Three question related to stick breaking. A stick is of length 1.
1) The stick drops and breaks at a random point distributed uniformly across the length. What is the expected length of the smaller part? (Source: DB interview)
2) In the above experiment, what is the expected length of the larger part? (Source: Made it up)
3) In the third experiment, the stick is dropped and breaks at two points. What is the probability that the three pieces could form a triangle? (Source: V.K.Bansal notes, MathOverflow.net)
Update (18/12/09): Solution: Partial solution by Aaditya Ramdas (CSE IITB 20052009) and Giridhar Addepalli (CSE, IITK alumnus and Yahoo! Sr. Software Engineer) and complete solution by me in comments.
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Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this art...

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Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...
ok, the first 2 parts are probably 0.25, and 0.75, and the third part seems to be some shady calculus thing (plot 3 inequalities n find the area of the region)...
ReplyDeleteu r right.. someone post the solution.. :)
ReplyDeleteSuppose stick is of length 1.
ReplyDeleteif all the 3 pieces are of length < 1/2 , subject to the constraint that sum of them is 1, then they form a triangle.
Two points are chosen uniformly,
Pr [ TF ] = Pr [TF  first point is in 1st half]
if a point belongs to [x , x+dx] , 0<=x<=1/2 , then the other point must belong to [1/2, 1/2+x] ( probability it belongs to this interval is x ).
[integral ( x dx ) 0 > 1/2 ] / (1/2)
= 1/4.
@giridhar.. nice solution.. Thanx
ReplyDelete1) The probability that the break point belongs to first half is 1/2. In this case, the expected length of the small part is L/4. Its L/4 in the other case too. Hence E = 1/2*L/4 + 1/2*L/4 = L/4 :) 0.25
ReplyDelete2) Similarly, E = 3L/4. 0.75 .
This can also be seen by the fact that E1+E2=1. Hence, E2=0.75
3) Let the three parts be of the length x, y, 1xy
Feasible region: x,y>0 and x+y<1
Favourable region: x+y>1/2, x<1/2, y<1/2
So, the probability is 1/4 (One can explicitly see it if one draws the diagram) 0.25
Another Geometric solution ( i don't remember its source) :
ReplyDeleteBased on following FACT  The sum of the length of altitudes (to all 3 sides) from any point inside a equilateral triangle is equal to the Height of it.
In the question under consideration too, we had 3 lengths whose sum is 1.
So,let us consider the equilateral ABC triangle of height 1.
Now consider the triangle formed by joining the midpoints of the sides of ABC,call it DEF ;
Now observe that length of each altitude from any point inside DEF is less than 1/2. Hence satisfying the conditions forming a triangle.
If we consider altitudes from any point outside triangle DEF, one of them will have length > 1/2 .Hence we cannot have triangle with those lengths.
Area( DEF )/Area(ABC) = 1/4.
The above mentioned FACT is used in many other places, basically whenever we have 3 variables with constant sum.
I guess it can serve as a starting point for the "Empty Bucket" problem on this blog.
Nice.. very nice solution..
ReplyDeleteThe proof of the FACT is trivial... did it in class VIII or IX.. :P
But the argument is strong.. thanx..
Although I am not able to see how this applies to empty bucket problem.. Help!!
join both ends of the stick to make a circle. Select 3 points on the circle (cutting points) . These 3 points would divide the stick in three parts. We are selecting 3 points at random. For that we first select 6 points (or 3 pairs of points with their antipodal points) and select a point from each pair. There are 2^3 ways to do this. Out of 8 only 2 will lead to the parts which can form triangle. If we choose 3 consecutive points out of 6 points we won't be able to make triangle.(and there are 6 ways to choose consecutive points). Hence the probability = 2/8 = 1/4. This argument can be used in many geometric probability puzzles like Probability of a random triangle on a circle (or in a disc) to lie in one half. Probability that center of a circle lies in a randomly chosen triangle on a circle. probability of a triangle on a circle being obtuse and many more.
ReplyDeleteFor the Giridhar solution, there is a one to one correspondence between the lengths of the parts of stick and altitudes of the triangle. This must be true but it needs mathematical proof.