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Let $U_1$ and $U_2$ be independent, uniformly distributed random variables on the interval $ I := [0,1] $. We decompose $I$ into three parts $I_1$, $I_2$ and $I_3$.
First I am supposed to give a proof, that the probability, that all lengths of $I_1$, $I_2$, $I_3$ are less than $\frac{1}{2}$, is $\frac{1}{4}$.

Then we consider three random points $V_1, V_2, V_3 \in \mathbb{R}^2$, that are independent, uniformly distributed on the unit circle.
Give the probability that these three points are the corners of an acute triangle (every angle less than 90°).

I have done a few approaches to this but they all ended in nothing.. I assume, I have to use the first part with the intervals to show the second part with the triangle, but I failed at that, too..

Gilligans
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2 Answers2

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The probability that $I_3$ is greater than $\frac12$ is clearly $\frac14$. Now convince yourself that the three interval lengths all have the same distribution by starting with a circle, uniformly choosing $3$ points on it, uniformly picking one of them to cut the circle, producing the unit interval, and noting that the resulting distribution is the same as in your situation; thus all cyclic permutations of the interval lengths are equiprobable.

So every interval has probability $\frac14$ to be greater than $\frac12$; that's three disjoint events whose probabilities add up to $\frac34$; that leaves $\frac14$ for the complementary event that none of them are greater than $\frac12$.

For the second part, use the fact that for points on a circle, a triangle over a diameter has a right angle, so that's the border case between obtuse and acute triangles.

joriki
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  • Thanks for the answer, but I can't quite follow you, how you concluded that every interval has probability $\frac{1}{4}$ to be greater than $\frac{1}{2}$ ? – Gilligans Jun 24 '18 at 13:59
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    @H.Sch: For $I_3$, this is true simply because both points have to be below $\frac12$ in order for $I_3$ to be greater than $\frac12$. An analogous argument also works for $I_1$. For $I_2$, it's not so obvious. To avoid having to perform an explicit integration for $I_2$ (and also to learn something more general and useful about this sort of problem), we can show that the three intervals in fact have the same distribution (and hence also the same probability of being greater than $\frac12$). This is done by chosing the points on a circle, where the cyclic invariance is manifest. – joriki Jun 24 '18 at 14:12
  • @H.Sch: A third point is then introduced (sorry, the original answer said $4$ points, perhaps that was why you couldn't follow, I've corrected it to $3$) to determine where the circle is broken up into an interval. Because each of the three points could equally likely have been this special point, each of the three cyclical permutations of the intervals is equally likely, which shows that the three intervals all have the same distribution and the central interval isn't as different from the outer ones as one might think at first sight. I hope that made it a bit clearer? – joriki Jun 24 '18 at 14:17
  • @H.Sch: The key here is that the uniform distribution is invariant under a cyclical shift of the interval, so it doesn't matter where the third special point is, the other two always end up uniformly distributed on the unit interval. – joriki Jun 24 '18 at 14:19
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    Yes, now it's clear, thank you :) – Gilligans Jun 24 '18 at 14:45
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Another way to solve this is.

Let $u_1$ and $u_2$ be the two cuts in a 0-1 line.

You have the condition

$u_1\lt \frac{1}{2}$

$u_2 - u_1\lt \frac{1}{2}\implies u_2\le \frac{1}{2} + u_1$

$u_2\gt u_1$

Putting all together

we get

The required probability $= \int_{0}^{\frac{1}{2}}\int_{u_1}^{\frac{1}{2}+u_1} du_2du_1 = \frac{1}{4}$

Satish Ramanathan
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