How can a problem be changed such that it forms a poisson rather than a binomial distribution?

Question

Say I have a nonbias die with sides1-200. Let X be the r.v counting the number of times 200 is rolled out of 500 trials. So, $X \approx bin(500,0.005)$.

How can one change this problem such that we actually need a poisson distribution to model the probability situation? What is the random variable whose distribution I'm looking at? How will the poisson distribution also be looking at the probability of success? I simply don't know how to create an analogy between it and the binomial distribution.

[Note: I am aware that the poisson distribution can be used to approximate the binomial distribution in scenarios such as the one pointed out above, since the number of trials and the probability of success is quite small. However, this is not my question.]

Someone please help.

Answer 1

As you said, for situations when $n$ is large and $p$ is small we can approximate the binomial distribution with a poisson distribution.

So let $Y \approx poi(\lambda)$ where $\lambda = np$ (the $np$ from your binomial distribution) where $n$ is the sample size of 500 and $p$ is the probability that we roll a 200 which is $1/200 = 0.005$ as you pointed out.

Hence, $$\lambda = 500\left(\frac{1}{200}\right) = 2.5$$ So if we wanted to compute the probability that 200 is rolled once out of the 500 times, we would need to compute $$P(Y = 1) = \frac{e^{-\lambda} \lambda^{1}}{1!} = \frac{e^{-2.5} 2.5^{1}}{1!} = \frac{2.5}{e^{2.5}}$$

We can generalize this with, $$P(Y = i) = \frac{e^{-\lambda}\lambda^{i}}{i!} = \frac{e^{-2.5}2.5^i }{i!}$$

Answer 2

It can be shown algebraically that the binomial distribution can be approximated by the poisson distribution if $n$ is large, $p$ is small and $\lambda=n\cdot p$

$\binom{n}{x}\cdot p^x\cdot (1-p)^{n-x}$

$=\frac{n!}{(n-x)!\cdot x!}\cdot \left(\frac{p}{1-p} \right)^x\cdot (1-p)^n$

Replacing $p$ by $\frac{\lambda}{n}$ and expanding $\frac{p}{1-p} $ by $n$

$=\frac{n!}{(n-x)!\cdot x!}\cdot \left(\frac{\lambda}{n-\lambda} \right)^x\cdot \left(1-\frac{\lambda}{n}\right)^n$

$$=\underbrace{\frac{n}{n-\lambda}}_{\to 1}\cdot \underbrace{\frac{n-1}{n-\lambda}}_{\to 1}\cdot \underbrace{\frac{n-2}{n-\lambda}}_{\to 1}\cdot \ldots \cdot \underbrace{\frac{n-x+1}{n-\lambda}}_{\to 1}\cdot \frac{\lambda^x}{x!}\cdot \underbrace{\left(1-\frac{\lambda}{n}\right)^n}_{\to e^{-\lambda}}$$

for $n\to \infty$

For a sufficiently good approximation $n$ must not go to infinity.