This sounds more like a brain teaser, but I had some kink to think it through :( Suppose you're parking at a non-parking zone, the probability to get a parking ticket is 80% in 1 hour, what is the probability to get a ticket in half an hour? Please show how you deduce the answer. Thanks!

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    $\begingroup$ As a Frenchman, 80% chance in 1 hour seems really high to me, would be rather 20% chance in 1 day. $\endgroup$
    – Benoit
    Jun 4, 2012 at 13:18
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    $\begingroup$ Where I'm from, between the hours of 3PM - 6PM, M-F, the probability of getting a ticket and towed is 100% within 10 minutes. $\endgroup$
    – user32990
    Jun 5, 2012 at 8:41

8 Answers 8


It really depends on what model is assumed. However, if the idea is that no matter how long you leave your car there, you have a $20$% chance of getting through any given hour unscathed, you can treat it as an exponential decay problem. Let $p(t)$ be the probability that you do not get a ticket in the first $t$ hours. Then $p(1)=0.2$, $p(2)=0.2^2$ (a $20$% chance of making it through the first hour times a $20$% chance of making it through the second), and in general $p(t)=0.2^t$. The probability of not getting a ticket in the first half hour is then $p(1/2)=0.2^{1/2}=\sqrt{0.2}\approx 0.4472$, and the probability that you do get a ticket in the first half hour is about $1-0.4472=0.5528$.

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    $\begingroup$ @Rock: How would you do it? You can’t say that your chance of getting a ticket in two hours is $0.8^2$, since it should obviously be more than $0.8$, and you certainly can’t say that it’s $1.6$, since that’s greater than $1$. It’s much easier to see how your chance of not getting one keeps shrinking. $\endgroup$ Jun 4, 2012 at 4:23
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    $\begingroup$ @Dave: You’re assuming something like full-time checkers with regular routes. Try assuming instead random checks. $\endgroup$ Jun 4, 2012 at 5:34
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    $\begingroup$ +1 for "It really depends on what model is assumed." Without making assumptions, all we can really say is that the probability of getting a ticket in 20 minutes is less than or equal to 80%. An alternative model is there is an approximately 80% chance that the ticket-writer is working today. If the ticket-writer is working there is an approximately 100% chance that you will get a ticket within 5 minutes. $\endgroup$
    – emory
    Jun 4, 2012 at 6:02
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    $\begingroup$ @smdrager, I don't understand. There are uncountable infinite number of models to choose from. Anyway you are not provided with any data to calculate likelihood. $\endgroup$
    – emory
    Jun 4, 2012 at 18:49
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    $\begingroup$ Parking attendants are not memoryless. $\endgroup$
    – Mark Adler
    Jun 5, 2012 at 4:56

If my probability of getting a parking ticket in a half hour is T, then


are the odds I got one in 1 hour. The first term says that I got it in the 1st half hour (so it doesn't matter what happens after that), and the second term says that I got it in the 2nd half hour (so not in the first). Solving and taking the sensible solution:

$$2T-T^2=.8\implies T^2-2T+.8$$


Which are roughly $55.3\%$ odds in a half hour.

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    $\begingroup$ @Rock you could instead choose this as your answer. $\endgroup$
    – user23320
    Jun 4, 2012 at 12:43
  • $\begingroup$ @Joshua Now I have it. The accepted answer gives a general solution which I guess is what the question potentially asked for. $\endgroup$
    – Rock
    Jun 4, 2012 at 18:26
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    $\begingroup$ You've made a Markov assumption, which may or may not be valid. $\endgroup$
    – Ben Voigt
    Jun 4, 2012 at 23:58
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    $\begingroup$ @BenVoigt fair enough, but it's either this, assume extra conditions that weren't specified, or be unable to do anything at all. You always have to assume some model anyway, and in the absence of other information a Markov assumption seems to be the natural choice. Of course not a statistician though, so I wouldn't be surprised if I'm begging to be corrected here. $\endgroup$ Jun 5, 2012 at 20:13
  • $\begingroup$ Hi @RobertMastragostino, just wanted to let you know that the "Italian Language & Usage" proposal has restarted. Since you committed to the previous proposal, maybe you are still interested. See you! (I hope this comment is not considered as spam: if so, I beg your pardon) $\endgroup$
    – Lucius
    Sep 11, 2012 at 19:01

Another way of looking at the exponential model: If $P$ is the probability you get lucky and don't get ticketed in the first 30 minutes, then $P^2$ is the chance you luck out twice in a row and don't get ticketed in the first hour. You know $P^2 = 0.2$. So $P$ is the square root of that or $0.447$. So the chance you are unlucky in the first 30 minutes and get a ticket is $1 - 0.447 = 0.553$ or $55.3$ percent.

