How does the cardinality of the set of all probability measure on a set $X$ change according to the cardinality of $X$? I was wondering concerning the following problem:  

Take $X$ as a parameter space endowed with its Borel $\sigma$-algebra.
   What is the cardinality of $\Delta (X)$, understood as the set of all probability measure over $X$? 

[When I write that $X$ is a parameter space, I am thinking about the cardinality of $X$ as finite, countable or uncountable]
If $X$ is a doubleton, we should already have $| \Delta (X) | = \mathfrak{c}$. 
What happens if we move on?
Also, how does this relate to the cardinality of the Borel sets?
Any feedback is most welcome.
Thank you for your time.
 A: This is only a partial answer (however, a complete question does not seem to be possible here).
Since you're mentioning Borel $\sigma$-algebra, I presume that $X$ is a topological space. 
Extending Ross Millikan's comment, If $X$ has a countable base $B$, then $|\Delta(X)|=\mathfrak{c}$ unless the only non-empty open set is $X$. Indeed, in this case the measure is completely determined by its values on sets of the form $A_1\cap\dots\cap A_k$, $A_i\in B$ (since it is a $\pi$-system generating the Borel $\sigma$-algebra). But there are only countably many such sets.  So the cardinality is at most $\mathfrak c$. On the other hand, it is at least $\mathfrak c$ by OP. In particular, for any separable metric space with at least two elements, $|\Delta(X)|=\mathfrak c$.
What can be said besides that, is an interesting question. Say, if $X$ is a non-separable metric space, then, obviously, $|\Delta(X)|\ge |X|$ (considering delta measures on singletons), and also $|\Delta(X)|\ge \mathfrak c$ (which would follow from the previous if we accept CH). 
However, it is hard to bound $|\Delta(X)|$ from above. A helpful idea is that there cannot be an uncountable collection of disjoint sets of positive probability. In particular, the cardinality of set of all discrete probability measures is at most $(X\times \mathbb R)^{\aleph_0}$ (to each such measure we associate a sequence of points from $X$ and their corresponding weights), which is $\max(\mathfrak c, |X|)$. 
Unfortunately, we cannot go much beyond this with such argument. We could write that each point from $X$ has a countable base of neighborhoods, and in total they make a base $B$ of topology. But the above argument will fail, since $B$ might not generate the whole Borel $\sigma$-algebra (since it is uncountable). It might still be that the values of this probability measure are completely determined by its values on some countable subfamily of $B$ (since, as I wrote, there can't be an uncountable disjoint family of sets with positive measure), but I am far from being sure in this. 
