How can I solve this joint probability problem? 
Two points are dropped at random onto the unit interval, creating
  three sub intervals this way. 
Find the probability that the central sub interval will be two times
  shorter than the right one.

I tried to solve it in the following way:
the length of the central interval $= y - x$
the length of the right interval $= 1 - y$
so, the favorable outcome would satisfy the following inequality,
$y-x \le \frac{1}{2} \cdot (1-y)$
$\Rightarrow 3y - 2x \le 1$

Now, how can I advance further to find the probability?
 A: You look at the random variable 
$$Z = 3Y - 2X \tag{1}$$
and you want to determine the probability 
$$P(Z\leq 1 \mid Y > X) \tag{2}$$
Using Bayes' formula we have
$$P(Z\leq 1 \mid Y > X)= \frac {P(Z\leq 1 , Y > X)}{P(Y > X)} \tag{3}$$
$Y$ and $X$ are independent continuous uniform $U(0,1)$ random variables, so 
$$P(Y > X) = \int_0^1\int_x^1dydx = \int_0^1(1-x)dx = 1-\frac 12 x^2\Big |^1_0 = 1/2 \tag{4}$$
For the numerator, we have
$$P(Z\leq 1 , Y > X) = P(3Y - 2X\leq 1 , Y > X) = P\left(Y \leq \frac {1+2X}{3} , Y > X\right)$$
$$= P\left (X \leq Y \leq \frac {1+2X}{3}\right)= \int_0^1\int_x^{\frac {1+2x}{3}}dydx = \int_0^1\left (\frac {1+2x}{3}-x\right)dx$$
$$=\frac 13 \int_0^1dx - \frac 13\int_0^1xdx = \frac 13 - \frac 16 = 1/6 \tag{5}$$
Inserting $(4)$ and $(5)$ into $(3)$ we get
$$P(Z\leq 1 \mid Y > X) = \frac {1/6}{1/2} = 1/3 \tag{6}$$
This is the mathematical approach that is equivalent to the geometric argument provided in @MikeP answer. 
A: I think the critical piece is to determine the universe of potential events, which isn't just [0,1] [0,1].  Since you've set the problem up with y being greater than x, the universe of possibilities is just the area above the y=x line (i.e. 1/2 unit area).  Of that, the area below the line you plotted is what satisfies the even of interest.  Since it starts at y=1/3, the area of the triangle from y=x to that is 1/2*1/3*1 = 1/6.  Dividing 1/6 by 1/2 gives you the answer of 1/3.
A: Another (!) way of solving it is as follows.  Note that the size of the leftmost interval is irrelevant; all you care about is the combined rightmost and central intervals.  This combined interval has some length $L$, and, by construction, is divided into two sub-intervals by having a uniform($0,L$) variable, label it $x$, "dropped" into it.  The leftmost of those two sub-intervals (which corresponds to the original middle interval) is $\leq$ half as long as the right interval (which corresponds to the original right interval) if $x \leq L/3$, which will obviously happen $1/3$ of the time. 
A: Somewhat hesitantly I suggest zero.
If we think of throwing the two markers (points) one after the other (a perfectly reasonable procedure), then once the first one has landed the second one must land in one out of two positions out of infinite possibilities (presuming we're dealing in fractional lengths) to make the ratio exactly 1 : 2

On the other hand, if we interpret the question to mean half or less than half, I, also hesitantly, suggest 5/12.
Thinking of the markers (points) landing one after the other again, let the first one land a distance X from the right hand end of the interval (i.e. from 1). The second one can land either side of the first one.
If it lands on the right hand side of the first one, the chance that the middle interval is a half or less than a half of the right hand one is 1/3. All in all the chance of this happening is therefore X/3.
If it lands on the left hand side of the first one, the chance that the middle interval is half or less than half the right hand one is X/2 divided by 1-X. But then the chance of the second one landing to the left of the first one is 1-X. Therefore the chance of landing to the left and leading to the desired outcome is X/2.
Adding together we get X/2 plus X/3, i.e. 5X/6. So, we need the expected value of this as X varies between 0 and 1. if we presume all values of X are equally likely we end up with 5/12
