Random cuts form a quadrilateral

[BMAD]

If you randomly cut a stick (straight cuts) into four pieces, what is the probability that the resulting pieces will make a quadrilateral? Pentagon? Octagon?

[bonanova]

Even in the simpler puzzle where two random cuts are to make a triangle, the probability depends on the manner that the cuts were made: (1) choose two random break points, (2) break at a random point, then break the larger piece, (3) there may be others. I take this puzzle to mean choice (1).

The first cut is made a distance f from the nearer end.

If neither of the next cuts falls in the portion of the stick from f to f+.5, we fail.

Both probabilities are 1/2.

For failure, both must happen, with a probability of 1/4.

Probability of success for a quadrilateral is thus 3/4.

For a pentagon, the next three cuts must fail to hit 1/2 of the stick with probability 1/8.

Probability of success for a pentagon is thus 7/8.

By extension, the probability of success for an n-gon is 1 - (1/2)^n-2

[m00li]

Even in the simpler puzzle where two random cuts are to make a triangle, the probability depends on the manner that the cuts were made: (1) choose two random break points, (2) break at a random point, then break the larger piece, (3) there may be others. I take this puzzle to mean choice (1).

The first cut is made a distance f from the nearer end.

If neither of the next cuts falls in the portion of the stick from f to f+.5, we fail.

Both probabilities are 1/2.

For failure, both must happen, with a probability of 1/4.

Probability of success for a quadrilateral is thus 3/4.

For a pentagon, the next three cuts must fail to hit 1/2 of the stick with probability 1/8.

Probability of success for a pentagon is thus 7/8.

By extension, the probability of success for an n-gon is 1 - (1/2)^n-2

but then the answer for triangle will be = 1/2. isnt it 1/4?