It would depend on the width of the stick. Assuming it to be a piece of string, 100% as all a triangle needs is 3 distinct sides with 3 corners which can be made from any 3 pieces of string!

I assume you're interested in a mathematical answer, not a practical one. What is the probability that two points chosen at random in the interval [0,1] will divide the unit into three segments which can then be rearranged to form the sides of a triangle. Right? The key insight is that no side of a triangle can be longer than the sum of the other two sides. In this case, no side can be greater than 1/2 in length, since the other two sides would then sum to less than 1/2, given that the total of the three sides is exactly 1. Let's say the two chosen points are P and Q. First place P. Now if P<1/2 then let point P' = P+1/2. Otherwise if P>=1/2, then let P' = P-1/2. Note that the segment P-P' will always be inside [0,1] and has a length of exactly 1/2. So now when you randomly place Q it has a 50-50 chance of falling inside or outside the segment P-P'. If Q is outside P-P', then the segment P-Q must be of length greater than 1/2, making it impossible to form a triangle. If Q is inside P-P', then all 3 segments will indeed form a triangle. The answer is 1/2. [Edit:] I think I found the error. My next to last statement, "If Q is inside P-P', then all 3 segments will indeed form a triangle" is wrong. That Q must be inside P-P' (which occurs half the time) is necessary but not sufficient. Defining point Q' in a manner similar to P', we find that P must also be inside segment Q-Q' (which occurs half the time). Since both conditions are independent and must be simultaneously true, the answer is (1/2)(1/2) = 1/4.

Let the random variables take the value P and Q. Draw a square with P to the right and Q upwards. Because the distributions are uniform, the probability of an event is given by the corresponding area within this square. Example: probability that P^2+Q^2 < 1 is given by the area of the quarter circle in which that is true. So it only remains to find the appropriate part of the square for which the conditions are true. First consider only the part of the square where P<Q - this is the part above the diagonal line P=Q. Let the three sides of the triangle in the question be A,B,C for a triangle we must have: longest side <= other two sides. add longest side to both sides: twice longest side <= sum of sides And the sum of the sides is 1m So longest side <= 0.5m The three sides are given by A = P B = Q-P C = 1-Q and the longest side is <= 0.5 ... therefore ALL are less than 0.5m I get P <= 0.5m Q-P <= 0.5 m and 1-Q <= 0.5 m These conditions correspond to throwing away the part of the square for which they aren't true. The first condition throws away the right hand half of the upper triangle. Shade out this part. The second to throwing away parts of the square where Q>=0.5m + P. Shade out this part. The third to throwing away parts where Q is less than 0.5m. Shade out this part. It is easy now to see that the probability that the conditions in the question are true when P<Q is 1/4 because the upper triangle has been divided into four equal triangles, three of which fail the conditions. The lower triangle is the same only with P and Q swapped (like putting a mirror on the line P=Q). Therefore the answer is that the probability is one quarter. Now... something like that is hard to come up with, and easy to make mistakes. So I checked the answer, as you should. I checked it by choosing all possible ways of picking P and Q on the 20cm marks on the ruler. There's 36 ways of doing that of which 6 satisfy the conditions. Thats 1 in 6. So I know I'm not too far off. It wasn't that close though ... there must be an easier way!