N number of coins tossed before first head appears on all of them This was a question that was asked in my distributed network algorithms assignment
Say I have n nodes in a distributed network and at every node I'm tossing a coin. Let $X_i$ be the independent random variables following geometric distribution, where $1 \leq i \leq n$, that denotes the number of coin tosses before a heads appear. let the parameter $ p =  1/2$. Then show with high probability that $Max\{X_i | 1 \leq i \leq n\} = O(logn)$.
Added the picture of question for more clarity:

This is the partial answer that I submitted.
E[x] = 1/P = 2
P($x_i = k_i$) = $(1-p)^{k_i-1}p$ where $k_i$ denotes the number of trails before heads appear in $i^{th}$ node
substituting p= 1/2, we get
P($x_i = k_i$) = $(1/2)^{k_i}$ for $ 1 \leq i \leq n$
How to procced after this ?
 A: Here $X_i\sim\text{Geom}\left(p=\frac{1}{2}\right)$ $\forall i=1(1)n$.
Now let us turn our attention towards $\mathcal{O}$:
$$\begin{align}
&\max\{X_i|1\leq i\leq n\}=\mathcal{O}\left(\log n\right)\\
\implies &\max\{X_i|1\leq i\leq n\}\leq \text{c}\cdot\log n \hspace{1em}\text{as }n\rightarrow\infty\hspace{1em}\text{where c is a positive constant}\\
\implies &X_i\leq\text{c}\cdot\log n \hspace{1em}\forall i=1(1)n \hspace{1em}\text{as }n\rightarrow\infty
\end{align}$$
Now we can find the probability as
$$\begin{align}
P\left(X_i\leq\text{c}\cdot\log n|1\leq i\leq n\right)&=\{P\left(X_1\leq\text{c}\cdot\log n\right)\}^n\hspace{5em}\text{as all are i.i.d}\\
&=\left\{\sum_{k=1}^{\lfloor\text{c}\cdot\log n\rfloor}\left(\frac{1}{2}\right)^k\right\}^n\\
&=\left\{1-\left(\frac{1}{2}\right)^{\lfloor\text{c}\cdot\log n\rfloor}\right\}^n
\end{align}$$
Now, do you understand why with high probability?

 Take $\lim_{n\rightarrow\infty}$ on the above probability. Then $\left(\frac{1}{2}\right)^{\lfloor\text{c}\cdot\log n\rfloor}\rightarrow0$ [Why?].

 So $1-\left(\frac{1}{2}\right)^{\lfloor\text{c}\cdot\log n\rfloor}\rightarrow1$.

 The rest is trivial...

A: To add to the previous answer. Why and when does $(1-(\frac12)^{\lfloor c \log n\rfloor})^n$ converge to $1$? We can calculate
$$\Big(1-\Big(\frac12\Big)^{\lfloor c \log n\rfloor}\Big)^n = \Big(1-\frac{n0.5^{\lfloor c \log n\rfloor}}{n}\Big)^n = \Big(1+\frac{-e^{\log n}0.5^{\lfloor c \log n\rfloor}}{n}\Big)^n$$
which converges to $e$ to the power of something depending on the numerator. If numerator converges to $0$ then the expression converges to $1$ and we are done. Thus, when just ignoring the floor function we need
$$e 0.5^c < 1 \Leftrightarrow c\log 0.5 < \log (1/e) \Leftrightarrow c > 1/\log 2$$
and it follows that for $c>1/\log 2$ with high probability $\max_{i=1,\dots,n} X_i \leq c\log n$.
But we also get that the above probability converges to $0$ if $c< 1/\log 2$. This means that in this case with high probability $\max_{i=1,\dots,n} X_i \geq c\log n$.
To summarize: With high probability
$$c_1\log n \leq \max_{i=1,\dots,n} X_i \leq c_2\log n$$
for $c_1<1/\log 2$ and $c_2>1/\log 2$.
