Two symmetrical coins - the difference of the tails We have two symmetric coins - A and B.
We throw a coin A and a coin B 1000 times (per coin).
What is the probability that the difference between the number of tails on coin A and the tails on coin B will be at least 100?
Coin throws are independent.
 A: Let $n_A,n_B,$ be the number of Tails on coins $A$ and $B$, and $p_A,p_B$ their probabilities respectively. Each coin is tossed $N$ times, and we want the probability of difference in number of Tails on $A$ being greater than number of Tails on $B$ by $\alpha\geq 0$, i.e. $P(n_A-n_B\geq\alpha)$. To fulfill the required condition, for any given value of $n_A$ such that $\alpha\leq n_A\leq N$, we must have $0\leq n_B\leq n_A-\alpha$. Since coin tosses are independent:
\begin{align}
P(n_A-n_B\geq\alpha) & = \sum_{n_A=\alpha}^{N}\sum_{n_B=0}^{n_A-\alpha}P(n_A)P(n_B)\\
& = \sum_{n_A=\alpha}^{N}\sum_{n_B=0}^{n_A-\alpha}C_{n_A}^Np_A^{n_A}(1-p_A)^{N-n_A}~C_{n_B}^Np_B^{n_B}(1-p_B)^{N-n_B}
\end{align}
Due to symmetry:
\begin{align}
P(|n_A-n_B|\geq\alpha) =\begin{cases}P(n_A-n_B\geq\alpha)+P(n_B-n_A\geq\alpha),&\quad\textrm{if }\alpha\neq 0\\
P(n_A-n_B\geq0)+P(n_B-n_A\geq0)-P(n_A-n_B=0),&\quad\textrm{if }\alpha=0
\end{cases}
\end{align}
in which
\begin{align}
P(n_A-n_B=0)&=\sum_{n_A=0}^{N}P(n_A)P(n_B=n_A)\\
& =\sum_{n_A=0}^{N} C_{n_A}^Np_A^{n_A}(1-p_A)^{N-n_A}~C_{n_A}^Np_B^{n_A}(1-p_B)^{N-n_A}\\
& =\sum_{n_A=0}^{N} \left( C_{n_A}^N\right)^2 (p_Ap_B)^{n_A}[(1-p_A)(1-p_B)]^{N-n_A}
\end{align}
A: Here is a hint. Suppose you throw both coins once and let $X_1$ be the random variable that represents how many more tails coin A yields. Then
$$
P(X_1=0) = 1/2,\quad P(X_1=1)=1/4\quad P(X_1=-1)=-1/4.
$$
More generally, if $X_i$ is how many more tails coin A yields after $i$ trials, then $X_0=0$ and
$$
P(X_i=X_{i-1}) = 1/2,\quad P(X_i=X_{i-1}+1)=1/4\quad P(X_i=X_{i-1}-1)=-1/4.
$$
(that is, $X_i-X_{i-1}$ has the same distribution as $X_1$).
This suggests a random walk and you want 
$$
P(|X_{1000}|\geq 100)
$$
Now you can google random walks for the computations (they use recursion).
