Question

 How many times do you need to toss a fair coin to get 100 heads with probability 90%?

Answer 1

Since a fair coin is considered, the probability of getting a head is $\frac12$ and the probability of getting a tail is $\frac12$. Assume that $X$ is a random variable, which denotes a number of heads after $n$ tosses. It has a binomial distribution. The aim is to find such $n$ that $P(X=100)=0.9.$ Since $X$ has a binomial distribution, we get: $P(X=100)=C_{n}^{100}(\frac12)^{n}$, where $C_n^{100}=\frac{n!}{100!(n-100)!}=\frac{(n-99)(n-98)\ldots n}{100!}$.

Denote $f(n)=\frac{(n-99)(n-98)\ldots n}{100!}(\frac12)^{n}$. The function is considered for $n\geq100$. We point out that $f(n)>0$. Consider $\frac{f(n+1)}{f(n)}=\frac{(n-98)(n-97)\ldots (n+1)}{(n-99)(n-98)\ldots n}\cdot\frac12=\frac12\cdot\frac{n+1}{n-99}$.Consider inequalities: $\frac12\cdot\frac{n+1}{n-99}<1\Longleftrightarrow n+1<2n-198\Longleftrightarrow n>199$. Thus, $f(n+1)<f(n)$ for $n>199$. It means that $f(200)>f(201)>f(202)\ldots$

$\frac12\cdot\frac{n+1}{n-99}>1\Longleftrightarrow n+1>2n-198\Longleftrightarrow n<199$. Thus, $f(n+1)>f(n)$ for $n<199$. It means that $f(199)>f(198)>f(197)>\ldots$. Thus, it is enough to compute $f(199)$ and $f(200).$

$f(199)\approx0.0563$ and $f(200)\approx0.0563$. Thus, the probability of getting $100$ heads is always less than $6\%$ irrespectively of the number of tosses.