxed{4825}$

We may also solve it by composition of ordinary generating functions:

Let $P(x)$ indicate the g.f. for a coin flip, where x represents the formal variable for a tail.

$P(x) = \left(\frac{1}{2}+\frac{1}{2} x\right)$ Since only the people who get a tail toss again, we can take the composition, and raise to power 6 for six people:

$P(P(x))^6 = \frac{1}{2^6} \left(1+\left(\frac{1+x}{2}\right)\right)^6$ and since all heads is the constant term, answer is $\dfrac{3^6}{2^{12}}$