[Duke]

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Part 1: Your camel's back can hold 1 unit. Straws have weights which are uniformly distributed from 0 to 1 unit, and independent of each other. You add one straw at a time. If the weights are 0.7, 0.1, 0.8..., then your camel's back breaks on the 3rd straw. On average, how many straws does it take to break your camel's back?

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There are many solutions. In white, I have written up the one I found without pencil or paper when I was 13 after an uncle posed the problem at a family Thanksgiving dinner, and at which blah_blah hinted.
<font color="white">
The expected value of a random variable X taking positive integer values is the sum of the probabilities P(X&gt;=1)+P(X&gt;=2) + P(X&gt;=3) + ... This useful trick is actually a change of the order of summation in something that looks like a single sum, but can be expanded:

E(X)
= P(X=1) + 2 P(X=2) + 3 P(X=3) + ...
= P(X=1) + P(X=2) + P(X=3) + ...
+ _________P(X=2) + P(X=3) + ...
+ _______________ + P(X=3) + ...

Summing rows, this is
P(X&gt;=1)+
P(X&gt;=2)+
P(X&gt;=3)+...

Anyway, P(X&gt;=n) = P(the first n-1 straws add up to less than 1, so with k=n-1 we want the sum of the probabilities that the first k straws add up to less than 1 for k=0,1,... The plane x1+...+xk =1 cuts off a simplex of volumn 1/k! from the unit cube, so the expected value is 1/0! + 1/1! + 1/2! + ... = e. </font>

Also at that dinner, we discussed the geometric mean of the numbers in [0,1], which is again related to e.

I'll post my solutions to Parts 2 and 3 later. I used a method which at least looks different from the suggestions of blah_black and bigpooch. However, I would not be surprised if they were all similar, from some perspective.

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Was your uncle the anomaly, or is your entire family filled with mathematicians?