Consider a game where you toss $N$ coins, and let $H$ denote the number of heads. Let's say you win the game if $|H - N/2| \geq K$, i.e. if the number of heads deviates east least $K$ from what you expect, with $K < N/2$.
Now, say you are allowed to cheat a bit in this game. You get to fix the outcome of some of the $N$ coins in the following sense. First you flip $T$ separate coins. Let $T_H$ and $T_T$ be the number of heads and tails respectively. If $T_H > T_T$, you fix $T_H$ heads in the game with $N$ coins, but if $T_T > T_H$ you fix $T_T$ tail coins in the game with $N$ coins.
What is the probability of winning?
My first observation is, that one can assume that $T_H > T_T$, and focus on the probability of winning, given that $T_H$ coins were fixed to heads in the game. One then multiplies by $2$ in the end, because the cases are symmetric.
The value of $T_H$ clearly follows a binomial distribution. Now, for the probability of winning, what I've basically come up with is an expression of the form $2 \cdot \sum_{t=T/2}^T Pr(T_H=t) \cdot \text{SOMETHING}$, where SOMETHING involves another binomial distribution depending on the value of the random variable $T_H$. This is OK, because I can approximate these using the normal approximation.
My problem is: $T$ is huge, like, in the order of $2^{90}$. So there is no way I am going to be able to evaluate this sum for concrete values of $N$ and $T$, which is precisely what I want. Dear community, can you help me out?