dareka asks:
If I give you a length of string and make two random cuts, what is the probability that the remaining three pieces of string can form a triangle?

More precisely, given the real space R and two random variables X, Y of continuous uniform distribution across (0, 1), what is the probability that the following three conditions will be satisufied:
1) min(X, Y) < 1/2
2) max(X, Y) - min(X, Y) < 1/2
3) 1 - max(X, Y) < 1/2

Thoughts:
So converting 1 and 3 into similar formats, we have 1) min(x, y) < 1/2 and 3) max(x, y) > 1/2. Condition 2 can be understood as |x-y| < 1/2. In other words, we need to find the probabilities of the variables x and y satisfying the conditions: one must be greater than 1/2, the other less than 1/2 and the must be no greater than 1/2 apart. Out of the four possiblities of x and y and greater or less than 1/2, two satisfy conditions 1 and 3 so we have a probability of 1/2 for combined conditions 1 and 3.

Given conditions 1 and 3, we can look at condition 2: if X < 1/2 and chosen, what's the probability that Y (which will be higher than 1/2) is within 1/2 of X? If X is close to 0, then the probability for a valid Y will be low and if X is close to 1/2, the probability for a valid Y will be high. At the center of it all is X=1/4, for which Y will have a 50/50 chance of staying within 1/2 of X. Since all the other probabilities for other choices of X are in inverse proportion, everything balances out and we get a value of 1/2 probability. So of all the times when conditions 1 and 3 are satisfied, the numbers will fall within 1/2 of each other half the time, bringing the final probability down to 1/4.

Also, given conditions 1 and 3 (which occurs half the time), we can think about condition 2 visually. If given a "bar" that represents the continuum from 0 to 1 and we sprinkle dots within this bar, what are the chances that the sprinkles will fall into a darker 1/2 length band? No matter where we move the band, the probability is 1/2 (fill the bar with dots and then slide the darker band around).

Makes sense to me anyway, though I may be forgetting some sort of dependence/independence relationships. But that's my line of thought.