Formula for Heads or tails task Someone offers you this:
"We toss a coin and we pick heads or tails. Whenever you are the winner - I give you \$1.1 (one dollar and ten cents).   Whenever I am the winner - you give me \$1 (one dollar).  However, we are obliged to toss exactly 300 times." 
Now the question is the following: what is the probability of you ending up with a loss (no matter big or small) after the 300 tosses and what is the formula to solve this?
Thanks in advance!
 A: Some hints to get you started: 
If I win $k$ times and you win $300-k$ times, then my net profit in dollars is $1.10k - 1.00(300-k)$. 
How big/small does $k$ need to be for this profit to be negative? 
The number of tosses that I win is a binomial distribution with $n = 300$ trials and probability of success $p = \dfrac{1}{2}$. Hence, the probability of me winning exactly $k$ times is $\dbinom{300}{k}\cdot 2^{-300}$. 
Since summing $\dbinom{300}{k}\cdot 2^{-300}$ over several values of $k$ is probably not an easy calculation, you might want to use the fact that for large $n$, the binomial distribution is close to a normal distribution with mean $np$ and standard deviation $\sqrt{np(1-p)}$. 
A: First of all, it does not matter whether we pick to win on heads or tails, as we (assume) the coin is unbiased and at each round it comes heads or tails equiprobably. Hence, without loss of generality, let us win on tails. 
Every time the coin comes tail, we get a payoff of $a = 1.1$ dollars. Every time we get a head we pay $b = 1$ dollar. 
Let $N_{T}$ be the number of tails in the sequence of $M=300$ tosses.
We end up with a loss exactly when
$$
a \cdot N_{T} < b \cdot (M-N_{T})
\quad
\Leftrightarrow
\quad
N_{T} < \frac{b}{a+b} M.
$$
What is the probability that $N_{T} = k$?
$$
P(N_{T} = k) = \binom{n}{k} \frac{1}{2^{k}}\frac{1}{2^{n-k}} = \binom{n}{k} \frac{1}{2^{n}}
$$
and hence the probability that $N_{T}  < \frac{b}{a+b} M$ is
$$
P(N_{T} < \frac{b}{a+b}M) 
= 
\sum_{i=1}^{\lfloor \frac{b}{a+b}M\rfloor}\binom{n}{i} \frac{1}{2^{n}}
=
 \frac{1}{2^{n}}\sum_{i=1}^{\lfloor \frac{b}{a+b}M\rfloor}\binom{n}{i}.
$$
