Lottery Sharing with Poisson Winners

Probability · Medium · Free problem

You win a lottery -- but you may not be the only winner. The number of other winners (besides you) is $N \sim \text{Poisson}(\lambda)$. Given $N = n$, the prize is awarded to one person selected uniformly at random from all $n+1$ winners (you plus the $n$ others).

  1. Find $P(\text{you receive the prize})$ as a closed-form function of $\lambda$.
  2. Evaluate numerically for $\lambda = 1$, rounded to the nearest thousandth.
  3. Verify the limiting behavior as $\lambda \to 0$ and $\lambda \to \infty$.

Hints

  1. Condition on $N$ using the Law of Total Probability: $P(W) = \sum_{n=0}^{\infty} P(W \mid N=n) \cdot P(N=n)$.
  2. After substituting $P(W \mid N=n) = 1/(n+1)$ and the Poisson PMF, you get a sum of the form $\sum_{n=0}^{\infty} \lambda^n / (n+1)!$. Multiply and divide by $\lambda$ to shift indices and recognize the Taylor series for $e^{\lambda}$.
  3. The key algebraic step: $\sum_{n=0}^{\infty} \lambda^{n+1}/(n+1)! = \sum_{m=1}^{\infty} \lambda^m/m! = e^{\lambda} - 1$.

Worked Solution

How to Think About It: Your chance of winning given $N = n$ is

/(n+1)$ -- you are one of $n+1$ equally likely recipients. The randomness comes from not knowing $n$ in advance, so you condition on $N$ and average over its Poisson distribution. The computation boils down to a power series that you need to recognize as a shifted exponential series.

Quick Estimate: For $\lambda = 1$, the Poisson mean is 1 other winner, so on average there are 2 total winners. Your share should be around

/2 = 0.5$. But when $N = 0$ (probability $e^{-1} \approx 0.368$), you win alone with probability 1. This pushes the answer above 0.5. A reasonable guess is $0.60$--$0.65$.

Formal Solution:

Let $W$ be the event you receive the prize. Condition on $N$:

$P(W) = \sum_{n=0}^{\infty} P(W \mid N=n) \cdot P(N=n) = \sum_{n=0}^{\infty} \frac{1}{n+1} \cdot \frac{\lambda^n e^{-\lambda}}{n!}$

Factor out $e^{-\lambda}$:

$P(W) = e^{-\lambda} \sum_{n=0}^{\infty} \frac{\lambda^n}{(n+1)!}$

To match the Taylor series of $e^{\lambda} = \sum_{m=0}^{\infty} \lambda^m / m!$, multiply and divide by $\lambda$:

$P(W) = \frac{e^{-\lambda}}{\lambda} \sum_{n=0}^{\infty} \frac{\lambda^{n+1}}{(n+1)!} = \frac{e^{-\lambda}}{\lambda} \sum_{m=1}^{\infty} \frac{\lambda^m}{m!} = \frac{e^{-\lambda}}{\lambda} \left(e^{\lambda} - 1\right)$

$\boxed{P(W) = \frac{1 - e^{-\lambda}}{\lambda}}$

Numerical answer for $\lambda = 1$: $P(W) = \frac{1 - e^{-1}}{1} = 1 - e^{-1} \approx 1 - 0.3679 = 0.632$

This is consistent with our quick estimate of $0.60$--$0.65$.

Limiting behavior: - As $\lambda \to 0$: $P(W) \to 1$. With very few other winners expected, you almost always win alone. - As $\lambda \to \infty$: $P(W) \to 0$. With many other winners, your share goes to zero (the

/\lambda$ factor dominates). - Note that $P(W) = (1-e^{-\lambda})/\lambda$ is a decreasing function of $\lambda$, which makes intuitive sense.

Answer: $P(W) = \dfrac{1-e^{-\lambda}}{\lambda}$, which equals

- e^{-1} \approx 0.632$ at $\lambda = 1$.

Intuition

The result $P(W) = (1-e^{-\lambda})/\lambda$ is the survival function of the Poisson distribution divided by $\lambda$, and it has a clean Poisson process interpretation. Imagine winners arriving as a Poisson process at rate $\lambda$ per unit time, and you are the winner at time 0. The probability you win is the probability that no other winner arrives before you -- in a sense, you are competing in a race. The $(1-e^{-\lambda})/\lambda$ form also appears in inventory and queueing models where you compute the average time until the first event, normalized by the rate.

The technical move worth memorizing is the index-shift trick for Poisson-type sums: when you see $\sum \lambda^n / (n+k)!$, multiply and divide by $\lambda^k$ to align the exponent with the factorial. This manipulaton comes up constantly in Poisson and gamma distribution calculations.

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