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? 
 A: Suppose the population of the world is $k$.
Then the probability they all have at least one tail in up to $n$ flips is $$\left(1-\frac1{2^n}\right)^k \approx \exp\left(- \dfrac{k}{2^n}\right)$$ so the probability that the longest string of heads is exactly $n$  $$\left(1-\frac1{2^{n+1}}\right)^k - \left(1-\frac1{2^{n}}\right)^k \approx \exp\left(- \dfrac{k}{2^{n+1}}\right)-\exp\left(- \dfrac{k}{2^{n}}\right)$$
giving the following figures for different $n$ when $k=7.347\times 10^{9}$:
n  probability
29 0.00106637
30 0.03160525
31 0.14808332
32 0.24439811
33 0.22688430
34 0.15545051
35 0.09111492
36 0.04934329
37 0.02567859
38 0.01309898
39 0.00661543
40 0.00332433
41 0.00166633

making the most likely outcome $32$ flips as the longest string of heads worldwide.  
The corresponding median would be $33$ and the expected value would be about $33.11$.  All three are reasonably close to the order-of-magnitude estimate of about $\log_2(7.347\times 10^{9}) \approx 32.8$ 
A: 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.
