Probability that Three Pieces Form a Triangle

MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
Instructor: Jimmy Li

License: Creative Commons BY-NC-SA

[mikekalish]

Simple, step by step logic. Love it! Well done!

[emindenizozkan]

Jimmy Li. You are the best TA. You are simply awesome!

[rabomeister]

For anyone looking, this is the same question in "Intermediate Course in Probability" by Gut - Chapter 4 Q10 - Order Statistic

[animmikamangaka]

Omg thank you, I've spent hours reading on forums how to figure this out, but this video made it all clear so thank you

[jacoboribilik]

(A, B, C) forms a triangle iff 2max(A, B, C)<A+B+C. In our problem then max(A, B, C)<1/2. The maximum might either come from the segments on the sides or from the one in the middle and the maximum must always be between 1/3 and 1. Then + Pr(MiddleMax<1/2|MiddleMax) Pr(Middle>1/3) + = 0.25x3x(1/3)= 1/4.

[rajatmodi]

we assumed x<y for this problem. so there is no need to explicitly think of the case that y>x. just imagine that the experiment is set up such that x<y [always]. then, we always lie in the upper triangular region of the square 0<=x, y<=1.. since the density is uniform in upper triangle, it has to be 2 at every point so that it integrates to 1. then the answer is simply the area of a concerned triangle multiplied by 2.

[ayaan]

Why does the order of X and Y matter? If we are looking at unique ways of cutting the initial line, won't we end up counting the same configurations twice?

[JamUsagi]

Commenting before the video:

If there are only 3 pieces, then the longest piece must be equal to or longer than 1/3 of the total length, otherwise one of the other pieces are longer. So there is always at least 1 piece longer than 1/3.

To make a triangle, the longest piece can’t be longer than 1/2 of the total length, otherwise the other two pieces aren’t long enough to form a triangle with it.

Of cases where the longest length is >1/3 long but <1 (i.e. all cases), only 1/4 of cases have the longest length <1/2, and are thus able to make a triangle.

So I think the answer is 1/4.

Idk if the working out makes sense, but I’m pretty confident in the answer.

[edmondmalepane]

I have two sticks, one of length one metre and the second of length two metres.
Suppose I break the two-metre stick randomly into two sticks so that a point
along its length is equally likely to be the breakpoint. Find the probability that
three sticks can be put together to form a triangle. is this question the same as the one you were doing?

[kirtiverma]

Uncle why uncle?
Consider point (x=1, y=1/2) that mean sides of triangle would be 1, 1/2, -1/2
Why considering -ve lengths?

[andresrossi]

I'd love to know how to make it formally working with density functions

[jonasschmidt]

So everybody seems to reach the 25% conclusion, but because of the following reason I just cant help but think that it is 50%.

So I will explain it in two different ways, for both cases the prequisite for making a triangle is that the two cuts are on opposing sides of the middle of the stick, also for both cases I will not look at the unique case where there is a cut in the exact middle, since the chance of that happening, mathematicaly is infinitely small.


1. You have the first cut, it is either to the left or the right of the middle, the chance for the second cut to be on the same side of the first is 50% so the chance for being able to make a triangle is 50%, case closed.

2. Think of the stick as a coin, left side from the middle is heads right side is tails, each cut is a flip of the coin, you either cut left of the middle or right of the middle (heads or tails). So basicly you have two coin toss which gives 4 different outcomes: Head-Head, Tails-Tails, Head-Tails or Tails-Head. Each outcome has a 25% chance to happen, which means you have 50% chance to have two cuts on the same side and 50% chance to get the two cuts on opposing sides. Still the chance to be able to make a triangle is 50%.

So anyone who can tell me what it is that im missing?

Also if you add in the unique cases where either one cut is in the exact middle or the two cuts are exactly on top of eachother you get that the chance of making a triangle is just shy of, but infinitely close to 50%.

[tayyabashaukat]

Informative ...what will be the probability if the three pieces are equal??..can you explain it please.

[geocarey]

Great video. I looked at the probability of a triangle if you specify that the second break has to be to the LEFT of the first break.
Following your reasoning it looks like you then exclude the lower shaded area so only X<Y and the probability is 1/8.
I tried a simulation in Excel, but get a probability near 1/5. Any thoughts?

[bowtangey]

What if you first make one break into two pieces, then randomly break the longer of your two pieces?

[siyaramkumar]

Zach star have given a much easier explanation on his channel
Though this one's more interesting

[phaniram]

awesome, but i wonder if people at MIT could just pop solution out of their minds or had to solve it for say upto an hour?

[simonandriessen]

You can also do this problem with help of an infinite loop in action. Take the line. Mark the 1/4 1/2 and 3/4 points. Cut x can now lie in four defined regions. So can cut y. If they lie in the most inward regions, but differen ones (x int the region from 1/4 to 1/2 and y in the region from 1/2 to 3/4 or reversed x and y) then you're good, as you can see no piece will be longer than half the line. This probability is 1/8. For x in the second region and y in the fourth (or x and y switched) and x in the first region and y in the third (or x and y switched) it isn't determined yet if that is all you know about the location. This probability is 1/4. The posibilities left are : both in the same region, or both in different outward regions (1 and 4) and it is clear that in that case you always have a part longer than half. Now we take the indeterminated cases. We can solve for region combo 1 and 3 because the other is analogical. You devide both regions in equal smaller regions. You can see that now the same pattern continues, only with 1/4 working immediately, 1/2 being inditerminate and 1/4 not working. This proces of yes maybe and no parts continues into infinity, eventually leading to getting half of the cases working and half not working. As such half of 1/4 is added to the 1/8 in the first part, and this is 1/8 +1/8=1/4 the same answer. This method is harder but I thought of it so I'm proud of it.

[mohamedsamsudeen]

Sum of two small pieces B+C must be larger than A. Break the stick into 2 pieces A and X. The probability of selecting the large piece is 50%. Break the large piece(X) into two B and C. Let say C is larger than B. So C is either larger than A or not. So the probability is 50%. Both of these events have to occur so multiply for total probability which is 25%

[karngyan]

what's the generalization of this .. for n sided polygon formation.