Probability of getting a parking ticket

Question

The city of Ithaca, New York, allows for two-hour parking in all downtown spaces. Methodical parking officials patrol the downtown area, passing the same point every two hours. When an official encounters a car, he marks it with chalk. If the car is still there two hours later, a ticket is written. Suppose that you park your car for a random amount of time that is uniformly distributed on (0, 4) hours. What is the probability you will get a ticket? (from Rick Durett)

When I park my car somewhere, it takes $2h-t_1$, until the official marks my car. ($t_1$ is the time, that has passed, since the official was the last time at this place.) I suppose, that $t_1$ is uniformly distributed on the interval [0,2]. You get a ticket, if your parking time $t_2$ is longer than $4h-t_1$, so I have to compute the probability $\mathbb{P}(t_2 \geq 4h-t_1)$. This should be $\mathbb{P}(t_1+t_2 \geq 4h) = 1-F(4h)$. F is 0 if $x \notin [0,6]$, $x/8$, if $x \in [0,2]$, 1/4 if $x \in [2,4]$ and $6/8-x/8$, if $x \in [4,6]$. So I get $\mathbb{P}(t_1+t_2 \geq 4h) = 1-1/4 = 3/4$ and this probability seems too much for me, so something must be missing. Where is my mistake? Thank you! :)

Answer 1

Like you said, lets have $t$ be the time since the officer's last visit, so $t\sim \mathcal{U}(0,2)$ Your parking time is $h\sim \mathcal{U}(0,4)$. Let $T:= \{4-t<h\}$ be the event that you get a ticket, as you correctly pointed out.Often the easiest way to tackle these problems is via conditioning.

$P(T|t=\tau)=P(h>4-\tau)=\frac{4-4+\tau}{4}=\frac{\tau}{4}$

This implies that $P(T)=\int\limits_{0}^{2}f_t(z)P(T|t=z)dz=\int\limits_{0}^{2}\left(\frac{1}{2}\right)\frac{z}{4}dz=\int\limits_{0}^{2}\frac{z}{8}dz=\frac{z^2}{16}|_{z=2} = \frac{1}{4}$

I think where you went awry is with calculating the pdf of the sum of two independent, dissimilar uniform distributions. Their joint pdf if simply $f_{h,t}=\frac{1}{8}$ over the domain $[0,4]\times[0,2]$, hence your probability is the area above the line $h+t=4$ on this rectangle times $\frac{1}{8}$, which is just the corner of the rectangle above the line connecting $(2,2)$ with $(4,0)$ which is just half the area of the rectangle formed by $[0,2]\times [2,4]$ which comes out to $2$, which when multiplied by $\frac{1}{8}$ you get the correct value of $\frac{1}{4}$

Answer 2

It looks to me like the probability of not getting a ticket is $\frac34$. (So the probability of getting a ticket is $\frac14$.

Here's an intuitive way of seeing this. Half the time you park for less than two hours and definitely don't get a ticket. The other half of the time, you park your car for more than two hours, and then you get a ticket if the official makes his (or her) second pass before you leave. Since both these events are randomly distributed over the two-hour subinterval, there is a $50\%$ chance that you leave first and get no ticket, and a $50\%$ chance that the official arrives first and you do get a ticket.

So your probability of a ticket is $0\cdot \frac12+\frac12\frac12=\frac14$; and correspondingly your chance of no ticket is $\frac34$.

Note: If you know the parking official's cycle, you should try to park right after s/he passes your space, and then you can safely get more time.