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    $\begingroup$ A very clear explanation! $\endgroup$ Jun 4, 2012 at 7:09
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    $\begingroup$ For me the smoothest solution ! $\endgroup$
    – berkay
    Jun 4, 2012 at 8:29
  • $\begingroup$ Thanks.. but to be clear I'm just explaining the exponential model in layman's terms, and I am not suggesting I agree this model is necessarily the "correct" one for the problem. $\endgroup$
    – Zarrax
    Jun 4, 2012 at 21:18

A simple but not unreasonable model is that the ticket person cycles through her/his round in roughly constant time. Suppose that you park illegally for exactly an hour at a randomly chosen spot in the round, and at a start time uniformly distributed with respect to the cycle. Since your probability of getting a ticket in an hour is $80\%$, the cycle length is $1.25$ hours. Thus your probability of getting a ticket if you park illegally for half an hour is $40\%$.

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    $\begingroup$ For the people who wonder how your model can give something different: The other answers assume that getting no ticket in the first half of an hour does not increase the chances of getting one in the second half of the hour. In a place where 80% of cars get a ticket within the hour, this is not a very realistic assumption. So, we can rejoice and expect that the probability is closer to 40% than to 55%. $\endgroup$
    – Phira
    Jun 4, 2012 at 12:35
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    $\begingroup$ Another explanation to this: Assume there are 100 illegal cars, so the parking warden can handle 80 tickets per hour (we assume linear distribution of tickets per time) and thus 40 tickets half an hour. If you are one of that cars during that half of an hour ... your risk getting caught is 40/100. $\endgroup$
    – Nappy
    Jun 4, 2012 at 14:52
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    $\begingroup$ IMO from a modeling perspective this is by far the most reasonable answer given. The exponential model seems unreasonable since clearly your hazard (aka force of mortality or failure rate) ought to be increasing over time. $\endgroup$
    – guy
    Jun 4, 2012 at 16:18
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    $\begingroup$ Agreed, @guy. With an exponential decay model of tickets, no matter how long you leave your car on the street, there is some nonzero chance that it will not get a ticket. That's not the behavior of any conceivable real-world method for writing parking tickets (which generally involve somebody walking down the street and ticketing every illegally parked car). $\endgroup$ Jun 4, 2012 at 17:22

There is another way to look at this problem.

Most of the answers of ~55.3% (including the currently accepted answer) assume the "meter reader" is checking cars COMPLETELY at random. In this model, there is an extremely small but non-zero chance you could leave a car in the same spot for a month and not get a ticket (although it's up there in "lottery" winner and struck-by-lightning chances). In fact, the chance of getting a ticket using this model is always less than 100% for a finite amount of time.

However, meter readers typically work in a linear pattern starting at one point and then looking at ALL the cars in a line down one side of a block. If we consider the realistic pattern that cars are checked in a linear fashion instead of cars being checked randomly, the answer becomes much simpler. Therefore, when they completely cover an area (in a finite amount of time), your car has a 100% chance of getting a ticket. If you are 100% likely to get a ticket within a finite amount of time, then the p(t)=1-0.2^t answer for getting a ticket is invalid.

So lets consider linear coverage.

If you have an 80% chance of getting a ticket in one hour, the meter readers in your city walk 4 out of 5 blocks per hour (in 60 minutes). Assuming they follow a pattern which has complete coverage and fairly evenly sweeps the whole parking area, they could have 100% coverage of all parking spots in 75 minutes. That means no matter where you park, within 75 minutes you will get a ticket. Under the linear coverage model, p(t)=0.8t. So in half an hour, you have a 40% chance of getting a ticket.

This is a case of not trying to get too mathematically clever because the simplest answer could actually be the correct one.

The real answer depends whether your meter people ticket cars randomly or sweep areas linearly with even coverage (and gets more complicated with real life where they sweep some areas much more often than others but still get 100% coverage over a finite amount of time).


Not enough data to come up with an answer. You would have to know the pattern that the parking warden takes to get the correct answer (first half hour, second half hour), also this depends on which part of the time unit the parking warden would take up, for example 1pm-2pm, 1.05pm-2.05pm etc

80% can reasonable be assumed for an hour, but for half an hour less so.

I would therefore conclude that there is an 80% chance that you will get a ticket as there is no way to distinguish the half hour from the full hour because of lack of data for the movements of the parking warden.

