Does $150$% chance exist? I seen an internet meme that was titled "Kids that did this sh&t in school have a 150% chance of being in jail right now." 
My understanding is that we are dealing with probabilities here and probabilities are limited by $0 \le P \le 1$ or $0$% $\le P \le$ $100$%.
Of course $100$% can exist in other areas of maths. For example $150$% of $100$ is $150$.
But does the statement in the title make sense?
 A: You are correct. Probabilities (or chance) can only range from $0$ to $100$%. So mathematically this statement makes no sense. 
It's really just a rhetorical way of saying that there is a 'really high' chance this happens ... where by 'really high' they probably don't even mean some absolute percentage like $90$%, but rather 'much higher than a random person (who don't do this sh&t in school) would have', i.e. it's more of a claim about the relative chance.
In fact, it is possible that the $150$% came from exactly such a comparison, e.g. That a person not doing this sh&t would have a $1$% change of being in jail, whereas a person doing this sh&t has a $2.5$% chance of being in jail, i.e. $150$% higher ... and subsequently someone misinterpreted or misused this number. It certainly wouldn't be the first time people misuse or misunderstand statistical claims.
But again, most likely it is just a made up number for rhetorical effect, just like when I say ' I have a billion and one things to do'.  You know all $87$% of all statistics are made up on the spot, right? :)
A: No, probability of something between 0 and 1 {or 0% and 100%}(inclusive).
It's only a joke.
A: As for probability, 150% chance does not exist. And if it was a teacher that 'did this sh&t', the teacher should be in jail instead.
However, there is also the 'expected number', that is somewhat related.
Half of them could be in jail for the second time..
A: It depends on how you interpret the word 'chance'.
If chance = probability, then no, as probabilities $\in [0,1]$
However, if chance = some kind of looser reference to likelihood, or odds, then yes.
An odds ratio $\frac{p}{1-p}$, $\in [0,\infty]$ and gives you an indication of the probability, larger values implying larger probability. In this case an odds of $1.5$ would correspond to a probability of $3/5$
A: In that particular example, as some users already said, probably the expression is purely rhetorical. However, sometimes in the news you hear sentences like that. While it's true that their literal meaning makes no sense mathematically, most of the times what they actually mean is something different, and it has to do with conditional probability:
Suppose that kids (say, in America) have some probability $p$ of being in jail right now. $p$ will be a number from $0\%$ to $100\%$. 
The kids that "do that kind of stuff" will have a higher probability, $q>p$, still a number from $0\%$ to $100\%$, but higher than before. 
Now what the message may say is that 150% is exactly "how much higher". That is: $q/p=150\%$ (equivalently, $=1.5$). 
In other words, mathematically:
$$
\dfrac{Prob(\mbox{being in jail}|\mbox{doing that stuff})}{Prob(\mbox{being in jail})} = 150\%\;,
$$
and that is mathematically perfectly acceptable.
Sometimes in the news the situation above is described as "having $50\%$ more chance of ending up in jail", and it can get confusing because you don't know if that $50\%$ refers to $q-p$ or to $(q-p)/p$.
