The Parking Ticket Expected Value: A Poisson Process Estimation

I was parking my car today and while I was enroute to proctor an exam for my probability class, I had a curious idea. I was wondering what an expected value argument would say about whether or not to pay for parking. I get 1 hour of parking for free, but all day parking is $25. I know from an absent minded friend that the ticket for having an expired parking pass is $60.

Assuming that the parking attendant checks the parking lot as a Poisson process with rate $\alpha$, what would $\alpha$ have to be for me to stay $t$ hours past my free parking and have my highest expected value being to not pay for parking?

The Poisson Process

A Poisson process is a mathematical model for a series of discrete events where the average time between events is known, but the exact timing of events is random. Key properties include:

Independence: Events occur independently of each other.
Constant Rate: The average rate $\alpha$ (events per unit of time) is constant.
Memorylessness: The time until the next event does not depend on how much time has already passed.

In our case, $\alpha$ represents the number of times per hour the parking attendant checks the lot.

The Math

If you stay $t$ hours past your free hour, the number of checks $N(t)$ that occur in that interval follows a Poisson distribution with mean $\lambda = \alpha t$:

\[P(N(t) = k) = \frac{(\alpha t)^k e^{-\alpha t}}{k!}\]

You will receive a ticket if the attendant checks the lot at least once during your stay ($N(t) \ge 1$). The probability of being ticketed is:

\[P(\text{ticketed}) = 1 - P(N(t) = 0) = 1 - e^{-\alpha t}\]

Expected Cost

The expected cost of not paying for parking is the cost of the ticket ($C_{\text{ticket}}$) multiplied by the probability of being ticketed:

\[E[\text{not paying}] = C_{\text{ticket}} \cdot (1 - e^{-\alpha t})\]

The cost of paying for parking is a flat fee $C_{\text{parking}}$. You should (mathematically) choose to not pay if:

\[C_{\text{ticket}} \cdot (1 - e^{-\alpha t}) < C_{\text{parking}}\]

Solving for the Critical Rate ($\alpha$)

To find the threshold rate $\alpha$ where it becomes worth it to pay, we set the costs equal:

\[1 - e^{-\alpha t} = \frac{C_{\text{parking}}}{C_{\text{ticket}}}\] \[e^{-\alpha t} = 1 - \frac{C_{\text{parking}}}{C_{\text{ticket}}}\] \[-\alpha t = \ln\left(1 - \frac{C_{\text{parking}}}{C_{\text{ticket}}}\right)\] \[\alpha_{\text{critical}} = -\frac{\ln\left(1 - \frac{C_{\text{parking}}}{C_{\text{ticket}}}\right)}{t}\]

If the attendant checks the lot more frequently than $\alpha_{\text{critical}}$, your expected cost of not paying exceeds the cost of the permit.

Interactive Estimation

Use the sliders below to adjust the parameters of your parking situation and see your “Expected Value” strategy.

Code

viewof t = Inputs.range([0.1, 12], {value: 3, step: 0.1, label: "Hours past free parking (t)"})
viewof cTicket = Inputs.number({value: 60, label: "Ticket Cost ($)"})
viewof cParking = Inputs.number({value: 25, label: "All-day Parking Cost ($)"})
viewof alpha = Inputs.range([0.01, 2], {value: 0.1, step: 0.01, label: "Checks per hour (α)"})

Code

probTicket = 1 - Math.exp(-alpha * t)
expectedCost = cTicket * probTicket
criticalAlpha = cParking < cTicket ? -Math.log(1 - (cParking / cTicket)) / t : 0

md`
### Results

At a check rate of **${alpha}** times per hour, your probability of getting ticketed over **${t}** hours is **${d3.format(".1%")(probTicket)}**.

- **Expected cost of not paying:** \$${d3.format(".2f")(expectedCost)}
- **Cost of paying:** \$${d3.format(".2f")(cParking)}

${expectedCost < cParking 
  ? "### 🟢 Strategy: Don't Pay" 
  : "### 🔴 Strategy: Pay for Parking"}

${cParking < cTicket 
  ? md`The **critical rate** for your stay is **${d3.format(".3f")(criticalAlpha)}** checks per hour. If you think the attendant rounds the lot more than once every **${d3.format(".1f")(1/criticalAlpha)}** hours, you should pay!`
  : md`Since the parking permit costs as much as or more than the ticket itself, **you should never pay** for parking from a purely expected-value perspective!`}
`

Caveats

This model assumes the attendant’s rounds are truly random (Poisson). In reality, attendants might follow a schedule, which would change the distribution. Additionally, most people are “risk-averse” and would pay even if the expected value is slightly in favor of not paying, to avoid the high-variance outcome of a $60 fine! A second estimation could be done to determine how much money is profitable to invest in discovering the attendants checking schedule, which would allow one to not pay for parking with higher accuracy.

The Poisson Process

The Math

Expected Cost

Solving for the Critical Rate (\(\alpha\))

Interactive Estimation

Caveats