[jay_shark]

I have a solution for problem 1 although I couldn't get the text in white .

Solution : Re-formulate the problem and select random numbers from the interval [0,1] .

With probability 0 the camels back will be broken on the first straw .

With probability 1/2 , the camels back will be broken first on the second straw . You may argue that the probability x2 <x1 is 1/2 and so the probability 1< x1+x2 <1+x1 is 1/2 .

With probability 1/2*2/3 the camel's back will be broken first on the third straw . This is only possible if the camel's back is not broken by the first two straws(1/2) multiplied by the probability it is broken on the third straw . The probability the camel's back is broken on the third straw given that it's not broken by the first two straws is 2/3 since x3 is equally likely to be in the interval [0,x1] , [x1,x1+x2] or [x1+x2,1] . So x3 belongs in the interval [0,x1+x2] with probability 2/3 .
Hence 1 < x1+x2+x3 <1+x1+x2 with probability 2/3 .

The camel's back will be broken by the ith straw with probability 1/(i-1)!*(i-1)/i

So if we compute the expectation , we should get
1*0 + 2*1/2 + 3*1/2*2/3 + 4*1/2*1/3*3/4 + 5*1/2*1/3*1/4*4/5 + .... + i*1/(i-1)!*(i-1)/i

As i approaches infinity , the above expression is equivalent to:

e= 1/0! +1/1! + 1/2! + 1/3! + 1/4! +....