Poisson Limit of Lottery Winners
A total of $2N$ people play a lottery each week. Each person is independently assigned a uniformly random integer from $\{1, 2, \dots, N\}$. One of the $N$ integers is then selected uniformly at random as the winning value. Let $W_N$ be the number of people who win in a given week.
Compute $\displaystyle \lim_{N \to \infty} P(W_N \leq 3)$ to the nearest thousandth.
Hints
- What is the distribution of $W_N$ for fixed $N$? What happens to the mean as $N$ grows?
- When the number of trials grows while the success probability shrinks and their product stays constant, think Poisson limit theorem.
- The limiting distribution is $\text{Poisson}(2)$. Compute $\sum_{k=0}^{3} e^{-2} \cdot 2^k / k!$ to get the answer.
Worked Solution
How to Think About It: Each of the $2N$ people independently wins with probability $1/N$, so $W_N \sim \text{Bin}(2N, 1/N)$. As $N$ grows, you have more and more trials with a smaller and smaller success probability, but the expected number of winners stays fixed at $2N \cdot 1/N = 2$. That is the textbook setup for a Poisson limit. In the first 10 seconds of the interview, you should say: "This converges to a Poisson(2), so I need the CDF up to 3."
Quick Estimate: For $X \sim \text{Poisson}(2)$, the PMF is $P(X = k) = e^{-2} \cdot 2^k / k!$. We need $P(X \leq 3)$. A quick mental computation: $e^{-2} \approx 0.135$. Then $P(X=0) \approx 0.135$, $P(X=1) \approx 0.271$, $P(X=2) \approx 0.271$, $P(X=3) \approx 0.180$. Summing: $0.135 + 0.271 + 0.271 + 0.180 \approx 0.857$. That is our answer.
Approach: Apply the Poisson limit theorem to the sequence of Binomial random variables, then evaluate the Poisson CDF exactly.
Formal Solution:
For fixed $N$, each person wins if their assigned number matches the winning value. This happens with probability $1/N$, independently across people. So
$$W_N \sim \text{Bin}(2N,\; 1/N).$$
As $N \to \infty$, the number of trials $2N \to \infty$ and the success probability $1/N \to 0$, while the mean is
$$E[W_N] = 2N \cdot \frac{1}{N} = 2.$$
By the Poisson limit theorem, $W_N \xrightarrow{d} X$ where $X \sim \text{Poisson}(2)$. Therefore
$$\lim_{N \to \infty} P(W_N \leq 3) = P(X \leq 3) = \sum_{k=0}^{3} \frac{e^{-2} \cdot 2^k}{k!}.$$
Evaluating term by term:
- $P(X = 0) = e^{-2}$
- $P(X = 1) = 2e^{-2}$
- $P(X = 2) = \frac{4}{2} e^{-2} = 2e^{-2}$
- $P(X = 3) = \frac{8}{6} e^{-2} = \frac{4}{3} e^{-2}$
Summing:
$$P(X \leq 3) = e^{-2}\left(1 + 2 + 2 + \frac{4}{3}\right) = \frac{19}{3} e^{-2} = \frac{19}{3e^2}.$$
Numerically, $e^2 \approx 7.389$, so $P(X \leq 3) \approx 19 / 22.167 \approx 0.857$.
Answer: $\displaystyle \lim_{N \to \infty} P(W_N \leq 3) = \frac{19}{3e^2} \approx 0.857$.
Intuition
This is the canonical example of the Poisson limit theorem (also called the law of rare events). Whenever you have a large number of independent trials, each with a small probability of success, and the expected count stays bounded, the Binomial converges to a Poisson. In quant interviews, this pattern appears constantly -- the number of defaults in a large portfolio, the number of extreme moves in a trading day, the number of arrivals in a short window. The moment you see "many independent things, each unlikely to fire, fixed expected count," jump straight to Poisson.
The subtle point people miss is that the Poisson approximation is not just a computational shortcut -- it tells you something structural. Once the limit is Poisson, the exact value of $N$ does not matter much. Whether there are 100 players or 10,000, the distribution of winners looks nearly the same, because the only thing that matters is the rate $\lambda = 2$. This is why Poisson models are so robust in practice: the fine details of the mechanism wash out, and only the rate survives.