Longest, First Attempt, Chain of Heads on a Coin Flip So taking the question of :
If everyone in the world flipped a coin until they got tails what is the most likely longest chain of heads?
Assuming population is 7.347*10^9
And everyone has an ideal coin and has the ability to flip a coin "randomly"
What would the answer be?
: thinking that out of two people one is likely to get heads, I applied the reasoning to larger quantities dividing the Population till I reached the smallest number <1 making it 32. 
Also how would you solve using compound distribution? 

The time $t_f$ to first tail for one person is distributed geometrically:
$$
P(t_f = \tau) = 2^{-\tau}
$$
For example, the probability that Mary will get a first tail on her second flip (thus a chain of one head) is $2^{-2}=\frac14$.
So your question asks about the distribution of the maximum of $N$ geometric variates, with $N$ large.  It is worth noting that for a singe person and a given chain length $L$, 
$$
P(t_f \geq L) = 2^{-L} \\
P(t_f = L) = 2^{-(L+1)} \\
P(t_f \leq L) = 1-2^{-(L+1)} \\
$$
Then the probability that the maximum chain is less than or equal to $L$, that is, the probability that all $N$ people will have chains less than or equal to $L$ is 
$$
\prod_{i=1}^N \left(  1-2^{-(L+1)} \right) = \left(  1-2^{-(L+1)} \right)^N
$$
If $N$ and $2^{L+1}$ are large, this is well approximated by
$$
\prod_{i=1}^N \left(  1-2^{-(L+1)} \right) 
= \left[ \left(  1-2^{-(L+1)} \right)^{2^{L+1}}\right]^{\frac{N}{2^{+1}}}
\approx e^{-{\frac{N}{2^{L+1}}}}
$$
The peak value in the p.d.f. will be the maximum in the slop of this c.d.f; that is, we need to solve for the $L$ where the second derivative is zero.
$$
\frac{d^2\left(e^{-{\frac{N}{2^{L+1}}}} \right)}{dL^2}= 
N(N-2^{L+1}) (\log 2)^2 e^{-2^{-(L+1)}N}2^{-2(L+1)}
$$
This is zero when $2^{L+1} = N$ or $L = \log_2N-1$ or with your $N$, $L=31.77$.
The peak will thus occur at a length of 32 heads.
