Relating the binomial probability distribution to the Poisson Distribution in an example Reading the textbook Mathematical Statistics and Data Analysis 3rd ed, by Rice. I've come up on an example that I'm trying to extend beyond the text:

So I am trying to obtain one of the stated Poisson probabilities, but using the binomial distribution instead. I'm not sure if I am interpreting things right to get my stated goal. For instance let's take trying to get $\text{Number of Deaths} = 0$. From the Poisson Probability this is given as $0.543$.
With the given information I am able to calculate a "probability" but I'm not sure what it means:
$$np = \lambda \\ \Rightarrow p = \frac{\lambda}{n}$$
So we know that $n = 200$ and $\lambda = 0.61$, meaning
$$p = \frac{0.61}{200} = 0.00305$$
I took this as meaning the "probability of dying from horse kick". Here is where I get stuck trying to convert the problem into a binomial distribution problem. I could see framing things in terms of deaths -no deaths and that may possibly look like:
$$\binom{200}{109}(0.00305)^{109}(0.99695)^{91}$$
But how would I go about things if I wanted to get  1 death, 2 deaths,...etc? How could I frame things to get the same (or close to) Poisson probabilities stated but with a binomial distribution instead ?
 A: The random variable to which Bortkiewicz attributes the Poisson distribution with expected value $0.61$ is the number of such deaths in each corps in each year. Thus if $n$ is the number of soldiers in each corps and $p$ is the probability that  a soldier is killed in this way during a year, then $np=\lambda = 0.61.$ So let $X$ be the number of such deaths in a particular corps in one year. Then we have
\begin{align}
& \Pr(X=3) = \binom n 3 p^3(1-p)^{n-3} \\[10pt]
= {} & \frac{n(n-1)(n-2)}{3!} p^3 (1-p)^{n-3} \\[10pt]
= {} & \frac{\big(np\big)^3 }{3!}\cdot {} \underbrace{ \frac{n(n-1)(n-2)}{n^3} \cdot \left( 1-\frac\lambda n \right)^{-3} }_\text{These approach 1 as $n\,\to\,\infty$} {} \cdot \left( 1 - \frac \lambda n \right)^n \\[12pt]
\to {} & \frac{\lambda^3}{3!}\cdot 1 \cdot 1 \cdot e^{-\lambda} = \frac{0.61^3 e^{-0.61}}{3\cdot2\cdot1} \quad \text{as } n\to\infty.
\end{align}
A: A binomial distribution with $n=200$ and $p=0.00305$ measures the number of "successes" in 200 independent trials, each with a probability of "success" of 0.00305.  If you want "success" to be "death" and "trial" to be "corps-year", you have a bit of a problem.  For each of these 200 corps-year "trials", each one either succeeds (one death "success", singular) or fails (no deaths), so you can't model multiple deaths (successes) per corps-year (trial).
If you really want to apply the binomial distribution here, you probably want to think of a "trial" as a person-horse encounter with a per-encounter probability of getting kicked to death equal to $p$.  How many encounters are there over 200 corps-years?  Well, we don't know, but it's probably a lot.  Let's suppose that there are $n=34000$ person-horse encounters in 200 corps-years.  What's the probability that a single encounter leads to a kick death?  Well, from the data above, there were $65\times1+22\times2+3\times3+1\times4 = 122$ deaths, so that's $p=122/34000=0.00359$.
Now, how do we use this $Binom(n=34000,p=0.00359)$ to get the probabilities we want?  Well, the number of kick deaths per corps-year will be distributed with binomial distribution where $p$ is the same as before but $n$ is the number of person-horse encounters per corps-year.  Since 34000 was the 200 corps-year total, the number of encounters per corps-year was $n=34000/200=170$.
Now, since the number of kick deaths $X$ in a year has distribution $Binom(n=170, p=0.00359)$, we can calculate:
\begin{align}
P(X=0) &= (1-0.00359)^{170} = 0.543 \\
P(X=1) &= 170(0.00359)(1-0.00359)^{169} = 0.332 \\
P(X=2) &= \left(170 \atop 2\right)(0.00359)^2(1-0.00359)^{168} = 0.101
\end{align}
How did I know that the right number of person-horse encounters to assume was 34000?  I didn't.  The number doesn't really matter.  Pick something else reasonably "big" (like 5000, say), and redo the math.  You'll get roughly the same answers.
Note: Taking a "trial" to be a person-horse encounter was also pretty arbitrary.  If you prefer, define a "trial" as a single person (who can only die once) or as a horse (who, if it kills a person, is probably going to be euthanized and won't get a chance to kill again).  Any unit that can result in either zero or one deaths such that separate units can be considered reasonably independent will do.
