How do I solve this probability problem with Probability?

Question

Consider a stick of length 1. Pick two points uniformly at random on the stick, and break the stick at those points. What is the probability that the three segments obtained in this way form a triangle?

I have implemented two versions. Version 1 works correctly, but version 2 doesn't. How can I fix the second version?

Version 1

Module[{min,  max},
  {min, max} = {Min@#, Max@#} & /@ RandomReal[1, {10^5, 2}] // Transpose;
  Mean@MapThread[
    Function[{x, y, z}, N@Boole[x + y > z && x + z > y && y + z > x]],
    {min, max - min, 1 - max}]]

Version 2

Block[{t1, t2, t3},
  t1 = {x > 0, y > 0, x + y < 1};
  t2 = {x + y > z, x + z > y, y + z > x} /. z -> 1 - x - y;
  Print[t3 = And @@ t1 ~ Join ~ t2 // FullSimplify];
  Probability[t3,
    Distributed[x,  UniformDistribution[{0, 1}]] &&
    Distributed[y,  UniformDistribution[{0, 1}]]]]

Answer 1

I think you are mixing up the "coordinate" of the breaking point with the length of the resulting stick segment. The length of the resulting stick segment is y-x, (if y>x) not y. If we want x and y to be uniformly distributed, it is is useful to think in terms of the length of the leftmost stick segment l1=Min[x,y] and the length of the center stick segment which is l2=Max[x,y]-Min[x,y] (=Abs[x-y])

This seems to be what you want

Block[{t1, t2, t3},
 t1 = {1 > x > 0, 1 > y > 0};
 t2 = {Min[x, y] + (Max[x, y] - Min[x, y]) > z, 
    Min[x, y] + z > (Max[x, y] - Min[x, y]), (Max[x, y] - Min[x, y]) +
       z > Min[x, y]} /. z -> 1 - Max[x, y];
 Print[t2];
 Print[t3 = And @@ t1~Join~t2 // FullSimplify];
 Probability[t3, 
  Distributed[x, UniformDistribution[{0, 1}]] && 
   Distributed[y, UniformDistribution[{0, 1}]]]]

-> 1/4

Possibly the following is nicer

Block[
 {t1, t2, t3, x, y, l1, l2, l3},
 t1 = {1 > x > 0, 1 > y > 0};
 t2 = {l1 + l2 > l3, l1 + l3 > l2, l2 + l3 > l1} /. {l1 -> Min[x, y], 
    l2 -> Abs[x - y], l3 -> 1 - Max[x, y]};
 Print[t3 = And @@ t1~Join~t2 // FullSimplify];
 Probability[t3, 
  Distributed[x, UniformDistribution[{0, 1}]] && 
   Distributed[y, UniformDistribution[{0, 1}]]]]

-> 1/4

Answer 2

Amazingly, the most direct and mindless possible approach works instantaneously.

Start by characterizing the valid side lengths: each side is the shortest distance between its endpoints; the path made by the other two sides cannot be any shorter.

triangleQ[x_, y_, z_] := x <= y + z && y <= z + x && z <= x + y;

Break the stick uniformly and independently at locations $u_1$ and $u_2$:

f = UniformDistribution[{{0, 1}, {0, 1}}];

Noting that the breaks (from left to right) create pieces of length $\min(u_1,u_2)$, $|u_2-u_1|$, and $1 - \max(u_1, u_2)$, request the probability that the pieces form a triangle:

Probability[triangleQ[Min[u1, u2], Abs[u2 - u1], 1 - Max[u1, u2]], {u1, u2} \[Distributed] f ]

$\frac{1}{4}$

Answer 3

Each uniform divides the stick into a smaller and a longer side. If the smaller side of both uniforms coincide (left-left, or right-right) then you won't have a triangle because the rightmost/leftmost division will be longer than 0.5. The odds of this happening is 1/2. If they don't coincide, then you will have a triangle only when the sum of the smallest sides is higher than 0.5. So,

1/2 Probability[
  x + y < 1/2, {x, y} \[Distributed] 
    TransformedDistribution[Min[z, 1 - z], 
     z \[Distributed] UniformDistribution[]] // Thread]

(* 1/4 *)

By the way

With[{min := Min[x, y], max := Max[x, y]}, 
 RegionPlot[
  And @@ Thread[0 < {x, y} < 1] && 
   Max[min, max - min, 1 - max ] < 1/2, {x, 0, 1}, {y, 0, 1}]]

Answer 4

You can also use OrderDistribution (to get the joint distribution of Min and Max in a sample of size 2 from a standard Uniform distribution) combined with a simpler condition:

dist = OrderDistribution[{UniformDistribution[], 2}, {1,2}]; 
Probability[y > 1/2 && 1/2 > y - x && 1/2 > x, Distributed[{x, y}, dist]]
(* 1/4 *)
Probability[Max[x, y - x, 1 - y] < 1/2, Distributed[{x, y}, dist]] (* thanks: Rojo *)
(* 1/4 *)

Answer 5

This is just a variant of Rojo's argument.

If one considers the lengths of the segments rather than the points then:

Probability[
 x + y > 1/2 && x < 1/2 && y < 1/2 \[Conditioned] 
  x + y < 1, {x, y} \[Distributed] UniformDistribution[2]] 

yielding 1/4

You can see this as follows:

FullSimplify@
 Reduce[x + y + z == 1 && x + y > z && x + z > y && y + z > x && 
   x > 0 && y > 0 && z > 0, {x, y, z}, Reals]

gives:

2 x < 1 && x + y > 1/2 && 2 y < 1 && x + y + z == 1

A little fun:

c = x + y > 1/2 && x < 1/2 && y < 1/2;
rp = RegionPlot[{x + y < 1, c}, {x, 0, 1}, {y, 0, 1}, 
   PlotStyle -> {LightRed, LightBlue}, BoundaryStyle -> None, 
   ImageSize -> 300];
tf[p_] := Module[{u, v, tg},
  {u, v} = p;
  tg = Quiet[SSSTriangle[u, v, 1 - u - v]] /. 
    SSSTriangle[__] :> Text["Does not form a triangle", {0.5, 0.5}];
  Graphics[tg, PlotRange -> Table[{0, 1}, 2], Frame -> True, 
   ImageSize -> 300]]
Manipulate[
 Row[{Show[rp, Graphics[Point[pt]]], tf@pt}],
 {{pt, {0.4, 0.4}}, Locator}]