Coin with claim#1: fair, claim#2: P(head)=0.6, refute at least one with 99% chance I've encountered the following problem, and would like to receive some help:
We have a coin. $A$ claims it's a $fair$ one, $B$ claims it has a $60\%$ probability of getting a $head$.
What is the minimum number of trials that should refute at least one of the claims with $99\%$ chance?
Let $P(head)=p$ to simplify things.
I calculated the expected values and variances for both cases but my main problem is that we don't even know whether $p \in \{0.5;0.6\}$  is true or not. Anyway, I tried using Chebyshev's inequality but didn't get anywhere.
 A: You need various assumptions including a normal approximation, but if you assume for simplicity that the standard deviation of the sample mean is $1/\sqrt{4n}$ (it may be less than that, so this is slightly conservative) and that you want the distance between $0.5$ and $0.6$ to be at least twice $2.576$ standard deviations (since $\Phi(2.576) \approx 0.995002$, which is what you want if you are using a two-sided test to reject) then you need to solve $$2 \times 2.576 \times \frac{1}{\sqrt{4n}} \ge 0.6-0.5$$ which will give you $n \ge (25.76)^2 \approx 664$.   
You could get a slightly lower figure if the individuals were confident about their claims. Suppose you took a sample size of $537$ and said that A is the winner if $295$ or fewer heads occurred and B was the winner if $296$ heads or more occurred. Then [using an exact binomial distribution]  A would expect to win with probability over $0.9901$ and B would expect to win with probability over $0.9903$, which they may both see as good enough.  Something similar is not possible with a sample size of $548$, but is possible with a sample size of $549$ or more.  This is analogous to using one-sided confidence intervals.
