What is the probability that a stick of length 1 divided randomly into 3 parts can form a triangle with the three parts?

& then, General Bonus asks u...Soldier, what's the probability that a polygon with (N+1) sides can be made from (N+1) segments obtained by randomly dividing a stick of length l into (N+1) parts?

You gave me the probability formula . Here’s my explanation --

Suppose we mark a circle’s 12 o’clock position with “0”, 3 o’clock with 90, 6 o’clock with 180 etc. So each position on the circle has a unique number from 0 to 359.999999999.

We put N+1 random dots (all at once) on the circle’s circumference, creating N+1 segments. (If any segment exceeds half the circle’s circumference, then we would fail to form a polygon.)

Rotate the clock to align 12 o’clock position to any random dot. Break the circle at that dot (and losing that dot). Now we have N dots and N+1 segments.

Now we color the segments. Say black segment is first from the 12 o’clock position, 2

^{nd}is red segment.
P (black segment exceeding half the circle) = P (all N dots landing beyond 6 o’clock position) = 2

^{-N}
P (red segment exceeding half the circle) ? Now keep all the dots and the colors. Rejoin the circle. Rotate it by the length of black segment. Break circle there. Now red segment is the new 1

^{st}segment. P (red segment exceeding half) = 2^{-N}
There are N+1 segments, so P (any segment exceeding half the circle) = (N+1)*2

^{-N}
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