Cool. I wonder, can you also calculate the mean lengths of the other n-1 pieces?

The size of the kth largest piece is given byI just noticed that this was a problem on IBM's brainteaser website:Ponder Thiswhich in particular offers a more elementary explanation of the result about Z_n that I derived by integration.

Right. It is rather amusing, isn't it, that harmonic numbers, being the subject of an active discussion in another thread, show up in a seemingly unrelated problem.Edit: I think your formula gives the mean length of the kth smallest piece.

No, it is the kth largest---the 1st largest has the harmonic sum from 1 to 1/n as a factor, which is larger than the 2nd largest, which sums only from 1/2 to 1/n, which is larger than the 3rd largest, which sums only from 1/3 to 1/n, and so on, which is what makes sense.

Yes, you are certainly right, my mistake.Here is yet another way to explain the formula for the smallest piece.If we start with n=3, we can note that the conditions and define a triangle of height 1/3. The average value of x over this triangle is obviously 1/9.For n=4, the conditions are and we get a pyramid of height 1/4.Similarly, in the general case (the conditions are already spelled out earlier), we get an (n-1)-simplex of height 1/n, its base being the (n-2)-dimensional face of the unit hypercube defined by x=0. The mean value of x over the simplex is

Hey all,I have defined a variable x representing the length of the bar on the left-hand side.The length y of the smallest bar is then defined with l the length of the original bar: Finding the expectation of y is fairly straightforward with integrals and I find: What a disappointing results this is....

Both solutions of sevvost and gentinex are very nice.Just came back to this forum now.

Answer is E(min) : 1/4, E(max) = 3/4.(edited ), sorry was a bit drunk last nite, in some crazy mood, embarass me no more!

QuoteOriginally posted by: pgeekThere is ONLY ONE SOLUTION to the first problem, that is 1/3 (not 1/4) , PM me if u want the solution. u need to find an expectation of a min function of 2 dependant random variables, X and Y, every other solution will have an error..SHOUTING certainly makes for a very compelling argument, especially when no others are presented. I guess if you used all caps throughout your eloquent post, you would be awarded victory by TKO.

QuoteOriginally posted by: pgeekThere is ONLY ONE SOLUTION to the first problem, that is 1/3 (not 1/4) , PM me if u want the solution. u need to find an expectation of a min function of 2 dependant random variables, X and Y, every other solution will have an error..1/3 is indeed the solution to a different question: given two independent random variables X and Y chosen uniformly from the interval (0,1), what is the expected value of min(X,Y)?But please, since your supposed answer to the actual question at hand contradicts the derivations of just about everyone who's posted on this topic, an explanation of your brilliant insight for the general public would be most appreciated. With the very notable exception of Fermat, most people who claim to have proven something don't only offer their solution by PM.I think it's funny that pgeek also posted a similar solution to more or less the same question here. pgeek, why don't you search through all the instances of this problem throughout this forum and disseminate your brilliant answer without a proof?

Are you sure Fermat offered his proof by PM?