Why is the probability the same when drawing two marbles in different order I have 10 red, 15 blue and 5 green marbles. I draw two times, one marble each time without replacement.
Why is the probability of "first red then blue" the same as "first blue then red" intuitively speaking?
$$p(\text{red then blue}) = \frac{1}{3} \frac{15}{29} = \frac{5}{29}$$
$$p(\text{blue then red}) = \frac{1}{2} \frac{10}{29} = \frac{5}{29}$$
Here is the tree diagram:

 A: "Draw first one, then another" can also be thought of as "draw two simultaneously, then order them". Clearly, if you've already drawn a red and a blue marble, but not ordered them yet, then either ordering is equally likely.
If we look at the math and don't simplify our fractions, we can see that in either case we get
$$
\frac{\text{Number of red marbles}\cdot \text{Number of blue marbles}}{\text{Number of marbles before first draw}\cdot \text{Number of marbles before second draw}}$$
so another way to answer to your question is "because multiplication is commutative"
A: A random process where the probability of $X_1,X_2,...,X_n$ does not depend upon the order of the realizations is called $exchangeable$.  This characteristic occurs for a lot of simple processes, like drawing balls from an urn.  Let me turn your question on its head, "When would order actually matter?"  Imagine a situation where a genie replaced every red ball you drew out with two more new red ones.  Then order would matter.  In the real world, if and only if the sequence of events does not depend upon the history of how one arrives at the $n-th$ event, then the process is exchangeable.  So one needs to worry about path-dependence of events in most realistic examples.    
