# What is the probability that an unfair coin's head appears less than 50 after 100 tosses?

I met a question about probability, it seems easy but I got stuck. The question is:

Suppose there is an unfair coin, the HEAD probability is $$p=0.7$$.
(Q1) If we toss the coin for 100 times, what is the expectation and the variance of this experiment?
(Q2) Answer with reason whether or not the probability is higher than $$1/10$$ that the number of HEAD appear times is less than $$50$$ as we toss the coin for $$100$$ times.

Q1 is easy, I know expectation is $$n*p=70$$ and variance is $$n*p*(1-p)=21$$. But for Q2 I have no idea.
At first I think it looks like... a sample distribution of sample mean used in statistics but... I don't know whether (or how) it will obey a normal distribution.
Then I also try to calculate the sum of $$P(H=0)+P(H=1)+...+P(H=50)$$, but the work is huge, even I use an approximation of Passion distribution...
So could you share some of your thought? Thank you!

• Well...you can use the normal approximation if you want, but just speaking roughly: We have $\sigma=\sqrt {21}=4.83$ so you are talking about a $4.36\sigma$ event, so the answer is effectively $0$. For that matter, doing it exactly, with the binomial distribution, isn't especially difficult either. – lulu Aug 21 at 11:09
• Another fast way to do it is to note that the answer is clearly less than $50\times P(50)$ and even that is way less than $.1$ – lulu Aug 21 at 11:13
• Thank you @lulu! I think both of these two methods you proposed are effective, but as you said 'with the binomial distribution, isn't especially difficult either', do you mean it isn't such difficult even we calculate the $P(H<50)$? If so, could you give me a rough procedure of it? – Peter Nova Aug 21 at 11:25
• Just use a spreadsheet or a program. here is Wolfram Alpha's version. – lulu Aug 21 at 11:27
• @PeterNova Here is the computation with Hypergeometrics Wolframalpha - Hypergeometrics – InterstellarProbe Aug 21 at 20:56

We can show that the answer to Q2 is "No" even without appealing to the Central Limit Theorem.

Let's say $$H$$ is the total number of heads. By the Chebyshev inequality (see below), $$P(|H-70| \ge 21) \le \frac{21}{21^2} \approx 0.048$$ But $$P(|H-70| \ge 21) = P(H \le 49) + P(H \ge 91)$$ so $$P(H \le 49) \le P(|H-70| \ge 21) \le 0.048$$

Chebyshev's inequality: If $$X$$ is a random variable with finite mean $$\mu$$ and variance $$\sigma^2$$, then for any value $$k>0$$, $$P(|X-\mu| \ge k) \le \frac{\sigma^2}{k^2}$$

Since $$H$$ is a sum of Bernoulli random variables, then $$H$$ is modeled as Binomial with $$N = 100$$ trials and $$p=0.7$$ probability of success. Indeed, it is exhaustive to compute the desired probability which is $$P(H < 50) = P(H=0) + \ldots P(H=49)$$

Instead what you could do is notice that $$N =100$$ is large enough, and hence we could approximate the Binomial with a Gaussian distribution, so $$H \sim N(Np,Np(1-p)) = N(70,21)$$ So $$\Pr(H<50) \simeq \Pr(\underbrace{\frac{H-70}{\sqrt{21}}}_{Z} < \frac{50-70}{\sqrt{21}}) \simeq \Pr(Z < -4.36)$$ where $$Z$$ is a centered Gaussian. According to the z-table, the above probability is way less than $$\frac{1}{10}$$.

• Thank you very much @Ahmad Bazzi ! I should think about this... I only thought this experiment should approximate this binomial to a possion distribution... Can I ask a stupid question that when can we approximate the binomial with a gaussian distribution? I guess like when experiment times n=10 or p=0.99 we could not do that, right? – Peter Nova Aug 21 at 11:29
• You are right, when we say $N$ is large, it should be compared to a threshold to be able to say "it is large enough". A rule of thumb is when $N > 9 \max(\frac{1-p}{p},\frac{p}{1-p})$. @PeterNova – Ahmad Bazzi Aug 21 at 11:58

Here's an argument that doesn't rely on approximating a normal distribution. The basic idea is to show that

$$\sum_{k=1}^{49}{100\choose50-k}p^{50-k}q^{50+k}\lt{1\over10}\sum_{k=0}^{50}{100\choose50+k}p^{50+k}q^{50-k}$$

where $$p=0.7$$ and $$q=1=p=0.3$$

Now $$(q/p)^2=9/49$$, so we have $$(q/p)^{2k}\lt1/10$$ for $$k\gt1$$. Since $${100\choose50-k}={100\choose50+k}$$, it follows that

$${100\choose50-k}p^{50-k}q^{50+k}\lt{1\over10}{100\choose50+k}p^{50+k}q^{50-k}$$

for $$2\le k\le50$$. It would be nice if $${100\choose49}p^{49}q^{51}$$ were less than $${1\over10}\left({100\choose50}p^{50}q^{50}+{100\choose51}p^{51}q^{49}\right)$$; unfortunately, it isn't. However, it suffices to show that

$${100\choose48}p^{48}q^{52}+{100\choose49}p^{49}q^{51}\lt{1\over10}\left({100\choose50}p^{50}q^{50}+{100\choose51}p^{51}q^{49}+{100\choose52}p^{52}q^{48} \right)$$

which simplifies to showing

$$50\cdot49q^4+52\cdot50pq^3\lt{1\over10}\left(52\cdot51p^2q^2+52\cdot50p^3q+50\cdot49p^4 \right)$$

This can be done with a direct calculation, but can also be verified with a couple further simplifications:

$$50\cdot49q^4+52\cdot50pq^3\lt50\cdot52(q^4+pq^3)=50\cdot52q^3(q+p)=50\cdot52q^3=70.2$$

and

$${1\over10}\left(52\cdot51p^2q^2+52\cdot50p^3q+50\cdot49p^4 \right)\gt{50\cdot49\over10}(p^2q^2+p^3q+p^4)=245p^2((p+q)^2-pq)=245p^2(1-pq)=94.8395$$

(There are undoubtedly additional ways to reduce the final amount of explicit computation for the comparison. I would appreciate any suggestions.)

Another way to do this is:

$$\sum_{n=0}^{49}\dbinom{100}{n}(0.7)^n(0.3)^{100-n} = 1-\dbinom{100}{50}(0.7)^{50}(0.3)^{50}{{_2}F_1\left(1,-50;51;-\dfrac{7}{3}\right)}\approx 10^{-5}$$