  • $\begingroup$ Why be pessimistic? Maybe it's your lucky day and there's a zero percent chance. :) $\endgroup$
    – Zarrax
    Jun 4, 2012 at 21:34
  • $\begingroup$ If your a pessimist, there are no unexpected nasty surprises, you will never feel the pain of disappointment, and reality will hold no fear. Abide by the law, there will be a 100% chance of not getting a ticket. :) $\endgroup$ Jun 6, 2012 at 11:32
  • $\begingroup$ You know that there is an 80% chance of getting a ticket in a hour. Since the full hour encompasses the half hour in question, then it is definably an 80% chance through set theory. Without further data about the parking wardens movements, it is impossible to suggest anything better. $\endgroup$ Jun 7, 2012 at 0:36

This sounds like a Poisson Process to me... so, assuming that:

Since we don't have additional data, I will asume that the expected probability of being spotted once per hour is $p=0.8$. This is a Bernoulli experiment (either you get spotted or not). I will model the probability of being spotted by a machine, person, or some other authority that can issue a ticket ;-)

If $\lambda$ is the expected times that you get spotted every hour, then $\theta=\frac{1}{\lambda}\ $ is the expected time between events.

The number of events given in $t$ can be modeled by a random variable $N \sim Poisson(\lambda·t)$ where $t$ is the amount of time (in hours) elapsed, and the time between events can be modeled by a random variable $X\sim Exponential(\theta)$.

So, the probability of getting $n$ tickets in $t$ hours, with an expected probability of $\lambda$ that you get a ticket in one hour is: $$Pr\{N=n\ |\ \lambda, t\}=\frac{e^{-\lambda·t}·(\lambda·t)^{n}}{n!} $$ Since we need to calculate the probability of being spotted at least once, we can calculate $1-Pr\{N=0\}$. Then, for $n=0$, $\lambda=0.8$ and $t=0.5$, then: $$1-Pr\{N=0\ |\ \lambda=0.8, t=0.5\}=1-\frac{e^{-0.4}·(0.4)^{0}}{0!}=1-e^{-0.4}\approx0.4918 $$

You should notice that, even if you "get spotted" more than once per day, this model still holds, since this process is memoryless.

Further reference: Ross, Sheldon, "Introduction to probability models", Ninth edition, Chapter 5.

(I really hope this is not your homework)

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    $\begingroup$ A Markov chain is a much better assumption, since I've never heard of multiple tickets being issued to a car (that remains parked in one place) in the same day. Therefore "got 1 ticket" is a terminal state, there is no chance of moving to two tickets. The problem stated that there is an 80% chance of getting a ticket, not that the mean number of tickets is 0.8. Your answer confuses these two quantities. $\endgroup$
    – Ben Voigt
    Jun 5, 2012 at 0:03
  • $\begingroup$ With no further information, the expected value of the tickets per hour is 0.8 (is a Bernoulli experiment). I think the Poisson Process hypothesis holds because all you know is that you may "get caught" with a given probability in a unit of time. Of course you can't be caught more than once a day, but that doesn't implies that you can't be "spotted" n times a day. $\endgroup$
    – Barranka
    Jun 5, 2012 at 18:35
  • $\begingroup$ That said, I'm correcting my answer, because I calculated the probability of "being spotted" once per interval, which is not what we are looking for. The probability we're looking for is "what is the probability of 'being spotted' at least one time", which is 1 - Pr{"Not 'being spotted'"}. $\endgroup$
    – Barranka
    Jun 5, 2012 at 18:37
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    $\begingroup$ @BenVoigt: Unfortunately, in some cities, you can be ticketed multiple times, in the same spot, if your car is still illegally parked. $\endgroup$
    – Dancrumb
    Jun 5, 2012 at 20:48
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    $\begingroup$ You did it wrong. $\lambda$ is not $0.8$. To get $\lambda$, you need to reverse the calculation that you did get right. If the probability of getting a ticket in an hour is $0.8$, then $1-e^{-\lambda t}=0.8$ where $t$ is one hour, so the rate $\lambda$ is $1.61$ tickets per hour. Then we use $1-e^{-\lambda t}$ with $t$ equal to half an hour. We get $1-e^{\lambda{1\over 2}hr}$ is $0.553$. The same as the exponential examples above, since it is the same memoryless assumption. $\endgroup$
    – Mark Adler
    Jun 6, 2012 at 5:09

I agree that it depends on the model. Otherwise some fair of independence is presume in suggesting either 'half hours are equi-likely'

Although if you specify that is its an arbitrary hr, and any arbitrary half hour segment (rather than the first half of every hr) that might be different. Otherwise in every hr one inspector might give a zero percent chance for each first half hour, and 80 percent for the second 'half hour'.

But once its specified that

' to be arbitrary hr, or any arbitrary half hr segment'or the average or expected probability value". Or that

" that the model or probability, value does not pertain a one shot case (ie one is not considering a singular case scenario)- where it only applies to a singular occasion for some specific hr ."

Then I presume that this, would not be be a counter-example to the claim (the second half - hour, is also the first half of the full next hour segment from t/2 to 1.5 t where t = 1 hrs), and that the conventional answer sounds reasonable.


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