Normalize output scores for binary classification I am handling a binary classification problem on an imbalanced dataset. 
The goal is to create a system able to insert the returned score (probability to be in the positive class) in a bins between 1 and 10, where 1 means low probability to be in the positive class and 10 high prob. 
The main problem is that I only have the training dataset, so I don't know any values in the test set. Moreover, the predictions will be done one by one, so I cannot analyze the whole test scores.
I tried many models, but in particular I use tree-based models (such as XG-Boost, RF). In these cases, considering also the imbalanced dataset, the output scores are in a very small range, much smaller than [0, 1]. The scores are necessary since I don't want to classify the instances directly into the 0-1 class, but I want to analyze the scores.
How should I build a method able to find the different thresholds in order to create the 10 classes?
 A: Well, if your models outputs probabilities for the positive class and your decision boundary is 0.5, i.e. p < 0.5 -> class 0 and p >= 0.5 -> class 1 then you can discretize the probability interval from $[0.5, 1.)$ into 10 bins and assign a class for each interval, e.g. $[0.5,0.55)$ => class 1, $[0.55,0.6)$ => class 2, ...
Now each class represents some confidence about the positive prediction.
A: First, the fact that your outputs are in only the very low end of $[0,1]$ strikes me as a feature, not a bug, of the outputs of a probability predictor in an imbalanced setting. It might typically be the case that a minority-class event is unlikely, even if the probability is higher than the prior probability (class ratio). If that is the case, then, yes, the predicted probabilities should be at the low end of $[0,1]$.
You want to then scale $[0,1]$ to $[1,10]$. That sounds like a job for $y=9x+1$: first multiply the value in $[0,1]$ by $9$, and then add $1$. If you are determined to have only the integers, you could consider rounding (though I do not see any statistical advantage to doing so). Yes, this approach will have a lot of values down towards one, but think about what it means if you manage to get a seven or a ten!
