Doubly Broken Stick

A stick of length 11 is broken simultaneously at two points chosen independently and uniformly at random along its length, producing three pieces. You are allowed to discard exactly one of the three pieces, chosen after observing their lengths, and keep the remaining two. Alternatively, you are allowed to just keep the three pieces you observed. If you decide to discard a piece, you then break the longer of the kept pieces at a uniformly random point along its length. What is the probability that the final three pieces can form a triangle under the assumption that you are looking to maximize this probability? Round your answer to three decimal places. It may help to have a calculator to solve a resulting integral.

A

About 0.7500.750: always discard the smallest piece and re-break the largest remaining piece.

B

About 0.6750.675: keep the original pieces if they already form a triangle; otherwise discard the extreme piece that makes the two kept pieces most balanced.

C

About 0.2500.250: never discard, because the classical broken-stick probability is optimal.

D

About 0.5000.500: always discard the largest piece and re-break the longer remaining piece.