Imagine you snap a straight stick randomly at two different points. What is the probability that you can make a triangle with the remaining pieces?
@trafalgar in his satirical job interview post says
And just when you're starting to get good at not being a wisecrack, the curve ball hits you smack in the face: Imagine you snap a straight stick randomly at two different points. What is the probability that you can make a triangle with the remaining pieces? What?? Are you kidding me?
Ok, I'm going to try and answer this (yes, I'm a fun guy at parties)
For a triangle to form, we need 3 sides ( line segments) such that the sum of any two sides is always greater than the third side.
This is the only condition one needs to consider in this case.
Now imagine the straight stick curved into a circle (for simplicity) and we have to pick 3 , not 2, points on the circle. The reason is that marking a point on a line gives you two separate line segments, while marking a point on a circle will not, since it is looped back on itself.
The key point is this : if the distance between any two points on the circle (in any one direction) exceeds that of the length of the semi circle, the triangle cannot be formed from those set of points
Then, we have following cases :
Distance between any two pair of points exceeds the semi circle length -
No triangle formation possible. Imagine cutting the stick at places such that one of the three pieces has length more than 1/2 length of the stick, call this side A.
Now, sum of B and C, the other sides, will always be < A.
(1- A) < A ; when A > 1/2
All three points lie within a semi circle's length - Triangle formation possible
Now, it essentially boils down to picking 3 points within the length of a semi circle.
First point can be anywhere, and is not subject to probabilities
Second and Third point must lie in the same semi circle as the first point. Each case has a probability of 1/2 ;
1/2 * 1/2 = 1/4
So, we have a 1/4 or 25% chance of having pieces that can be arranged to form a triangle.
Note that we're not physically forming triangles from these sets of points, but merely calculating probabilities from an equivalent case.
@trafalgar is the Asian, I'd expect him to know this in his sleep!!
It must be 100%
right?
Nope, imagine the stick goes from 0 to 1. You broke the stick at 0.1 and 0.2, how will you make a triangle from three sides of length 0.7, 0.2, 0.1?
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The odds are almost exactly 50/50
To form a triangle no one piece can be equal to nor longer than one half the length of the stick. Both breaks must be on opposite sides of the center of the stick. The first break can be anywhere, except the center. The second break must be on the other side of the center of the original stick. The odds are 50/50 exactly if the center is not allowed.
What are the odds that one of the breaks is exactly in the center of the stick? That's a tough question. With a deck of 52 cards, you have 1/52 odds to pick a certain card. If you have 2 decks, the odds are 1/104, etc. How many possible break points on the stick that are not the center? It's not infinite because you probably won't be break apart atoms. But it's such a high number that the odds that one of the two random breaks are exactly in the center is very near to, but not exactly, zero.
So the total is very close to 50/50 odds.
Thank you for the thought provoking puzzle.
Very interesting your article. I will follow you because I love math and you are intelligent. Bravo