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Let $\mathcal{T}$ be the set of three-vertex subsets and let $I_A$ indicate that $A$ forms a triangle. ThenIn particular $p\sim (6\alpha {)}^{1/3}/n$. We will use the Stein-Chen method with edge-sharing dependency neighborhoods, rather than wrongly treating triangle indicators as mutually independent.

Here is the needed Poisson approximation with dependency neighborhoods, together with its proof. For indicators $I_A$ with means $p_A$, let $B_A$ contain $A$ and suppose $I_A$ is independent of the entire family of indicators outside $B_A$. With $\mu ={\sum }_A{p}_A$ and $W={\sum }_A{I}_A$, defineThe total variation distance convention is $d_{TV}(P,Q)={\sup }_E|P(E)-Q(E)|$. We claimFor an indicator test function $g$ on the nonnegative integers, solve the Poisson Stein equationwith $\Vert \Delta f{\Vert }_{\infty }\le 1$. This elementary Stein factor can be proved as follows. Take the immigration--death process with immigration rate $\mu$ and per-particle death rate one. Starting at $k$, its time-$t$ distribution iswith independent summands. Its stationary law is that of $Z$. Coupling the initial particles with a stationary Poisson initial population makes the centered expectations decay at least as a constant times $(k+\mu )e^{-t}$, sois finite. Its generator equation is $\mu (u(k+1)-u(k))+k(u(k-1)-u(k))=g(k)-\mathbb{E}g(Z)$, obtained by integrating the semigroup derivative. Set $f(k)=u(k)-u(k-1)$ for $k\ge 1$; choose $f(0)=f(1)$, which does not affect the equation. To bound $\Delta f(k)$, couple the processes starting at $k-1,k,k+1$ by adding two independent initial particles. Both survive to time $t$ with probability $e^{-2t}$, and otherwise their second difference cancels in expectation. Since $|g(j+2)-2g(j+1)+g(j)|\le 2$,This proves the Stein factor for $k\ge 1$, and the chosen $f(0)$ handles $k=0$.

Put $W_A=W-{\sum }_{C\in B_A}{I}_C$. Independence gives $\mathbb{E}[I_{A}f(W_A+1)]=p_A\mathbb{E}f(W_A+1)$. Inserting this identity into the Stein equation yieldsThe first sum in absolute value is at most $b_1$, since the two arguments differ by ${\sum }_{C\in B_A}{I}_C$. In the second sum, on $I_A=1$ the arguments differ by ${\sum }_{C\in B_A\setminus \{A\}}{I}_C$, giving bound $b_2$. Taking the supremum over event indicators $g$ proves the claimed total variation distance bound.

For triangle $A$, choose $B_A$ to consist of $A$ itself and all triangles sharing an edge with $A$. There are exactly $1+3(n-3)$ such triangles. Every outside triangle uses none of the three edges of $A$, even if it shares one vertex, so $I_A$ is independent of their entire indicator family. For distinct neighboring triangles, their union has five edges and $\mathbb{E}(I_A{I}_C)=p^5$. ThereforeThe Stein-Chen method gives $d_{TV}(\mathcal{L}(X),\mathrm{Poisson}({\mu }_n))\to 0$. Coupling Poisson distributions of two means by adding an independent Poisson variable givesThis proves the Poisson limit for triangles in a binomial random graph:Applying the same bound to the event $\{1,2,\dots \}$ givesFor finite $n$ this is an approximation, not an exact formula: the error from $1-e^{-{\mu }_n}$ is at most $b_1+b_2$, and the error from $1-e^{-\alpha }$ is at most $b_1+b_2+|{\mu }_n-\alpha |$.

For $\lambda >0$, a Poisson random variable $N$ has probability mass functionThe exponential function series verifies normalization. At $\lambda =0$, define $N=0$ with certainty. The equality of mean and variance is a distinguishing model property, though it does not by itself characterize the law.

When counting events in a homogeneous Poisson process, distinguish its rate $r$ from the distribution parameter: an interval of duration $t$ has $\lambda =rt$. A rate measured per minute times a duration measured in minutes gives a dimensionless expected count. This process assumes independent increments, a constant rate, and a negligible chance of multiple events relative to interval length in a sufficiently short interval. A rate alone does not establish a Poisson distribution. See Bertsekas and Tsitsiklis, chapter 6 summary, pages 46-47.

For a constructed arrival model with $r=2$ events per minute over $t=1.5$ minutes, $\lambda =3$. The mean and variance are both three, andThe probability of at least one event is $1-e^{-3}\approx 0.9502129$. These results describe the specified process, not every arrival mechanism with the same average rate.

The Poisson limit theorem, often called the law of rare events, connects this law to the binomial distribution. Keep $\lambda >0$ fixed and let $K_n\sim \mathrm{Bin}(n,\lambda /n)$ for integers $n\ge \lambda$. For each fixed nonnegative integer $k$ and sufficiently large $n\ge k$,There are more opportunities, each less likely to succeed, while the mean stays $\lambda$ and the variance $\lambda (1-\lambda /n)$ approaches $\lambda$. For $\lambda =3$, the probability of three successes is $0.2668279$ at $n=10$, $0.2310569$ at $n=50$, and $0.2257398$ at $n=200$, approaching the Poisson distribution value $0.2240418$.

The panels show the same count scale and the same Poisson distribution reference. At $n=10$, $p=0.3$ is not especially small and the difference is visible. No universal trial-count cutoff guarantees an adequate approximation for every event.

The independence assumption matters even when every success is rare. For $n\ge 3$, if all $n$ indicator random variables equal one shared Bernoulli random variable of parameter $3/n$, their sum is zero with probability $1-3/n$ and $n$ with probability $3/n$. Its mean is three, yet its zero-count probability tends to one, not $e^{-3}$. Some dependent counts still admit approximation through the Stein-Chen method, but this requires additional control of dependence.

For historical orientation, the BnF records Poisson's original probability treatise, Recherches sur la probabilité des jugements, as published in Paris in 1837. This bibliographic fact alone does not establish when the modern distribution terminology arose. See BnF record of Poisson's 1837 treatise.