I have a 1-meter long stick. Let $x,y$ be i.i.d with uniform ([0,1]), representing the cut in the stick. What is the probability that the 3 segment form a triangle?
Attempt:
Pr{triangle} = Pr(triangle | x < y) Pr(x < y) + Pr(triangle | x > y) Pr( x > y).
Let's look at the first term. Pr(triangle | x < y) requires:
1) x + y-x > 1-y
2) x + 1 -y > y-x
3) y-x + 1-y > x
Which is equivalent as
1.y >0.5,
2. x < 0.5
3. y - x < 0.5.
This is a triangle which has the area of 1/8.
Due to symmetry, Pr(triangle | x > y) = 1/8.
Now. Pr{triangle} = Pr(triangle | x < y) Pr(x < y) + Pr(triangle | x > y) Pr( x > y)
= 1/8 *0.5 + 1/8 *0.5 = 1/8 $\neq$ 1/4 (as posted here)
Confusion: Did I do something wrong?
Formally, a triangle can be formed iff no piece exceeds 1/2 in length. This is correctly summarized by your equivalence.
Your calculation of the probability of this event is slightly off, since you forgot to assume that x>y is given. The triangle has area 1/8, unconditional on $x>y$. Notice that your equivalence is a subset of the event $x<y$. So:
$P(y>0.5,x<0.5,y-x<0.5|x<y)=\frac{P(y>0.5,x<0.5,y-x<0.5 , x<y)}{P(x<y)}=\frac{1/8}{1/2}=1/4.$