# The Mathematical Ninja and the Poisson Distribution

“What are the ch…”

“About 11.7%,” said the Mathematical Ninja. “Assuming $X$ is drawn from a Poisson distribution with a mean of 9 and we want the probability that $X=7$.”

“That’s a fair assumption, sensei,” pointed out the student, “given that that’s what the sodding question says.” A wiser student may not have pushed their luck. “So, where does 12% come from?”

“It’s simple,” said the Mathematical Ninja. “You know, of course, that $P(X=n) = e^{-\lambda} \frac{\lambda^n}{n!}$?”

“But of course! I remember it because it’s the $n$th term of a Maclaurin series for $e^{-\lambda} e^{x}$ evaluated at $x=\lambda$.”

Even the Mathematical Ninja doesn’t like a smart-arse. The student’s interjection was ignored. “So, we need $e^{-9} \frac{9^7}{7!}$.”

“If you were a normal tutor, you’d say ‘put it into a calculator.’ But…”

“But.”

“Let’s see,” said the student. “I know $e^3$ is about 20, so $e^9$ is about 8,000.”

“An adequate start, although an expert would note the error in $e^3$ is about half a percent, so $e^9$ is around 8,100. *Actually*.”

The student said nothing about the Mathematical Ninja’s ego, correctly reasoning that there would be no good outcome to this gambit. “I don’t know $9^7$. But I *do* have $9^2$ on the bottom from your, uh, excellent correction. I can make it $\frac{1}{100} \times \frac{9^5}{5040}$.”

An infinitesimal nod of approval.

“And there’s even a factor of 9 left on the bottom - I’ve got $\frac{9^4}{56,000}$. And I *do* know $9^4$, it’s 6,561.”

“Mhm.”

“I feel like knocking it down to 6,560, knowing that that’s $80\times 82$ by difference of two squares. That gives me $\frac{82}{700}$.” A pause for breath. “Then $\frac{82}{7}$ is a smidge below 12… 11.7, I suppose, giving 11.7%.”

“Not bad,” said the Mathematical Ninja. “Not bad at all. Now try it for $X=30$.”

* Edited 2017-04-24 to fix LaTeX and category. Thanks, @dragon_dodo!