# Toss coin probabilities (Bino vs Gauss approx)

1. May 24, 2015

### ChrisVer

Hi, I calculated the probability to this scenario:
getting between 3 and 6 Heads after tossing a coin for n=10 trials...

The binomial probability for this is:
$P(3 \le k \le 6) = \sum_{k=3}^6 Bi(k;p,n)= \sum_{k=3}^6 \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} =\sum_{k=3}^6 \frac{10!}{k!(10-k)!} (0.5)^{10} = 0.7734375$

Whereas for the Gaussian approximation to the binomial:
$G(x) = \frac{1}{\sqrt{2 \pi n p (1-p) }} \exp\Big( - \frac{ (x-np)^2}{2np(1-p)} \Big)$

I get:
$P(3 \le x \le 6) = \int_3^6 dx~ G(x) = ... = 0.633505$

My question is why are the two probabilities so off from each other? Is it because $n$ cannot be considered as "large" ? I think when extracting the Gauss approximation, the additional higher orders go as $\frac{1}{n}= 0.1$ but that's for the gaussian distribution $G$ and not the integral of it.
Any feedback?

2. May 24, 2015

### Simon Bridge

3. May 24, 2015

### ChrisVer

So the difference is mainly due to the fact that $np= n(1-p)=5$ is not large enoguh? So Gauss can go a bit off in its predictions (here ~10%)...

The strange thing is that this is the only rule the approximation break, whereas the other two rules give:
$\Big| \frac{1}{\sqrt{n}} \Big( \sqrt{\frac{q}{p}} -\sqrt{\frac{p}{q}} \Big) \Big| = 0 <0.3$
and
$np \pm 3 \sqrt{npq} \in [0.26 , 9.74 ]$ which is in [0,n=10]...

For the calculations I try to build my own source codes just for a practice (or check with the formulas in wolframalpha), and so far my numerical methods work fine.

4. May 30, 2015

### MrAnchovy

You are forgetting the continuity correction: $$P(3 \le x \le 6) \approx \int_{2.5}^{6.5} G(x)~dx \approx 0.7717$$