Poisson Eats Cheeseburgers

A few years ago I read this piece about meat purveyor Pat LaFrieda and since then I’ve wanted to visit one of his restaurants. While at LaGuardia Airport awaiting a flight, all outbound flights from the airport were delayed due to inclement weather and only one outbound runway was operational. This was fortuitous—the heavens opened and I began my siege upon one of Pat’s restaurants—Custom Burgers.

No one seemed to know exactly when my flight would depart and I was concerned that if I chose to eat at Custom Burgers I might miss my departing flight. There were only 5 flights ahead of me in the queue and I didn’t want to get stranded in LGA overnight nor did I want to miss eating at Custom Burger. This provoked an interesting question—how to estimate the probability that I could eat at Pat’s restaurant without missing my flight?

Since only one runway was open, flight departures were disjoint events—no two planes could takeoff at the same time. By discretizing time into minutes, seconds, and ever increasingly smaller intervals, it follows that the probability of a flight departing in a given time interval approaches zero as the number of time intervals, $ n $ increases. Therefore, the probability of precisely $ k $ flights departing in $ n $ small time intervals is given by the Binomial distribution:

$$ \Pr(k) = \dbinom{n}{k} \text{ } p^k(1 - p)^{n - k} \text{ where,} $$

$$ p = \frac{x}{n}$$

The unknown $ p $ represents the probability of the mean number of flights, $ x $ per $ n $ interval. I didn’t have a way to get the true average number of flights leaving in a given time interval, but I could generate a reasonable approximation by gathering the departure times of outbound flights from Flight Tracker Pro on my iPhone.

Once I had this data, using a crude maximum likelihood estimation I approximated $ x $ to be 8.16 flights per hour out of LGA.1 Solving for $ p $ given $ \hat{x} = np $ allowed the rate of departing flights to be approximated. Substituting the expression $ \frac{x}{n} $ into the Poisson distribution facilitated the approximation of the binomial probability where $ n $ is large and $ p $ is small, giving the probability density function of the Poisson:

$$ \Pr(k) = \frac{x^k e^{-x}}{k!} $$

I estimated that it would take me a minimum of thirty minutes to eat at Custom Burgers, so I wanted to find the probability of 6 or more flights leaving in the next thirty minutes. Since I was 6th in the queue, if 6 or more flights departed, I would miss my flight. To determine this probability, I first calculated the probability of 5 or fewer flights leaving within the next thirty minutes:

$$ \Pr(\leq k) = \sum_{i=1}^{N} \Pr(k_i) \text{ where}, $$

$$ \Pr(k_i) = \frac{e^{-4.08} e^{k_i}}{!k_i} $$

Using this calculation, it was possible to determine the probability of missing my flight.

$$ \Pr(k > 5) = 1 - \Pr(k \leq 5) = 1 - 0.7725 = 0.2275 $$

There was an $ \approx 23\% $ chance I would miss my flight if I went to Custom Burger for thirty minutes. These odds seemed like an acceptable risk, so I left to eat—and I’m glad I did, the meal was great and I made my flight.

  1. I assume $ p $ is constant across the time interval. ↩