Flip coin Problem Combinatorics Let's say we are flipping a coin $n$ times. What is the probability that we always have more heads than tails. Meaning that if we are counting the number of times we have had heads and the number of time we have had tails, what is the probability that throughout the $n$ throws we continuously have more or equal heads.
 A: If, at each flip there must be at least as many heads as tails then the required probability at $n$ flips is given by:
$$\Pr(\text{# Hs always }\ge\text{ # Ts})=\frac{1}{2^n}\binom{n}{\lfloor n/2 \rfloor}\tag{Answer}$$
This follows directly from summing the terms in Catalan's Triangle for constant $n=N+k$:
$$C(N,k)=\binom{N+k}{k}-\binom{N+k}{k-1}$$
This is the number of monotonic lattice paths starting at the origin with $k$ vertical steps and $N$ horizontal steps with $N\ge k$, in other words: those that don't exceed the diagonal. 
There is a natural bijection between successful outcomes (never more tails than heads in $n$ coin flips) and total monotonic lattice paths starting at the origin, having $n=N+k$ steps and that remain weakly below the diagonal. 
The bijection involves mapping horizontal steps to heads and vertical steps to tails. 
The required sum is: 
$$\sum_{k=0}^{\lfloor n/2 \rfloor} C(n-k,k)=\sum_{k=0}^{\lfloor n/2 \rfloor} \left(\binom{n}{k}-\binom{n}{k-1}\right)$$
which is telescopic i.e. all terms cancel except $\binom{n}{\lfloor n/2 \rfloor}$. These paths represent our set of successes out of $2^n$ total possible coin flip strings, hence the result leading to the answer above:
$$\text{successes}=\sum_{k=0}^{\lfloor n/2 \rfloor} C(n-k,k)=\binom{n}{\lfloor n/2\rfloor}$$
