1. ## Triangle probability puzzle

I'm stuck with a question: Two points are selected at random on a straight line segment of length 1. What is the probability that a triangle can be constructed from the results line segments.

So far all I can come up with is:
• Points p1 and p2 have equal probability of 1.
• Lengths a, b and c can be formed from two cases; if p2 > p1 then a = p1, b = p2 - p1 and c = 1 - p2. For the other case, p1 and p2 are reversed.
• The inequality $\displaystyle |(a^2 + b^2 - c^2)/(2ab)| <= 1$ must be satisfied. I can substitute p1 and p2 in, and imagine some double integral over p1 and p2 for each case, but I don't know how to deal with the inequality.

2. ## Re: Triangle probability puzzle

Originally Posted by entropyslave
I'm stuck with a question: Two points are selected at random on a straight line segment of length 1. What is the probability that a triangle can be constructed from the results line segments.
You can think of ordered pairs: $\displaystyle \{(x,y):~0\le x <y\le 1\}$.
There are further restriction:
$\displaystyle \begin{array}{l} y>1-y\\ x+1-y>y-x\\ y-x+1-y>x \end{array}$

Now plot that area in a unit square. The probability is area.

3. ## Re: Triangle probability puzzle

Don't you also need the probability that the 2 points will be in same "spot"?

4. ## Re: Triangle probability puzzle

Originally Posted by Wilmer
Don't you also need the probability that the 2 points will be in same "spot"?
I have no idea what that could possibly mean!
This is a standard problem. A good reference is Jim Pitman's Probability~.
He calls this the the uniform triangle space.

In general, "the probability that the 2 points will be in same "spot"" has probability zero for continuous distributions.

5. ## Re: Triangle probability puzzle

No need to get excited, Mr Plato; I didn't know before, now I do; thank you.

6. ## Re: Triangle probability puzzle

Another way to approach this problem is like this:

1. Assume the first point selected in the interval (0,1) is called point A. Consider the case A < 1/2. In order to have a triangle the second point to be positioned (call it B) must be placed somewhere in the interval between (1/2,1/2 + A). If B falls outside that interval then the line segments can't form a triangle. For example: if point A = 0.3 then point B must lie between 0.5 and 0.8. The width of that interval is A, and hence the probability that B will be positioned in that interval is also A.
2. Now consider if A>1/2. In this case B must land in the interval between A-1/2 and 1/2. The probability of that happeng is 1-A.
3. So you have a continuous function P(triangle) = A for A between 0 and 1. The resulting integral is:

$\displaystyle \int_0^{1/2} A dA + \int_{1/2}^{1} A dA = \frac 1 4$

7. ## Re: Triangle probability puzzle

Thanks for all the help folks. The way I (hopefully) solved it is as follows (probably a bit long-winded):
Two points, x1 and x2. Two (symmetric) cases, so considering $\displaystyle x_2>x_1$:
To make a triangle from a unit length, all sides must be < 1/2, i.e. $\displaystyle x_2>1/2$ and $\displaystyle x_2-x1<1/2$

$\displaystyle p(x_2-x_1<1/2, x_2>1/2 | x_2>x_1) = \int_{1/2}^1p(x_2-x_1<1/2|x_2>1/2,x_2>x_1,x2)p(x_2>1/2|x_2>x_1,x2)p(x_2|x_2>x_1)dx_2$

where $\displaystyle p(x_2|x_2>x_1)=x_2$ and $\displaystyle p(x_2-x_1<1/2|x_2>1/2,x_2>x_1,x2)=1/2x_2$ so the final integral is:
$\displaystyle = \int_{1/2}^{1}1/2dx_2=x_2/2|_{1/2}^{1}=1/4$

For $\displaystyle x_1>x_2$, the situation is the same with the variables switched, so the final probability is twice the above, i.e. 1/2

8. ## Re: Triangle probability puzzle

I think you're off by a factor of 2. Your last step was to double the value you calculated, so as to cover the case x1< x2 and x1> x2. But in doing that you need to multiply the first by the probability that x1<x2 and the second by the probability that x1>x2, both of which = 1/2.