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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.

Aurelio
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    Do we know that the coins are fair? (probability of heads is $1/2$?) If the coins are not necessary fair and coins $A$ and $B$ don't even have to be unfair in the same way, then it should not be possible to solve the problem. – Zubin Mukerjee Oct 07 '17 at 15:30
  • Let's assume they are fair. However, it would be nice to see the solution also for not fair coins. Coins are certainly fair in the same way. – Aurelio Oct 07 '17 at 15:32
  • A normal approximation would have a very good fit here, though it is technically not exact. – RideTheWavelet Oct 07 '17 at 16:20
  • What have you tried specifically? Not that the question is easy or anything, I just thought it might prompt the free flow of ideas – gen-ℤ ready to perish Oct 07 '17 at 16:28
  • I can try to approximate it. But I wanted to ask mathematicians first whether there is no other (more certain) way. Tomorrow I'll try to approximate it. – Aurelio Oct 07 '17 at 16:30
  • Perhaps answered here. – BruceET Oct 07 '17 at 18:32

2 Answers2

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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}

Deep
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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).

A.G.
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