What is the chance to get a parking ticket in half an hour if the chance to get a ticket is 80% in 1 hour?

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!

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 $1-0.4472=0.5528$.

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

$$T+(1-T)T$$

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:

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

$$T=1-\sqrt{.2}$$

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

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.