Can I use an SVM for labeling data as more than one class I'm trying to classify e-mails using Mallet. If the classifier is too unsure about a new e-mail I would like a user to do the classification instead.
I figured I could use the Mallet Labeling output for this:
Labeling labeling = classifier.classify(instance).getLabeling();

My problem is that when I use an SVM my labels are all of certainty 1 or 0.
CLASS1: 1,0         CLASS2: 0,0         CLASS3: 0,0         CLASS4: 0,0         CLASS5: 0,0 

I would like the output to be more like this:
CLASS1: 0,9945          CLASS2: 0,0047          CLASS3: 0,0005          CLASS4: 0,0001          CLASS5: 0,0001  

The second output was generated using the Maximum Entropy classifier.
My question is:
Can an SVM give me estimates between 1 and 0? Is this behavior typical of an SVM or is it just the implementation I'm using.
Follow up question: Is the Maximum Entropy classifier generally just as good as an SVM when it comes to text?
 A: SVM are not designed to estimate probability of a class. Nevertheless, there is a way to estimate such probabilities, using Platt scaling.
There is SVM-like probabilistic method, namely, Relevance Vector Machines. Unfortunately, they're patented by Microsoft, though, apparently, free to use in academia.
Finally, if you want to find these xs where SVM is unsure about correct labeling, you can do it even without probability. The decision rule of SVM is
$$
class(x) = \begin{cases}
1, & \mathbf w^T x + b > 0 \\
0, & \mathbf w^T x + b < 0 
\end{cases}
$$
Where $\mathbf w^T x + b$ is SVM's decision function, effectively, a signed distance to the separating hyperplane. If this distance (modulo of it) is too small — SVM is less sure about $x$. You can come up with some kind of threshold on decision functions to find vectors of interest.

About MaxEnt vs. (linear) SVM:
It's mostly opinion-based, but their loss functions are quite similar

SVM's loss is called "Hinge loss", MaxEnt's — "Logistic".
Asymptotically (as signed distance goes to $\pm \infty$) they're the same (modulo scaling constant because of arbitrary log base in Logistic loss), so if you data is well-separable, both should perform fine. Both methods will find a separating hyperplane, though, SVM's one will have a bigger margin (because it's a max-margin classifier).
When data is not linear separable, SVM will try to push as many points behind the boundary of distance = $\pm$ 1, thus having 0 loss on them, while logistic loss may tolerate some errors to become even more confident in already correctly labeled samples.
This is my opinion, it is not necessarily true :-)
A: Just to add to what @Barmaley.exe said (+1), if you want probabilities, then use an algorithm that was designed for that purpose, such as the Relevance Vector Machine, Gaussian Process classifier or Kernel Logistic Regression (I like KLR and use it a lot, e.g. here which gives some benchmark results compared to SVM and Gaussian process classification etc.).  Platt scaling will give probability estimates, but they are not likely to be as good as those from a model designed to give a probabilistic output.
One of the ideas behind the SVM is that it is better to solve the problem directly, rather than solving a more general problem and simplifying the answer.  For instance if you are only interested in discrete yes/no classification, then it is better to try and infer the decision boundary directly, rather than estimate the posterior probability of class membership (a more general problem) and then threshold at 0.5 (the simplification step).  The reason for this is that model resources are not allocated to improve the performance of the model away from the decision boundary and the problem of estimating the parameters is simplified, and less data are likely to be required.  The SVM then is likely to do a good job of estimating the decision boundary (or p = 0.5 contour of the probability), but the raw output of the SVM will not necessarily be a good basis for estimating the probability away from the decision boundary. 
