So this was a question that I sought to solve after its similarly phrased Triangle version. I found two answers on: here and a more general case here. These yield two different answers, with the former approx $39$%, whereas the second $50$%. I found the second one far more intuitive, and was hoping someone on here could clarify the difference.
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In both What is the probability that the resulting four line segments are the sides of a quadrilateral? and the probability of those n broken parts of sticks to form a closed polygon?, it is possible to use a model in which we make one break in the stick at a time.
In both models, we start by picking a point randomly and uniformly between the two ends of the stick, and breaking the stick there. The difference between the two models is how we make the other two breaks.
In the first model, we are required to break each of the two pieces again. So we can first break the piece on the left somewhere with a uniform random distribution between its two ends, and we can then break the piece on the right with a uniform random distribution between its two ends.
In the second model, there is no requirement that each piece be broken again. We just put the pieces back in line as if they had not been broken, again pick a point uniformly at random between the two ends of the original stick, and break the stick at that point.
In the second model, if the first break in the stick breaks it into pieces in a $3$-to-$1$ ratio, that is, the larger piece is $75\% = \frac34$ of the original stick, each of the next two breaks has an independent $\frac34$ chance to occur on the larger piece. That is, there is a $\frac34\times\frac34 = \frac9{16} > \frac12$ chance that we will break the larger piece twice and not break the smaller piece again. That is, the most likely result for the last two breaks is that both occur on the larger piece. But in the first model this is impossible.
On the other hand, if we initially break the stick in a $3$-to-$1$ ratio, there is a $\frac1{16}$ chance in the second model that both the other breaks will occur in the smaller piece, whereas the first model guarantees that the larger piece gets broken again.
In order to have a quadrilateral we need to break the larger piece enough that no remaining piece has length more than $\frac12.$ Breaking it twice tends to make this more likely. So this effect tends to favor a quadrilateral more in the second model.
On the other hand, not breaking the larger piece at all makes the quadrilateral impossible. This effect tends to favor a quadrilateral more in the first model.
But the two effects are not equal. The best way to see this is to follow each of the solutions to the other questions carefully and see why one comes out to $39\%$ and the other comes out to $50\%.$ (I also find the $50\%$ solution neater and more intuitive, but I believe that's because it is solving what is fundamentally a simpler and more symmetric problem.)
David K
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